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*Your ConquerHSC Notes are consistently being revised throughout the 2019 HSC Year to ensure quality!

Week 7 Content

Overview of Week 7 Inquiry Question

Learning Objective #1 – Conduct practical investigations to analyse the concentration of unknown acid or base by titration

Learning Objective #2 – Investigate titration curves and conductivity graphs to analyse data to indicate characteristic reaction profiles

Learning Objective #3 – Calculate and apply the dissociation constant (Ka) and pKa to determine the difference between strong and weak acids

Learning Objective #4 – Describe the importance of buffers in natural systems

Learning Objective #5 – Conduct a practical investigation to prepare a buffer and demonstrate its properties

  • Using Henderson-Hasselbalch Equation 

Learning Objective #6 – Explore acid/base analysis techniques that are applied:

  • In industries
  • By Aboriginal and Torres Strait Islander Peoples
  • Using digital probes and instruments

Learning Objective #7 – Conduct a chemical analysis for a range of common household substances for its acidity or basicity, for example:

  • Soft drink
  • Wine
  • Juice
  • Medicine
Learning Objective #8 – Model the neutralisation of strong and weak acids and bases using a variety of media

Week 7 HSC Chemistry Syllabus Video – Titration and Quantitative Analysis

Week 7 Homework Problem Set (Essential for Band 5)

Week 7 Curveball Questions (Moving from Band 5 to Band 6!)

Week 7 Extension Questions – Regularly posted additional exam-style questions

Solutions to Week 7 Questions

Overview of Week 7 Inquiry Question

Welcome back to Week 7 of your HSC Chemistry Syllabus Notes! 

In the previous week, we have looked at the different between strong and weak acids & bases. 

In this week’s notes, we will attempt to examine the relationship between strong acids, strong base, weak acids and weak bases. 

We will do this by having a look at titration curves, conductometric curves, Ka & Kb equilibrium values and other analytical techniques.

We will also explore buffers which are systems that are made from weak acids or bases and their respective conjugate base/acid. A buffer system or solution is able to resist significant changes in pH when small amounts of strong acids or base are added into it. 

Learning Objective #1: Conduct practical investigation to analyse the concentration of an unknown acid or base

The purpose of titration, a type of volumetric analysis, is to determine the unknown concentration of an acid or base. This can be done by titrating it against a base or acid with known concentration respectively.  

  • It is also possible to determine the unknown molecular formula of the acid or base of unknown concentration after titration, if given sufficient information. There will be homework questions & solutions to these type of questions at the end of this week’s notes.

Without further ado, let’s get into the crux of the titration procedure then we will look into the calculations & techniques afterwards.

Titration aim in our example: Determine the unknown concentration of nitric acid by titrating it from known concentrations of sodium carbonate. 

  • Note: You will have to consolidate few of the following steps during your exam. I expanded the steps each by a bit to make it more easy to follow. 

Step 1: Weigh 3.000 grams of anhydrous sodium carbonate in a clean, water-free beaker. [1]

Step 2: Dissolve the sodium carbonate solid with as minimal water as possible, about 50 ml. [2]

Step 3: Transfer the sodium carbonate solution in the beaker into an clean, empty 250ml volumetric flask. [3]

Step 4: Use the wash bottle with distilled water to dissolve and transfer remaining sodium carbonate in the beaker into the volumetric flask. [4]

Step 5: Using a wash bottle, add distilled water to fill up the standard solution’s meniscus up to the graduation mark. Use a dropper to add the final drops of water to fill standard solution’s meniscus to graduation mark. [5]

Step 6: Invert and swirl the volumetric flask five times to ensure even mixing of the now-standardised solution. [6]

Step 7: Rinse the 25ml pipette with distilled water followed by small quantities of standardised solution. [7]

Step 8: Rinse your burette with distilled water followed by small quantities of your titrant (the solution you are adding into the burette, in this case, it is the nitric acid of unknown concentration).

Step 9: Insert your 25ml pipette into your standardised solution to obtain a 25ml aliquot (known volume) of standardised solution (sodium carbonate) where the meniscus touches the 25ml pipette graduation mark.

Step 10: Transfer your 25ml of sodium carbonate solution into a clean, conical flask. [8]

Step 11: Attach your burette to your clamp on retort stand. 

Step 12: Transfer approximately 70ml of HNO3 of unknown concentration as your titrant into your burette and record the initial titrant volume that you added into burette (if you added exactly 70ml, then your initiate titrant will be 70ml). Ensure that there is acid and no air bubbles in the space below your burette’s tap. [9]

Step 13: Attach your burette to the clamps connected the retort stand. 

Step 14: Add three drops of the appropriate indicator into your conical flask and lower your burette’s tip so that it is just inside the conical flask. [10]

Step 15: Open the stopcock or tap to allow the titrant to react with the solution in the conical flask while swirling the conical flask with your other hand. [11]

Step 16: Reduce the flow of the titrant into the conical flask when you begin to see a colour change. [12]

Step 17: Stop adding titrant when there is permanent colour change and calculate your titre (volume of titrant used)

Step 18: Discard your salt solution in the conical flask to an empty beaker and wash your conical flask thoroughly with distilled water (tap water is fine though).[13]

Step 19: Repeat steps 9 – 15 to obtain three sets titre results that are + or – 0.1ml difference apart and calculate the average titre.

Step 20: Calculate the unknown concentration of your acid or base. In our example, it will be the unknown concentration of HNO3.

Titration Tips

[1] An anhydrous substance is one that does not absorb water from the air, increasing the mass of your weighed primary standard. Therefore, some of your primary’s mass will be due to water and eventually resulting in your calculated concentration of your unknown acid/base to be higher than in reality.

[2] You will be transferring all this solution into a volumetric flask. If you fill up the beaker with too much water, you may exceed the graduation mark in the volumetric flask which requires you to remake your standard solution.

[3] Different size of volumetric flask allows you to prepare different volumes of your stand solution. You will be given one volumetric flask on the day of your titration. Feel free to rinse the volumetric flask with distilled water to remove any particles that may be present. You can have distilled water remaining in the volumetric flask as you will be adding water after adding your standard solution to fill up to the graduation mark. 

[4] You may use a stirring rod to aid the dissolution process and the transfer of standard solution into the volumetric flask. Remember to however transfer any rinsed solution on the stirring rod to the volumetric flask. This is because the stirring rod may contain some of your standard which may be loss if not transferred. 

[5] Make sure that your eye level is on the same level as the graduation mark when comparing the meniscus and the graduation mark. If not, you may be adding more or less distilled water than you need. This will affect the concentration of your standardised solution and therefore affect your results of the unknown concentration of acid or base that you want to determine. 

[6] I would mix it five times just to give myself certainty that the standardised solution is mixed thoroughly enough. You do not want your aliquot in each titration to have different concentrations affecting your average titre.

[7] Clean your pipette and burette with distilled water to remove any tiny solids (such as dust) that may be present. Afterward rinsing with water, you need to rinsing the pipette with small amounts of the solution that you are pipetting and, for the burette, rinse it with your titrant. This is to ensure that the water you rinsed the instrument is not present in your instrument so it not dilute your solutions. If you don’t rinse you instrument, there may be water present and that will lower the actual volume of solution that you used in titration as you added few mL of water instead. When rinsing, do not use large amounts as you will run out of your solution! You will have limited amount of standard and titrant on the day. You need to do the titration at least 3 times to get three sets of similar results! 

[8] Clean your conical flask with distilled water to remove any particles that may be present. You can leave your conical flask wet, this is because your moles of aliquot solution is known. It does not change with volume because the acid-base reaction is dependent on the moles of H+ and OH- ions. This is why it is important to ensure you rinse your pipette and burette with the solution you are adding into them after rinsing with water. Pour out excess water though lol, don’t your aliquot with like large volumes of water in the conical flask since you will be adding your titrant in the conical flask too. It makes swirling easier too with less volume of total solution in the conical flask which helps with obtaining accurate results.

  • Since you know both the volume of the known solution and unknown solution that will be used at the end of titration, it does not matter whether your put your known solution in the burette or conical flask. 

[9] Sometimes there may be dust trapped in the tip that hinder your titrant from leaving the burette. Sometimes, there may be air bubbles trapped in the tip. Either way, release the tap and leave some of the titrant flow out of the burette into a separate beaker to release the air bubbles. Sometimes, it may be hard for you to remove trapped dust or solids in the tip. In that case, let your lab instructor know and go grab a different burette on the day. 

[10] If your burette is too high up (from your conical flask) there may be splashes so lowering it down so that it’s just inside the conical flask is good distance. 

[11] Make sure you are actively swirling the conical flask throughout the titration process to ensure even mixing or neutralisation of the acid and base.

[12] Once you see a new colour appearing (usually a dot in the centre of your solution), start to reduce the rate at which you are adding into the conical flask by adjusting the stopcock or tap of the burette. While doing this, continue swirling because sometimes you may thing it is a permanent colour change but then after some swirling the new colour fades away. This is because the temporary colour change is due to uneven mixing of the acid and bases and there are still free H+ or OH- ions (depending if your solution in conical flask is an acid or base) in solution.

[13] After you have discarded the salt solution into a separate beaker. Flush your conical flask with water all the way to the top and then pour it out into the sink and do this for at least three times. This will ensure that there will not be any remaining acids or base solutions in your conical flask that will affect your subsequent titration’s results. Yes, if your titration involves a strong acid and a strong base, you can just discard the salt solution into the sink as pH = 7. If not (e.g. strong acid with weak base), transfer it to a separate beaker and dispose it in a waste container that your teacher will have in place for you at the end of your titration session.

When stop titrating or adding the titrant from the burette into your conical flask?

You should stop titrating, or adding in the titrant, when you observe a permanent colour change. 

  • This is called the end point

The end point is the pH range in which the indicator that you selected changes its colour. As we have touched on in Week 5’s notes, different indicators changes colour over a different pH range which means that different indicators have different end points.

The equivalence point is the point at which the total number of moles of H+ ions is EQUAL to the total number of moles of OH ions being reacted to form pure water. 

This means that if you have an acid in your conical flask, the equivalence point is the point where ALL of the hydrogen ions has been neutralised and vice versa if you have a base in your conical flask. 

Selecting the right indicator to use in your titration experiment

You need to select the right indicator by closely matching the pH range in which your indicator changes over with the equivalence point of your acid-base titration. 

For example, suppose your acid-base titration have an equivalence point when the solution has a pH = 7. 

The best choice of indicator for such titration would be an indicator that changes colour at pH = 7 (i.e. a titration that has an ‘endpoint’ at pH = 7).

  • For example, suppose we have an indicator that is blue in solution if the pH is 0 – 6 and it turns yellow when the pH is 7 – 14
  • This means that when the indicator changes from blue to yellow, you will know that the equivalence point has been reached.

However, at school, you will mostly likely NOT get such an accurate indicator that has an endpoint exactly equal to the equivalence point. That is, you are unlikely to be given an indicator that has such a sharp ‘endpoint’, i.e. an indicator that changes colour over a narrow pH range that we have explored just above. 

So what do you do? 

Well, in this case, the alternative choice of indicator is that it should change colour over a pH range that at least includes the equivalence point. 

For example, again, if our equivalence point has a pH = 7, then a suitable indicator could be one that changes colour over the pH range of 6 – 8. 

  • Perhaps, the indicator may be blue when the pH of the solution is 0 – 6 and changes its colour to red when the pH of the solution is 8 – 14 (or any other colours xD). 
  • In this case, if the colour turns permanently red, we know that our equivalence point (where pH = 7) has been reached. 

You may or may not ask that using this alternative method of selecting an alternative indicator would mean that we need to add in more titrant and deviate from the equivalence point (pH = 7) since our endpoint is pH = 8. Well, that’s is true but we will explore why this method is still appropriate. 

Soon, in the learning objective #2, we will take a look as to why this alternative method of ‘selecting an indicator with a pH range includes the equivalence point’ is good enough and will not significantly affect our titration results (for our purposes). 

Criteria for selecting and using an appropriate primary standard to make standardised solution

A primary standard is a substance that have fixed chemical composition, high purity, chemical stability and solubility such that it can be used to prepare a standardised solution with an accurately, known concentration.

High molecular mass to minimise percentage error when measuring the mass of the primary standard.

High solubility in distilled water so that the standard completely dissolves in water and thus its  calculated known concentration of the standard is accurate

High level of purity so that the calculated known concentration of the standard is accurate.

Definite and fixed chemical composition so that the mole ratio of the dissociated ions from the primary standard can be accurately calculated. 

The primary standard should be chemically stable so that it does not undergo violent chemical reactions with distilled water or with moisture in the air.  

The primary standard should be non-efflorescent meaning that it does NOT release water into the surrounding environment. This makes its chemical composition not definite and unfixed. Therefore, its concentration after standardising would be uncertain. 

  • HCl is efflorescent or hydrated compounds (with water molecules in their chemical formula) such as hydrated sodium carbonate.

The primary standard should be anhydrous meaning it DOES NOT absorb water from the air and so it does not form bonds with water. A hydrous standard will result in lower concentration of dissociated ions when dissolved in the water in the volumetric flask. 

  • Sodium hydroxide absorbs water from the air and so it is hydrous, i.e. have deliquescence property. 

The primary standard should be NOT BE hygroscopic so that it DOES NOT absorb moisture from the air (carbon dioxide and water). This is because a hygroscopic substance will react with carbon dioxide and thus lower the concentration of ions when dissolved in the water in the volumetric flask. 

  • For example, sodium hydroxide is hygroscopic and thus it will absorb moisture in the air (containing carbon dioxide). The following reaction will occur:
  • 2NaOH(s) + CO2(g) —> Na2CO3 (aq) + H2O(l)
  • This will lower the concentration of NaOH and [OH] when dissolved in water in the volumetric flask when attempting to standardise the sodium hydroxide. 

Some potential sources of error in titration or discrepancy between students’ results

  • Different people have different perception and sensitivity to colour changes. Therefore, it is possible to misjudge when the endpoint of the indicator has been reached during titration. 

Minimise error: Place a piece of white paper underneath your conical flask to allow easier judgement of colour change.

  • Misreading the volume of titrant used – parallax error. The eye level must be at the same level as the meniscus to accurately determine the volume of the standardised solution that is pipetted and amount of titrant used in burette. 

Minimise error: Place a piece of white paper behind the burette or volumetric flask to allow easier comparison of the graduation mark and meniscus.

  • Not transferring all solid/liquid when preparing standardised solution – it may happen that some of the primary standard was still left in the beaker or stirring rod. 

Minimise error: Thoroughly wash and transfer your wash solution into the volumetric flask multiple times to ensure all of the primary standard is transferred to the volumetric flask.

List of some common indicators used in HSC Chemistry

Before moving to the next learning objective, I recommend you to memorise the names of the common pH indicators below, their pH range and their colours at certain pH values for HSC exams.

Also, I recommend you to memorise them HSC practical exams as your teacher may supply you with multiple indicators on the titration day. On the day, you therefore may need to choose the most appropriate one for your titration out of many indicators. 

Yes, the appropriateness of your indicator will depend on the equivalence point of your titration. We will explore how you can determine the equivalence point for different types of titration that you may encounter in the next learning objective.

HSC-Chemistry-Titration-Indicators
HSC-Chemistry-Titration-Indicators-List

Learning Objective #2 - Investigate titration curves and conductivity graphs to analyse data to indicate characteristic reaction profiles

Strong acid – Strong Base Titration

Generic strong acid – strong base titration curves 

HSC-Chemistry-Strong-Acid-Strong-Base-Titration-Curve

Note the difference between the two curves due to the word ‘with‘. 

  • Acid titrated ‘with’ a base means that you are adding a base into the conical flask (containing acid) and therefore the base is your titrant (solution in burette).
  • Base titrated ‘with’ an acid means that you are adding an acid into the conical flask (containing base) and therefore the acid is your titrant (solution in burette).
Another thing to point out is that the above curve is just a generic titration curve between a strong acid and a strong base. The volume of acid/base added from the burette into the conical flask will vary depending on the concentration & volume of your unknown solution as well as your standardised solution. 
 

The key idea here is that, at the equivalence point for all strong acid and strong base titration, the pH of the solution is always equal to 7. 

You may ask why? Well, here is the explanation. 

Suppose we have a strong acid – strong base titration such as:

HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)

Similar to the equivalence point of all other forms of acid – base reactions (e.g. weak acid – strong base, weak acid – weak base, etc), the equivalence point for strong acid – strong base titration refers to the point where all (total moles) of H+ and OH ions are neutralised (reacted) to form pure water.

Net Ionic Equation: H+ (aq) + OH (aq) -> H2O(l)

The conjugate acid of a strong base formed as product in titration is a very weak acid.

  • For instance, the conjugate acid in our example for the strong base, NaOH, would be water and Na+ is a spectator ion for the example reaction above. Water has a pH of 7, similar to the pH at equivalence point for a strong acid – strong base titration. If you can recall, the autoionisation of water will result in a pH of 7 at room temperature due to equal concentration of H+ and OH- produced as a result of auto-ionisation.
  • Sodium ion is unable to react with water via hydrolysis to result in any significant changes in pH.

Similarly, the conjugate base of a strong acid as product in titration is very a weak basic and is unable to react with water to act as a proton acceptor (i.e. unable to accept a H+ from water)

  • For instance, the conjugate base in our example would be the Cl ion. 
  • Since the chloride ion is the conjugate base of the strong acid, HCl, it is unable to react with water as a base, i.e. chloride ion cannot accept a H+ from water

This means that only water has an effect on the pH of the solution at equivalence point. Since pure water is neutral (pH = 7 from autoionisation of water) and the conjugate acid/base components of the salt product (e.g. NaCl) DO NOT react with water and, thus does not affect the pH, this means that the pH of the solution at equivalence is equal to 7. 

Yes, of course, if you keep adding excess acid/base after the equivalence point, the pH will change as shown in the titration curve diagram as the strong acid/base will react with water and affecting the pH. 

Choosing the right indicator for strong acid – strong base titrations

Remember from last learning objective we said there are two suitable options when selecting the suitable indicator for your titration? We also said that we will take a look at why those two options are equally good enough. 

Option A: Match the pH range over which your indicator changes colour exactly with your acid-base titration’s equivalence point

Option B: Select an indicator with a pH range that includes the equivalence point of your acid-base titration in which the colour will change colour. 

We will use the above strong acid – strong base titration to explore why both options are good enough when selecting an appropriate indicator for your titration.

We have explored why, at equivalence point, the pH of the solution for a strong acid – strong base titration is equal to 7. 

So, if we use option A as a guide to select our appropriate indicator, an example of an appropriate indicator, Indicator A,  could be:

  • Indicator A will appear Blue from pH 0 – 6 and
  • Indicator A will appear Red from pH 7 – 14.

This indicator is appropriate because we are able to visually observe the point in time when the equivalence point for our strong acid – strong base titration has been reached (i.e. pH = 7). This is because at equivalence point, the indicator will change colour from blue to red. 

Once the solution in the conical flask turn from blue to a permanent red colour, we stop titrating because we know equivalence point has been reached. 

Now, let’s explore using option B as a guide to selecting our appropriate indicator and examine why it is good enough criteria in assisting us in selecting the appropriate indicator. 

Option B says we can choose an indicator that changes colour over a pH range that includes the equivalence point for our acid – base titration. 

In this case, we are dealing with a strong acid – strong base titration, so the pH range in which the indicator has to include a pH = 7. 

So an appropriate indicator, e.g. ‘Indicator B’, according to option B could be:

  • Indicator B will appear Blue from pH = 0 – 5.5 and
  • Indicator B will appear Red from pH = 11 – 14

Notice that in this case, the pH of solution at equivalence point (pH = 7) is included inside the pH range over which the indicator changes colour (from 5.5 to 11). 

You may now ask, doesn’t this mean by the time you observe the indicator changes colour from Blue to Red, you have already exceeded the equivalence point by adding excess acid/base into the conical flask?? 

Therefore in your calculation, your unknown concentration of your acid/base (whichever you need to calculate) will be incorrect? Thus making option B of selecting your indicator is invalid?

The answer to your questions is yes and no. Why?

Well, take a look at the following graph.

HSC-Chemistry-Equivalence-Point-and-End-Point

As you can see from the graph above, for strong acid – strong base titration, if you continue to add excess acid/base after equivalence point (pH = 7) has been reached, you DO NOT need to add a lot of acid/base (volume wise) to reach the end point of your indicator. 

For example, as shown in the diagram above, the end-point of our indicator is at pH = 11, i.e. the pH at which our indicator changes colour. Notice that the extra volume (2ml) of acid/base difference between our equivalence point and end-point for our strong acid – strong base titration is negligible

Therefore, option B is a good enough criteria in helping us selecting the right indicator. Therefore, you can choose an indicator that changes colour over a small pH range (as small as possible, the above example actually changes colour over quite a large pH range). 

Also, the indicator’s pH range over which it changes colour should include the pH of your titration’s equivalence point.

  • Note: At school, it is unlikely you will be supplied an indicator that allows you to use option A (where indicator that you use will have an endpoint that is exactly equal to your titration’s equivalence point). 
  • Therefore you will mostly likely be using option B as your guide in selecting the most appropriate indicator on titration day and in HSC Chemistry questions.

Strong Acid - Weak Base Titration

HSC-Chemistry-Weak-Acid-Strong-Acid-Titration-Curve

Strong Acid – Weak Base Titration: Why Equivalence Point LESS than 7?

Similar to the equivalence point of all other forms of acid-base reactions, the equivalence point for a weak base being titrated with a strong acid refers to the point where the total number of moles of H+ ions equal to the total number of moles of OH neutralised (reacted) to form pure water. 

As explained before, the conjugate base of a strong acid is a very weak base and is unable to react with water to act as a proton acceptor (i.e. unable to accept a H+ from water).

That being said, the conjugate acid of a weak base is remains to be weakly acidic and is able to react with water to act as a proton donor (i.e. donate a H+ to water) to form hydronium ions in solution at the equivalence point. 

This means that, at equivalence point, all titration involving a weak base and a strong acid have a solution with a pH less than 7 at their equivalence point due to the presence of H3O+ ions. 

The equivalence point is no longer equal to 7 as both the pure water that is formed as product of titration AND hydronium ions (from conjugate acid of weak base reacting with water) BOTH determine the overall pH of the solution at equivalence point. 

Strong Acid – Weak Base Titration: pKb and Buffer Region

We will explore pKb and buffers in the next learning objective.

As shown in the graph, for strong acid – strong base titration, there is a buffer region. Within the buffer region, if you add more titrant (in this case it is strong acid), the change in pH is small. This is because buffers are able to resist significant changes in pH when small amounts of acid or bases are added. We will explore how buffers are able to have such chemical property in the next learning objective. 

Another point that is worth noting is that the pKb of the strong base in a strong acid with weak base titration is exactly the half way mark towards the equivalence point. 

Therefore, if you know the volume of equivalence point, you will be able to determine the pKb by dividing it by two.

Notice that since there is a pH value for the solution at equivalence point, there is also a pOH value for pKb. We use pOH for pKb because we are dealing with Kb (dissociation of base constant) which we will explore more of in the next learning objective. 

  • Fun Fact – The pKb of the strong base is in the midpoint of the buffer region.
  • Fun Fact – Notice that there is no pKa or pKb shown in the strong acid – strong base titration curves because the pKa of the strong acid is below 0 (negative pKa). The pKb of strong bases is also negative.

Weak acid – Strong Base Titration

HSC-Chemistry-Weak-Acid-Strong-Base-Titration-Curve

Weak Acid – Strong Base Titration: Why Equivalence Point greater than 7?

Similar to the equivalence point of all other forms of acid-base reactions, the equivalence point for a weak acid being titrated with a strong base refers to the point where the total number of moles of H+ ions equal to the total number of moles of OH neutralised (reacted) to form pure water. 

As explained already, the conjugate acid of a strong acid is a very weak acid and is unable to react with water to act as a proton donor (i.e. unable to donate a H+ to water).

That being said, the conjugate base of a weak acid is remains to be weakly basic and is able to react with water to act as a proton acceptor (i.e. accept a H+ from water) to form hydroxide ions in solution at the equivalence point. 

This means that, at equivalence point, all titration involving a strong base and a weak acid have a solution with a pH more than 7 at their equivalence point due to the presence of OH ions. 

The equivalence point is no longer equal to 7 as both the pure water that is formed as product of titration AND hydroxide ions (from conjugate base of weak acid reacting with water) BOTH determine the overall pH of the solution at equivalence point. 

Weak Acid – Strong Base Titration: Buffer Region

Again, for weak acid – strong base titration, there is a buffer region. 

Similar the previous case, the pKa is exactly half way towards the equivalence point. In quantitative terms, pKa is half the volume of titrant required to reach equivalence point.

The pKa of the weak acid is located in the midpoint of the buffer region. As promised previously, we will explore pKa and buffers in the next learning objective. 

Weak acid – Weak Base Titration

Notice that, compared to previous types of acid-base titration, in a weak acid – weak base titration, the pH does not change a lot when you continue adding acid/base in a weak acid – weak base titration after the equivalence point. 

That is, there is no rapid change in pH at the equivalence point or for small volumes of titrant added after equivalence point. Due to this, it is hard to find an indicator that has such as a sharp endpoint (changes colour over a small pH range) that matches the equivalence point exactly for weak base – weak acid titrations. 

If you choose to use option 2 (mentioned earlier in the notes as a guide in selecting the appropriate indicator), you will need to add a lot of excess acid/base after the equivalence point before the end-point of the indicator will be reach in weak acid – weak base titration. 

Thus, the end point of the indicator will have surpassed the equivalence point by a lot (in terms of the titrant that is required). This would lead to inaccurate results of total titre used which will affect the accuracy of the value you calculated for your unknown acid/base. 

For such reason, your teacher will not be asking you to do such titration at school.

Equivalence Point for titration involving weak base and weak acid could occur at various pH. That is, equal, greater or less than 7.

  • If the Ka of weak acid > Kb of weak base, then equivalence point occurs at pH < 7. 
  • If the Ka of the weak acid = Kb of weak base then equivalence point occurs at pH = 7.
  • If the Ka of weak acid < Kb of weak base then the equivalence point occurs at pH > 7.

Conductometric Titration Curve: Strong Bases

HSC-Chemistry-Titration-Electrical-Conductivity-Curves-Acids

Explanation for the (left) curve involving strong acid being titrated with strong base:

Suppose: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(aq)

Initially the high electricity conductivity of the solution is due to the presence of high number of dissociated hydrogen ions in solution from strong acid molecules that is dissolved. 

As strong base is added and dissociates in the solution (by reacting with water), the neutralisation reaction between H+ and OH ions will reduce total number of free ions in solution. This would therefore reduce the electrical conductivity of the solution. At equivalence point, there is still some electrical conductivity due to the the presence of ions of neutral salt that is formed (e.g. NaCl dissociated into Na+ and Cl if the strong acid is HCl and strong base is NaOH).

When excess base is added into the solution after the equivalence point, there will be an increase number of OH- ions in solution and thus the electrical conductivity increases (as number of total ions in solution increases). 

Fun Fact: Since hydrogen ions have higher molar conductivity than hydroxide ions, if you add 0.1M of NaOH into a solution and 0.1M of HCl into another solution, the solution with HCl will have higher electrical conductivity.

Explanation for the (right) curve involving weak acid being titrated with strong base:

Suppose: CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)

If you recall that strong acids have higher degree of dissociation compared to weak acids, you will come to understand that the electrical conductivity of the solution consisting of weak acid initially (right graph) is lower than the previous scenario where we had a strong acid initially in solution. This is assuming that the initial concentration of the both the strong and weak acids are the same. 

Initially, as we add the strong base in solution, the OH ions dissociated from the base molecules will react with the H+ dissociated from the acid molecules to produce pure, undissociated water molecules. Thus, the total ion concentration in solution would decrease. This means that the conductance of the solution will decrease. 

However, after the initial decrease in conductance, the electrical conductivity of the solution will increase until the equivalence point due to the formation of the CH3COONa molecules and their dissociation into solution to form CH3COO and Na+ ions. 

Due to the presence of these ions, the conductivity of the solution will increase slightly (graph not to scale). When equivalence point is reached, the conductivity is stabilised as no more CH3COONa is formed and dissociated. Thus, no more increase in [CH3COO] and [Na+] ions in solution.

  • Note that the steepness or gradient of the positive slope prior to reaching the equivalence point will be dependent on the strength of the acid (i.e. the amount of CH3COO ions in solution). This is because all Na+ ions from the strong base will always be 100% ionised in solution (as per definition of strong base). Therefore, only the strength of the acid will change the steepness of the curve.

However, if we continue further adding a strong base such as sodium hydroxide after the equivalence point, since the OH ion have higher molar conductivity than the acetate (CH3COO) ion and sodium ions, the solution conductivity will further  increase. 

  • Note: Sometimes there will be NO increase in conductance after the initial decrease in conductance because of the strength of the acid is very weak hence the formation of the conjugate base is minimal (low concentration). 
  • Therefore, in that case, the equivalence point right situated between the decrease in conductivity all H+ reacts with OH ions and the increase on the conductivity due to the addition of excess strong base (OH) ions. In that the curve on the top right would look like the curve below.
HSC-Chemistry-Conductometric-Titration-Curve-Weak-Acid-Strong-Base-Conductivity

Conductometric Titration Curve: Weak Bases

HSC-Chemistry-Titration-Electrical-Conductivity-Curves-Bases

Explanation for the (left) curve involving strong acid being titrated with weak base: 

Suppose the reaction: HCl(aq) + NH4OH(aq) -> NH4Cl(aq) + H2O(l)

Initially, the electrical conductivity is high due to the dissociation of the strong acid into H+ ions and its conjugate base ions. 

Anyways, back to the left graph, as you add the weak base, the OH dissociated from the base molecules will react with the H+ dissociated from the acid molecules to form undissociated water molecules. This means that the overall electrical conductivity of the solution will decrease due to the decreased total concentration of free ions in solution. The minimum concentration of total ions is when equivalence point has been reached where the OH ions neutralised all the H+ ions that is in solution. 

Upon adding excess weak base after equivalence point has been reached, the excess weak base will react with water to form OH ions in solution (by accepting H+) from water. However, as seen from the graph, the total electrical conductivity of the solution will NOT increase upon adding excess weak base after the equivalence point. The reason for this is due to the common ion effect. 

When equivalence point has been reached, there is high concentration of NH4+ (and Cl) ions in solution. Upon adding more weak base (NH4OH) into the solution, there will be an increase in [NH4+] ion in solution. As we have learnt in Module 5, according to Le Chatelier’s Principle, due to the addition of the common ion in solution, the system will shift the equilibrium position of the following reaction to the left to minimise the increase [NH4+] ion in the solution.

NH4OH(aq) <-> NH4+(aq) + Cl(aq). 

Therefore, more undissociated NH4OH molecules will remain in solution and therefore lowering the solubility of NH4+ ions being added into solution. Due to common ion effect, it effectively suppresses the dissociation of NH4OH molecules that are being added into solution. This therefore means that electrical conductivity won’t increase (at least not to significant extent)

  • NOTE: NH4OH molecules do not contribute to electrical conductivity if they are not dissociated. 
  • NOTE: The common ion effect does NOT occur in strong acid – strong base titration as there is no equilibrium formed but rather only complete dissociation! Recall that equilibrium only exists for weak acids or weak bases.

Explanation for the (right) curve involving weak acid being titrated with weak base:

Suppose the reaction: CH3COOH(aq) + NH4OH(aq) -> CH3COONH4(aq) + H2O(l)

Initially there is a decrease in conductance of the solution as the dissociated H+ ions from the weak acid reacts with the OH ions that is dissociated from the weak base. However, as the CH3COONH4 is formed and is dissociated into its constituent ions in solution (CH3COO and NH4+), the electrical conductivity of the solution increases upon the equivalence point is reached where no more CH3COONH4 ions are formed and dissociated.

Notice that if you add more of the weak acid after the equivalence point, the conductivity WILL NOT increase due to the common ion effect as explained in the strong acid – weak base titration scenario.

However, if you add more of the strong acid after the equivalence point, such as NaOH, then the conductance of the solution will increase!

Learning Objective #3 - Calculate and apply the acid dissociation constant (Ka) and pKa to determine the difference between strong and weak acids

We have briefly introduced on Ka in Module 5 under different types of Keq or equilibrium constants. In this learning objective, we will dive into greater detail and explore Ka, i.e. the acid dissociation constant. 

So, Ka is one type of equilibrium constant that is used to determine the extent of dissociation of an acid in solution (e.g. water). 

Like other equilibrium constant, Ka is basically the concentration of the products divided by the concentration of the reactants. For a generic acid with the formula HA, its Ka can be written as:

Ka = [H+] [A] / [HA] ; where A- is the conjugate base of acid HA.

Last week, we have already explored the difference between strong and weak acids. 

Strong acids ionise completely into its constituent ions in water whereas weak acids have less than 100% degree of ionisation. 

For strong acids, this would mean that the concentrations of [H+] and [A] would much much higher the concentration of [HA], where the [HA] is very small. This means that a strong acid have a very high Ka value compared to weak acid. 

If you recall, from the previous week’s note, the ‘p’ in front of H in ‘pH’, effectively means -log10. 

So, pKa means -log10 x Ka. So, 

  • pKa = -log10Ka
  • Similarly, pKb = -log10Kb. 

Base dissociation constant = Kb = [BH+] [OH] / [B] ; where B+ is the conjugate acid of base B

Due to the nature of logarithmic functions, as the value of Ka increases, the value of pKa decreases. You can substitute different Ka values into your calculator to test this inverse relationship out. 

Since strong acids have large Ka values, they have a smaller pKa value that weak acids. 

We will have homework questions at the end of this week’s notes for you to calculate pKa when given Ka and vice versa. 

But here is the a formula that you need to know when you are given pKa and you need to determine Kb. 

  • Ka = 10-pKa
  • Kb = 10-pKb
 

We talked about the autoionisation of water, Kw, last week. Now that we have learnt, Ka and Kb, we can also express Kw as:

Kw = Ka x Kb = 1.0 x 10-14 (ONLY at 25 degrees celsius as Kw changes with temperature like all other equilibrium constant which we have explanation in the derivation of Keq section in Module 5 notes)

  • If you wish, feel free to derive the above formula by multiplying Ka with Kb formulas with a common weak acid and you will see everything will cancel out so you will be left with [H+] x [OH] which is the formula for Kw that we have explored in last week’s notes. 

Yes, there is also a pKw

  • pKw = -log10Kw 
  • pKw = -log10Kw = 14 (as Kw = 1.0 x 10-14 at 25 degrees celsius)
  • Kw = 10-pKw 
  • pKw = pKa + pKb = 14 

Learning Objective #4 - Describe the importance of buffers in natural systems

A buffer system is a solution comprised of a equal amounts of weak acid and its conjugate base OR equal amounts of a weak base with its conjugate acid.

The importance and use of buffer are derived from its ability to resist changes in pH when small amounts (volume) of strong acids or base are added into the buffered system. 

Only small amounts of strong acid or base can be added in order for the buffer to resist changes in pH. That is, the strong acid or base that is being added into the buffer is a limiting reagent. This is because if excess amounts strong or base are added, all of the weak acid/base and its conjugate base/acid will be consumed, effectively exceeding the buffer capacity to resist changes in pH. 

Buffer capacity refers to the total amount of acid or base added into the buffer and where any significant changes in pH is resisted using the weak acid/base and its conjugate base/acid components in the buffer. 

  • NOTE: One way to increase buffer capacity is to increase the concentration of the components of the buffer. If the concentrations of the buffer components are increased, then it can ‘neutralise’ more acid or base that are added into buffer system. 

Sometimes you see that buffer capacity also being defined as the number of moles of H+ or OH- required to be added into the buffer system to increase or decrease the buffer’s pH by 1 for a one litre of the buffer solution.

  • NOTE: I would use this definition in your HSC response as it incorporates a quantitative element to it. The previous definition for ‘buffer capacity’ may be useful for you to initially grasp what it means.

Using Le Chatelier’s Principle, it is possible to account for how buffers are able to to resist changes in pH when small amounts of acid or base are added into the buffer system (or buffer solution).

Let’s explore two non-specific examples before looking into a specific buffer system in our blood. 

Weak acid - Conjugate Base Buffer

HA(aq) + H2O (l) <-> H3O+ (aq) + A- (aq)

Situation 1: Adding strong acid

Suppose if you add small amounts of strong acid into the weak acid-conjugate base buffer, the added hydronium ions will react with the conjugate base in the buffer (A-) to shift the equilibrium position to the left to minimise the increased concentration of hydronium ions, forming more weak acid (HA(aq)) as per Le Chatelier’s Principle. This effectively means that if you add a strong acid into the weak acid buffer solution, you will not see the pH rise as much as you would have expected. 

Situation 2: Adding strong base

Suppose if you add small amounts of strong base into the weak acid-conjugate base buffer, the added hydroxide ions will react with the hydronium ions which results in the increase in pH of the buffer system as there is less hydronium ions in solution. This is what you will expect when you add a base into any solution containing an acid which results in a increase in pH by consuming the hydronium ions. 

  • Well then, how does this weak acid buffer resist changes in pH in this situation when strong base is added? Well, as per Le Chatelier’s Principle, the system will shift to the right to minimise the decrease in [H3O+] where more weak acid (HA) will ionise in water and increase the [H3O+], effectively resisting the increase in pH of the buffer system when the strong base is added. 

Weak base – conjugate acid buffers

B(aq) + H2O (l) <-> BH+ (aq) + OH(aq)

Situation 1: Adding strong base

Suppose if you add small amounts of strong base into the weak base-conjugate acid buffer, the added hydroxide ions will react with the conjugate acid (BH+) to shift the equilibrium position to the left to minimise the increased concentration of hydroxide ions, forming more weak base (B aq) as per Le Chatelier’s Principle. This effectively means that if you add a strong base into the weak base buffer solution, you will not see the pH rise as much as you would have expected.

Situation 2: Adding strong acid

Suppose if you add small amounts strong acid into the weak base-conjugate acid buffer, the added hydronium ions will react with the hydroxide ions which results in the decrease in pH of the buffer system as there is less hydroxide ions in solution. This is what you will expect when you add an acid into any solution containing a base which results in a decrease in pH by consuming the hydroxide ions. 

  • Well then, how does this weak buffer resist changes in pH in this situation when strong acid is added? Well, as per Le Chatelier’s Principle, the system will shift to the right to minimise the decrease in [OH] where more weak base (B) will ionise in water and increase the [OH], effectively resisting the decrease in pH of the buffer system when a strong acid is added.

Natural Buffer: Carbonic acid - hydrogen carbonate buffer

The carbonic acid – hydrogen carbonate buffer occurs in a natural system which is in our blood. This buffer allows the pH of our blood to remain in a constant range between 7.3 – 7.4. 

The chemical expression of this buffer is:

H2CO3(aq) + H2O(l) <-> H3O+(aq) + HCO3-(aq)

Note that carbonic acid is a weak acid and its conjugate base is the hydrogen carbonate ion. 

When cells perform cellular respiration, carbon dioxide is a waste product that is produced. This carbon dioxide diffuses through the cells’ membranes and into the blood plasma (or blood stream). Since blood plasma is mainly water, the carbon dioxide that dissolved in the blood plasma will react with water to carbonic acid where the chemical reaction can be expressed as: CO2(g) + H2O(l) <-> H2CO3(aq). 

Shifting the carbonic acid – hydrogen carbonate buffer 

The kidney has the function of filtering the blood for nitrogenous waste which are mostly acidic. The kidney absorbs hydronium ions from the blood and secretes hydrogen carbonate (HCO3-) ions back into the blood to be absorb as the hydrogen carbonate ion is important in assisting the organism’s digestion processes. 

Comparatively, the hydrogen ion (or hydronium ion) is absorbed from the blood and secreted out the blood as a component of nitrogenous waste (urea).

When the kidney absorbs hydronium ions from the blood, it will shift the buffer system to the right as per Le Chatelier’s Principle to minimise the decrease in [H3O+]. This would mean that more carbonic acid (weak acid) will ionise in water. 

The overall effect of this buffer allows the change in the blood pH to be minimised and be maintain within a narrow range of 7.3 – 7.4.

When the kidney reabsorbs hydrogen carbonate ions by secreting it into the blood, it will shift the buffer system to the left as per Chatelier’s Principle to minimise the increase in [HCO3-] where the absorbed hydrogen carbonate ions will react with hydronium ions to form carbonic acid. 

The carbonic acid molecules will dissociate into carbon dioxide and water where the carbon dioxide will be secreted out of the body when exhaled as increasing the [H2CO3] will shift the equilibrium position of CO2(g) + H2O(l) <-> H2CO3(aq) to the left as per Le Chatelier’s Principle. The overall effect of this buffer allows the change in the blood pH to be minimised and be maintain within a narrow range of 7.3 – 7.4. 

Importance of maintaining the pH within a narrow range

Enzymes are examples of biological catalyst which are in the human body. Enzymes are operate most optimally around a specific pH range and, if the pH of the surrounding environment which it occupies in deviates too much from its specific pH range, it will denature. This means that the enzymes’ shape and active site will be be altered such that it will no longer be able to catalyst metabolic processes in the body. 

Failure to catalyst metabolic processes can have many consequences such as accumulation of wastes, inability to transport nutrients in and out of cells, etc. If the pH of the blood deviates from 7.3-7.4 for prolonged performs of time, the inability for the required chemical reactions to be catalysed could result in death. 

  • NOTE: Most chemical reactions in our body needs to be catalysed by enzymes!

Thus the blood’s pH should be maintain around 7.4 where the enzymes can function optimally and catalyse the appropriate (and necessary) chemical reactions. 

Therefore, the carbonic acid-hydrogen carbonate buffer is essential in the blood to maintain the wellbeing of the organism (e.g. human).

Learning Objective #5 - Conduct a practical investigation to prepare a buffer and demonstrate its properties

To prepare a buffer, the Henderson-Hasselbalch Equation is useful in assisting us in calculating the concentration of weak acid/base and its conjugate base/acid to make an effective buffer with high buffering capacity at a given pH level.

Originally, Henderson-Hasselbalch equation was used to help chemist approximate the pH of a buffer which can be expressed as:

pH =  pKa + log10 [conjugate base] / [weak acid]

  • pH = pKa + log10 [A-] / [HA]

pOH = pKb + log10 [conjugate acid] / [weak base]

  • pOH = pKb + log10 [BH+]/[B] or (doesn’t really matter, it’s a generic formula)
  • pOH = pKb + log10 [B+] / [BOH]
  • You will be okay if you input initial concentrations of conjugate base or conjugate acid into the equation rather than calculating their equilibrium concentrations as their concentration after dissociation from their weak acid/weak base is small compared to if they are dissociated from a salt compound. So, no ICE table required. 
  • Of course, if you are given equilibrium concentration of the conjugate acid/base, use them. 

Creating a buffer with the pH of 7.4

Since we have explored the carbonic acid – hydrogen carbonate buffer that exist in a pH of 7.4, let’s keep things in our way and attempt to create a buffer with a pH of 7.4.

Every buffer systems functions most effectively when its buffering range is +/- 1 of the weak acid’s pKa. For example, we cannot create an effective acetic acid – acetate buffer at a pH of 7.4 as the Ka of acetic acid is 1.75 x 10^-5 at 25 degrees celsius. This means that its pKa of acetic acid is 4.77. This means an effective acetic acid-acetate buffer can only be prepared at a pH of 3.77 – 5.77 and not at a pH of 7.4. 

The most effective buffer is one where its pH = pKa. This means that, from the Henderson-Hasselbalch equation, the most effective buffer is when the concentration of the conjugate base is equal to the concentration of weak acid. This is because log[1] = 0. This means pH = pKa

  • Carbonic acid is diprotic so there is two stages of dissociation.
  • At 25 degrees celsius, the pKa1 for carbonic acid is 6.4 (first dissociation)
  • At 25 degrees celsius, the pKa2 for carbonic acid is 10.33 (second dissociation)
An effective buffer is when pH is +/- 1 of the pKa. Therefore, the pKa1 for carbonic acid chosen to make the buffer with the pH of 7.4.
 

pH = pKa1 + log[HCO3-]/[H2CO3

7.4 = 6.4 + log[HCO3-]/[H2CO3]

log[HCO3-]/[H2CO3] = 1.0

[HCO3-]/[H2CO3] = 10^(1) = 10

NOTE: Notice that the above ratio DOES NOT tell you the volume of carbonic acid you need to use to make the buffer solution. Please have a look at the Solution to Question #10 of this week’s homework set (located after this week’s notes) to see an example of how you can prepare a buffer in practice by calculating the exact volumes of solution that you need to add to make a buffer solution at school.

So, we can make a carbonic acid – carbonate buffer with a pH of 7.4 by making a solution that has 1M of hydrogen carbonate ions and 0.1M of carbonic acid. Since the volume cancels out, we can say that we can make the buffer using 1 mole of hydrogen carbonate ions and 0.1 moles of carbonic acid. This means that pH does not depend on volume so diluting the buffer solution will not affect pH as both the conjugate’s concentration and weak acid/base are affected (changed/diluted) equally*

* This is true in theory when examining the Henderson-Hasselbatch equation but not in reality. This is because, in practice, as dilution increases (adding more and more distilled water), more of the pH will be determined by the autoionisation of water instead by the [conjugates] and [weak acid] or [weak base]. This is because you are adding in more water into the solution and the solution is mostly water. Water will autoionise into [H+] and [OH-] in equal amounts so, over time, as the solution gets more and more dilute, the buffer solution will approach a pH of 7. This means that the Henderson Hasselbatch equation on the pH of buffers do not hold when the solution is very dilute or very strong (strong acid/base). 

NOTE: Dilution will have less effect on buffer solutions than solutions that are not buffers.

NOTE:  if the solution is NOT a buffer, then dilution will decrease the concentration of the hydronium ions and increase the pH as pH is dependent on the concentration of hydronium ions. If you continue adding distilled water, the pH will approach 7 but never exceed 7 because distilled water used for the dilution is neutral (pH = 7). The diluted solution will also never be equal to exactly 7 as there are acids present in the solution. 

Learning Objective #6 - Explore acid/base analysis techniques that are applied to:
- In Industries
- By Aboriginal and Torres Strait Islander Peoples
- Using digital probes and instruments

In Industries

Wine Industry

Most of the wine solutions that you encounter comprises of dilute, weak organic acids. These acids in wine can be divided into two areas, those being fixed and volatile acids. These two division of acids determines total acidity of a wine solution. 

  • Fixed acids in wine are mainly malic acid and tartaric acid. 
  • Volatile acids include acetic acid and formic acid present in wine.

During the fermentation process of manufacturing wine, volatile acids (such as acetic acid) can be formed as a result of bacteria present in the wine. The amount of volatile acid in wine is called volatile acidity. 

It is important to analyse the volatile, organic acid content in wine as their concentration is proportional to the susceptibility of the wine being spoiled. This is because if bacteria enters the wine, the higher the concentration of volatile organic acids that is present in the wine, the more easily the bacteria can convert alcohol into acetic acid (example of volatile acid). This is called the spoiling of wine

  • Therefore, as consumers, we prefer our wine to have as less volatile acid as possible.

Altogether, the volatile and non-volatile acids give the tartiness or sharpness of the wine’s taste. The higher the acid concentration in the wine, the more sharp the wine will taste.

Acetic acid and other volatile acids can be analysed for their concentration in wine by using steam distillation (technique).

After steam distillation, the volatile acids will be the distillate. The volatile acids can be to titrated against a standardised NaOH solution to determine the concentration of the volatile acids. Typically, three drops of phenolphthalein indicator is used in the titration.

The remaining solution can be then analysed to determine the components of fixed acids in wine using thin layer chromatography (technique) or paper chromatography (technique). 

The alcohol content is required to be specified on the wine as per legalisation. Alcohol content such as ethanol present in wine can be determined using liquid gas chromatography (technique) or by using hydrometry.  Using a hydrometer, different wine with known ethanol densities (i.e. wine with different known amount of ethanol) are measured against their boiling point. 

This means that a density calibration curve can be constructed which consists of boiling (Y-axis) against of the densities for a range of wine with known ethanol densities, % by mass (x-axis). From this, the calibration curve can be used to determine the unknown ethanol concentration in a wine sample.

Wines also have added sulfur dioxide used as a preservative to prevent the growth of bacteria and yeast. Sulfur dioxide also inhibit the oxidation of wine by acting as an antioxidant to keep the wine smelling fresh. If oxidation of wine occurs, it would lower the quality of wine and this is often known as a wine fault in the industry.

When sulfur dioxide is added into the wine, most of them will react with organic aldehyde and ketone acid molecules present in the wine, which you will learn in Module 7. The rest of the sulfur dioxide will react with water in the wine to form bisulfite, sufite and hydrogen ions.

SO2(g) + H2O(l) <-> H+ (aq) + HSO3-(aq) <-> 2H+(aq) + SO32-(aq)

HSO3= bisulfite ion

SO32-= sulfite ion

It is important to determine the concentration of sulfur dioxide in wine because there is a limit for amount allowed in wines. As sulfur dioxide produces a strong, undesirable aroma that covers the smell and flavour of the wine, the reduction of sulfur dioxide improves the smell and taste of the wine. Furthermore, some of the sulfur dioxide can dissolve in the wine to produce sulfite ions which cause skin rashes. Therefore, by lowering the content of sulfur dioxide in the wine production process, there is also health benefits for wine consumers.

To determine the total [SO2] in the wine, titration is used. First, the sodium hydroxide is added into the wine sample to decompose bisulfite molecules (freeing SO2). 

Next, the solution is acidified with sulfuric acid. Finally, redox titration (technique) is performed with iodine as a titrant and starch as an indicator.

Description of the mentioned acid-base analysis techniques

Steam distillation – involves adding wine sample to boiling water in a flask that is being heated (i.e. using a heating mantle, water bath, etc) to evaporate volatile acids which are then collected as distillate using a condenser. 

Thin layer chromatography – involves the separation of substances according to their solubility as they distribute themselves throughout the stationary phase by travelling in a mobile phase. A silica gel is typically used as the stationary phase. The non-volatile acids that we want to test are placed on the silica gel whereby the gel is then partially submerged in an appropriate solvent (e.g. water). The more soluble the acid is, the higher up the silica gel in which the acid will travel. From this, the different components in the sample can be separated based off solubility.

Paper chromatography – The procedure is the same as thin layer chromatography but the stationary phase is not silica gel but a chromatography paper is used instead. The solvent for paper chromatography is typically water.

Liquid gas chromatography –The procedure is the same as the two chromatography techniques mentioned above. However, the stationary phase is typically a liquid inside a chamber and the mobile phase is a gas. The  sample is inserted into a separated heated chamber whereby the sample’s component can be separated according to their mass. This is because the components that are heavier will travel at a slower speed thus will be detected by the detector at a later point in time compared to the component in the sample that are of lower molecular mass. This is a quantitative measure of the components so compared to the two chromatography techniques mentioned above which are qualitative.

Hydrometry – procedure already discussed in the above learning objective involving the use of hydrometer and calibration curve.

Titration – already discussed in Learning Objective #1 in this week’s notes.

Redox Titration – Normal titration, however, the titrant and analyte has an oxidant and reductant relationship such that a redox reaction can occur. The titrant does not necessarily need to be the oxidant or reductant. 

Juice Industry

Another example of acid-base analysis techniques used in juice industry in the manufacture of orange juice. Titration can also used to determine the acid content in juices such as ascorbic acid in orange juice by titrating it against a strong base such as hydrochloric acid. 

Once the concentration of the ingredients are identified the correct nutritional information can be printed on the product’s nutrition label, allowing customers to select according to their nutritional demand. 

This is important because consumers expect that the stated ingredients that are either contained or removed from the product and their concentration to accurate. If not, the manufacturer may face legislative risk from their customers. 

By Aboriginal and Torres Strait Islander Peoples

Aboriginal and Torres Strait Islander Peoples demonstrated their technique of using acids and bases through their choice of selecting specific fruits and plants to meet their specific needs.

By performing volumetric analysis such as titration, chemists are able to determine the nature and concentration of various substances (acids and bases) that are contained within the fruits and plants used by Aboriginal and Torres Strait Islander Peoples. 

The Davidson plum is a natural Australian fruit that approximately 100 times more ascorbic acid (vitamin C) than contained in an orange. For such reason, it has a very sour taste due to the high concentration of acid. Aboriginal and Torres Strait Islander Peoples consumed the Davidson plum as way to boost their body’s nutrient level which reduced their chance of having scurvy disease.

When exposed to water, the soap tree’s leaves is able to lather or produce a soap solution that have antibacterial properties and thus act as an antiseptic. The reason for this is that the leaves contain saponin acid which has the ability to suppress bacteria growth. Aboriginal and Torres Strait Islander Peoples used the soap tree leaves as a way to heal cuts on their skin. 

Aboriginal and Torres Strait Islander Peoples also used yellow ochre (hydrated iron hydroxide) to treat stomach upsets. The chemistry behind is that the yellow ochre is basic and thus can react and neutralise with any excess hydrochloric acid in the stomach. This served as a way for Aboriginal and Torres Strait Islander People to remove any heartburns or stomach upsets. The yellow ochre has base similar to antacids which is an example of medicine used in everyday life as mentioned earlier in Module 6’s notes. 

The Grey Mangroves are used to treat stingray injury by preventing infection and neutralise the mildly acidic stingray venom. This is done by smashing the Grey Mangroves’s leaves and adding water to create a mixture that is a base which can be applied to the wound caused by the stingray.

Digital probes and instruments

Please refer to previous week’s notes on using pH probe and meter to determine the acidity of a range of solutions 

Learning Objective #7 - Conduct a chemical analysis for a range of common household substances for its acidity or basicity, for example:

- Soft drink
- Wine
- Juice
- Medicine

Titration can be produced on soft drinks, wine, juice and medicine (e.g. aspirin tablets) to determine its composition and pH. 

In this learning objective, we will briefly examine the acid or base content in each of the four substances. 

A youtube video will be released in Term 1 showing the titration of each of the four substances to determine their composition and pH. 

Soft Drinks

The carbon dioxide inside soft drinks react with water to form carbonic acid. It is the carbonic acid that gives soda drinks their tartiness or sharpness. Flavours are then added into this carbonated water, giving soft drinks their range of tangy flavours. 

The higher the carbonic acid concentration, the sharper the soft drink will taste. 

Since carbonic acid is acidic, soft drinks are acidic. 

In order to dissolve carbon dioxide in water to occur, or carbonation, it is accompanied with low temperature and high pressure. The high temperature allows carbon dioxide to be dissolved in water and low temperature enhances the solubility of gases such as carbon dioxide. 

You can relate this back to Le Chatelier’s Principle, a principle that we explored in Module 5. 

CO2(g) <-> CO2(aq) [Equation 1]

CO2(aq) + H2O(l) <-> H2CO3(aq) [Equation 2]

H2CO3(aq) <-> H+(aq) + HCO3-(aq) [Equation 3]

The forward reaction for both equation 2 and equation 1 are both exothermic. Therefore, increasing temperature will shift both Equation 1 and 2 to the left. This will decrease the concentration of carbonic acid in solution and therefore resulting in the soda going flat (losing its sharp taste). 

Increasing pressure will cause the equilibrium reaction in equation 1 to shift to the right. This would mean that the carbon dioxide gas will dissolve in water. The increase in [CO2] will shift equation 1 to dissolve the carbon dioxide in water which shifts equation 2 to the right to form carbonic acid. 

Vice versa, if you open the can of soda, it will cause equation 1 to shift the left to order to increase gas concentration and thus shift both equation 2 and 3 to the left. This lowers the concentration of carbonic acid and, overtime, the soda can will taste flat. 

Wine

See previous learning objective about acid-base analysis techniques in Industry – we talked about wines.

Juice

Some examples of juices are orange juice which has citric acid and apple juice has malic acid. 

Again, these acid gives the tartiness or sharpness of the juice. The higher the acid concentration in the juice, the more sharp or tangy flavour the juice will have. 

These acids are able to donate their proton(s) and thus act as Bronsted-Lowry acids.

Medicine

Medicines have phenol group (C6H5OH) which gives it their acidic properties.

Furthermore, medicine such as aspirin have a carboxylic acid group (COOH) which can act as an acid by donating a proton. It also enhances the water solubility of the drug. 

Aspirin is also called acetylsalicylic acid and can form acetylsalicylate ion as a conjugate base when it donates a proton. 

Many drugs, such as panadol, have a phenyl group (C6H5) which you see as a benzene ring but it has one less hydrogen as the phenyl group is attached to a the rest of the drug compound. Attached to this phenyl group is a hydroxyl group which makes the functional group called a phenol group (R-C6H5OH), where R is the rest of the organic drug compound which the phenol group is attached to. 

The hydrogen attached to the oxygen atom in the phenol group (C6H5OH) can be donated to a base such as water to form hydronium ions. 

Learning Objective #8 - Model the neutralisation of strong and weak acids and bases using a variety of media

There are many ways in you can model the reaction between:

Strong acid with Strong base -> Salt + Water

  • Example: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)

Strong acid with Weak base -> Salt + Water

  • Example: HCl(aq) + NH4OH(aq) -> NH4Cl(aq) + H2O(l)

Weak Acid with Strong base -> Salt + Water

  • Example: CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)

Weak Acid with Weak base -> Salt + Water

  • Example: CH3COOH(aq) + NH4OH(aq) -> CH3COONH4(aq) + H2O(l)

Previously, we have seen the titration curves and examples of each type of the above acid-base reaction in the conductometric titration curves section.

At school, you can model such reactions by watching a video, using the ball-and-stick modelling kit, drawing chemical structures, etc for the different types of acid-base reactions listed above.

NEW HSC Chemistry Syllabus Video - Titration and Conductometry Quantitative Analysis

[Coming this Summer!]

Week 7 Homework Problem Set

Week 7 Homework Question #1

List five criteria you would use when assessing the appropriateness of a substance used as  primary standard in titration and explain why such factors are important to be considered. 

Week 7 Homework Question #2

  • (a) Define a buffer 
  • (b) Explain a natural factor that affects a named buffer and how the buffer is able to resist significant changes in pH. 
  • (c) Explain what is meant by buffer capacity and the importance of the buffer’s function in its environment.

Week 7 Homework Question #3

Draw titration curves for the following acid-base titrations:

  • Strong acid titrated with strong base
  • Strong base titrated with weak Base
  • Weak acid titrated against strong Base
  • Weak base titrated against weak acid 
  • NOTE: Recall what is meant by ‘titrated with’ earlier in this week’s notes. It affects the shape/direction of your titration curve. 

Week 7 Homework Question #4

Describe the use of acid-base techniques used by Aboriginal and Torres Strait Islander Peoples.

Week 7 Homework Question #5

Describe the use and importance of acid-base analysis techniques used in Industry

Week 7 Homework Question #6

Describe the use of acid-base analysis techniques using digital probes. 

Week 7 Homework Question #7

Explain the difference in both Ka and pKa values for strong and weak acids. Similarly, explain the difference in both Kb and pKb values for strong and weak bases. 

Week 7 Homework Question #8

Explain why is it difficult to titrate a weak acid with a weak base (or vice versa)

Week 7 Homework Question #9

Describe the procedure you would use to determine the unknown concentration of HCl by titrating it with a known concentration of anhydrous sodium carbonate.

Week 7 Homework Question #10

Suppose you have 0.05M of CH3COOH and sodium hydroxide stock solutions and you want to prepare an acetate buffer at pH = 4.5 with a concentration of 0.20M and a volume of 0.01 litres. You given that the Ka of acetic acid is 1.8 x 10^-5. Explain your method in preparing such buffer.

Week 7 Homework Question #11

25mL of standardised NaOH at a concentration of 0.20 was titrated with H2SO4. The following are the titre of H2SO4 used was 29.0mL. Calculate the concentration of H2SO4 AND explain why as well as how you would standardised the sodium hydroxide. 

Week 7 Curveball Questions

Week 7 Curveball Question #1

Explain the reason to why the pH of the solution at the equivalence point of a strong acid – weak base titration is less than 7?

Week 7 Curveball Question #2

Explain the reason to why the pH of the solution at the equivalence point of a strong base – weak acid titration is more than 7?

Week 7 Curveball Question #3

Explain the reason to why the pH of the solution at equivalence point of a strong acid – strong base titration is equal to 7?

Week 7 Curveball Question #4

Suppose that you want to add 15mL of your standardised base solution into the conical base and titrate it with an acid with unknown concentration. 

(a) Explain why it is important to rinse the pipette with water and then the standardised solution after? If not, how will it affect your titre of acid used and calculated concentration of the acid in your results?

(b) Explain why it is important to rinse the burette with water and then with the acid after? If not, how will it affect your titre of acid used and concentration and calculated concentration of the acid in your results? 

Week 7 Curveball Question #5

Determine the molar mass and name of the diprotic acid (H2A) when 5.00 grams of the acid was standardised using a 250mL volumetric flask. The acid was titrated against potassium hydroxide to reach equivalence point. It was found that at equivalence point, a total of 11.1mL of 1.0M KOH was required to react with 25.0 mL of the acid. 

Week 7 Curveball Question #6

You compared your titration results with your peers and you notice some discrepancy in the results of your calculated concentration for your unknown acid. Propose three possible sources of that may have led to such discrepancy.

NOTE: The following curveball questions (Curveball questions 7 – 13) requires your knowledge from Module 5, more specifically, ICE tables. So if you are starting off with Module 6, you need to learn Module 5’s notes before you will be able to do the following questions. 

Week 7 Curveball Question #7

Calculate the pH of the solution when 100mL of 0.1M CH3COOH was titrated ith 60mL of NaOH at 0.1M. You are given that the Ka for acetic acid is 1.8 x 10^-5. 

Week 7 Curveball Question #8

Calculate the pH of acetic acid at 0.1M. You are given that the Ka for acetic acid is 1.8 x 10^-5. 

Week 7 Curveball Question #9

Calculate the pH of the solution when 0,1M of HNO2 was mixed with 0.2M of CH3COOH. You are given that the Ka for HNO2 is 4.5 x 10^-4 and the Ka for acetic acid is 1.8 x 10^-5.

Week 7 Curveball Question #10

Calculate the pH of the solution at equivalence point when 25mL of C6H5COOH at 0.165M was titrated with potassium hydroxide at 0.185M. You are given that the Ka for C6H5COOH is 6.6 x 10^-5 M

Week 7 Curveball Question #11

Calculate the pH for sodium carbonate solution at a concentration of 0.15 moles per litre. You are given that the Kb for carbonate ion to be 3 x 10^-4

Week 7 Curveball Question #12

Calculate the initial concentration of nitrous acid given that it has an acid dissociation constant of 6.0 x 10^-4 and where, when dissolved in water, the solution has pH of 3.65.

Week 7 Curveball Question #13

(a) Calculate the degree of ionisation for 0.05M acetic acid solution with a pH of 3.25

(b) Calculate the Ka of acetic acid solution at 0.05M with a pH of 3.25

Week 7 Curveball Question #14

Explain why a buffer cannot be made using HCl and NaOH? 

Week 7 Extension Questions

More exam-style questions for Week 7 content coming soon!

Solutions to Week 7 Questions

Solution to Question 1

An appropriate primary standard substance should have a high molecular mass. This criteria is important in reducing weighing errors when standardising the primary standard, thus reducing errors in calculated known concentration of the standard.  

An appropriate primary standard substance should have high solubility in distilled water. This criteria is important in ensuring the exact moles weighed for primary standard is dissolved in a fixed volume of water in the volumetric flask and hence ensures that the calculated known concentration is accurate. 

An appropriate primary standard substance should be non-hygroscopic.  This criteria is important in ensuring that the primary standard does not absorb moisture from the air which changes its mass which would otherwise affect the concentration of the unknown acid or base. 

  • For instance, sodium hydroxide is hygroscopic so it absorbs and reacts with carbon dioxide in the air and thus lowers the lower the concentration of OH- dissociated in solution and less moles of acid of unknown concentration is required to neutralise the NaOH standard. This would mean your calculated concentration of the acid would be higher than in reality. 

An appropriate primary standard substance should be non-efflorescent. This criteria is important in ensuring that the mass of the primary standard does not change by releasing water molecules into its surroundings. This would ensure that the calculated known concentration of the primary standard is accurate. 

An appropriate primary standard substance should have high purity. This criteria is important because it ensures that the calculated known concentration of the primary standard is accurate and its concentration is not lowered by impurities which would otherwise affect the concentration of the unknown acid or base.

*Refer to the relevant section of this week’s notes to see other selection criteria for an appropriate primary standard.

Solution to Question 2

Response to part (a): A buffer is a system (e.g. solution) made up of a weak acid and its conjugate base (or weak base and its conjugate acid) that is able to resist significant changes in pH when small amounts of strong acid or base are added into the buffer. 

Response to part (b): An example of a natural buffer is the carbonic acid – hydrogen carbonate buffer. occurring inside the body. The buffer can be chemically expressed as:

H2CO3(aq) <-> H+(aq) + HCO3-(aq)

A natural factor that affects the above buffer is the kidney’s function of filtering blood which absorbs H+ ions from the blood stream as it is a component of urea which would be secreted out of the body. This has the effect of decreasing the [H+] in the blood.

When the [H+] decreases in the blood stream it would mean that the pH of the blood would have increase (>7.3-7.4), the system will in response shift the equilibrium position to the right to increase the [H+], minimising the change as per Le Chatelier’s Principle. Therefore, more carbonic acid molecules will dissociate in the blood and the pH of the blood would decrease and return to the normal pH level of approximately 7.3-7.4. 

Response to part (c): Buffer capacity as the number of moles of H+ or OH- required to be added into the buffer system to increase or decrease the buffer’s pH by 1 for one litre of buffer solution. If you increase the buffer capacity, the buffer will be to increase the total moles of H+ and OH- ‘neutralised’ for the change the buffer system’s pH by one.

It is important for the blood to maintain a pH between 7.3-7.4 because enzymes in the blood are sensitive to changes in pH. Upon high fluctuation in pH, the enzymes in the blood can be denatured resulting the inability to catalyse the required chemical reactions which may result in death if the pH deviates from 7.3-7.4 for prolonged periods of time. Therefore, the carbonic acid-hydrogen carbonate buffer is essential in the blood to maintain the wellbeing of the organism (e.g. human).

Solution to Homework Question #3

[Insert Titration Curves Here]

Solution to Homework Question #6

Aboriginal and Torres Strait Islander Peoples consumed the Davidson plum, containing 100 times more ascorbic acid than vitamin, as way to boost their body’s nutrient level which reduced their chance of having scurvy disease.

Aboriginal and Torres Strait Islander Peoples used the soap tree leaves as a way to heal cuts on their skin. When exposed to water, the soap’s tree leaves is able to lather or produce a soap solution that have antibacterial properties and thus act as an antiseptic. The reason for this is that the leaves contain saponin acid which has the ability to suppress bacteria growth.

Aboriginal and Torres Strait Islander Peoples also used yellow ochre (hydrated iron hydroxide) to treat any stomach upsets or heartburns. This is because that the yellow ochre is basic and thus can react and neutralise with any excess hydrochloric acid in the stomach.

Solution to Homework Question #5

Wine contains fixed and volatile organic acids. It is important to analyse the volatile acid content in wine as their concentration is proportional to the susceptibility of the wine being spoiled. 

Acetic acid is a major component of volatile acids in wine and steam distillation is used to analyse the concentration of acetic acid. 

The distillate produced after steam distillation can be analysed to determine the components and concentration of fixed acids in the wine using thin layer chromatography.

The alcohol content (ethanol) in wine can also be analysed using hydrometry so the manufacturer can report the correct amount of alcohol in the wine. Using a hydrometer, different densities of ethanol solutions of known densities are measured against their boiling point. This means that a density calibration curve can be constructed which consists of temperature (y-axis) against a range of wine with varying around of total ethanol and their known ethanol densities (x-axis). From this, the calibration curve can be used to determine the unknown ethanol concentration in a wine sample.

*Also other components in wine – e.g. redox titration for free sulfur dioxide concentration in wine. See relevant section of the notes to remind yourself about that.

Solution to Homework Question #6

By attaching a pH meter with a pH probe and data logger, very precise information of acid and base solutions can be determined. 

For example, during titration between a known strong base and a unknown strong acid, the probe can be used to feed information to the data logger where the equivalence point of the titration can be accurately pinpointed. This can be used to determine the moles and volume of acid required to neutralise the strong base of known concentration. From there, the concentration of the unknown acid can be determined. 

A pH meter attached with a pH probe can also be used to determine the pH of a solution so that the relative acidity of the solution can be compared with others. This means that not only a solution can be determined if it is acidic or basic but also the strength of acidity or basicity.

Solution to Homework Question #7

The Ka expression for a generic acid, HA, is equal to [H+] [A-] / [HA]. As a strong acid dissociates completely in water, the [H+] and [A-] is much greater than [HA]. Comparatively, as weak acid partially ionise in water, the [H+] and [A-] is smaller than the [HA] and there will be less H+ and A- in solution for a weak acid compared to a strong acid. This means that strong acids have a higher Ka than weak acids. 

The Kb expression for a generic base, B, is equal to [BH+] [OH-] / [B]. As a strong acid dissociates completely in water, the [BH+] and [OH-] is much greater than [B]. Comparatively, as weak base partially ionise in water, the [BH+] and [OH-] is smaller than the [B] and there will be less BH+ and OH- in a solution of weak base compared to a strong base. This means that strong bases have a higher Kb than weak bases. 

Since pKa = -logKa, a higher Ka would mean a lower pKa. This means that strong acids have lower pKa than weak acids.

Since pKb = -logKb, a higher Kb would mean a lower pKb. This means that strong bases have lower pKb than weak acids.

Solution to Homework Question #8

As there is no sharp rise in pH at the equivalence point or small volumes of titrant added after equivalence point, the endpoint of the indicator may not be reached until excess titrant is added after the equivalence point. In such scenario, it is not possible to titrate the weak acid with a weak base (or vice versa). 

For a successful titration between a weak acid with a weak base (or vice versa), an indicator with a sharp endpoint (colours change over a small pH range) is required so that the endpoint can be seen close with the equivalence point without requiring to add significant amounts of titrant after the equivalence point. This will allow the titre of the titrant be acceptable, yielding an acceptance concentration for the unknown acid or base.

Solution to Homework Question #9

Please refer to this week’s notes for Inquiry Question #1. However, combine multiple steps into one to write the titration procedure more concisely and replace ‘HNO3’ with ‘HCl’ to match this question. 

Solution to Homework Question #10

HSC-Chemistry-Syllabus-Notes-Preparing-a-Buffer
HSC-Chemistry-Syllabus-Notes-Creating-a-Buffer
HSC-Chemistry-Syllabus-Notes-Creating-a-Buffer-2
HSC-Chemistry-Syllabus-Notes-Buffer-Solutions

Solution to Homework Question #11

HSC-Chemistry-Titration-Calculation-Questions

It is important to standardise sodium hydroxide with another acid before using it to titrate against the the sulfuric acid with unknown concentration. This is because sodium hydroxide is hygroscopic. This means, if NaOH is not standardised with another acid, the concentration of OH- ion available to react with H2SO4 would be lower.  This is because sodium hydroxide absorbs moisture in the air and reacts with carbon dioxide to form sodium carbonate. 

Thus, sodium hydroxide’s exact concentration can be standardised by titrating it against a chemically stable acid such as solid oxalic acid or potassium hydrogen phthalate. It is possible to accurate weigh oxalic acid or potassium hydrogen phthalate due to their chemical stability (non-hygroscopic), high solubility in water and high molecular mass. Thus, the concentration of oxalic and potassium hydrogen phthalate can be accurately determined by transferring it into a volumetric flask and titrating it against sodium hydroxide to find the known concentration of sodium hydroxide. 

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