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Version 2*

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

Week 6 Content

Overview of Week 6 Inquiry Question – What is the role of water in solutions of acids and bases?

Learning Objective #1 – Conduct practical investigation to measure the pH of a range of acids and bases

Learning Objective #2 – Calculate pH, pOH, [H+], [OH-] for a range of solutions

Learning Objective #3 – Construct models to communicate the differences between strong, weak, concentrations and dilute acids and bases.

Learning Objective #4 – Investigate the use of pH to indicate the differences between the strength of acids and bases

Learning Objective #5 – Calculate the pH of the resultant solutions when solutions of acids and/or bases are diluted or mixed.

Learning Objective #6 – Write ionic equations for dissociation of acids and bases in water, conjugate acid/base pairs and amphiprotic salts 

Week 6 Lecture Video – Using Bronsted-Lowry Theory

Week 6 Homework Set (Essential for Band 5)

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

Week 6 Extension Questions

Solutions to Week 6 Questions

Overview of Week 6 Inquiry Question 

Welcome back to Week 6 of your HSC Chemistry Module 6 Notes!

In this week’s notes, we will explore the application of pH values further. 

Last week, we have examined Bronsted-Lowry reactions. In this week notes, we will explore how we can determine the pH of the solution when we mix an acid with a base. 

We will also examine in detail what is meant by a strong and weak acid. We will look into the difference between a concentrated and dilute acid. Vice versa, we will also examine strong, weak, concentrated and dilute bases. 

We touched on pOH earlier in Module 5. We will explore how we use pOH to calculate the basicity of a substance. 

Lastly, we will finish off this week’s notes by examining what amphiprotic substances are. They are substances that are capable of acting as an acid and a base! 

Learning Objective #1 - Conduct a practical investigation to measure the pH of a range of acids and bases

This is a simple practical.

The aim of this practical is to determine the pH of a range of acidic and basic substances. 

Since we did a similar practical last week by making a natural indicator using red cabbage, we won’t use an indicator to test acidity/basicity of substances for this learning objective. 

We will use a pH probe to measure the acidity/basicity of a substance instead so that we have some variety in our responses although variety is not always necessary to get the job done. 

The point basically is that you can still make the red cabbage indicator and use it to determine pH of the substances if you wish. 

Anyhow, let’s see how we can use a pH probe and pH meter to determine the pH of various acids & bases!

Procedure

Step 1: Attach the pH probe and adaptor to the pH meter. 

  • NOTE: On the pH meter, there is a electronic display screen that shows the pH value which you can read off from and determine the pH of the substance.

Step 2: Attach the adaptor to the electrical power supply

Step 3: Wash the lower part of the probe with distilled water.*

Step 4: Pour 25ml of 0.1M HCl, 0.1 NaOH, 0.1M ammonia, orange juice, vinegar, red wine, distilled water, soapy water, oven cleaner into separate test tubes and label as test tubes 1-9 respectively along with the name of solution it contains. 

Step 5: Submerge the pH probe into test tube 1 and record the pH shown on the pH meter. 

Step 6: Wash the pH thoroughly with distilled water.*

Step 7: Repeat steps 5-6 to measure the pH of a different substance in another test tube. 

*NOTE: You can also insert the pH probe in a buffer solution with a pH of 7 (i.e. neutral buffer) in a separate beaker to reset the pH reference point to 7 rather than rinsing the probe with distilled water. Note that distilled water have a pH of 7 which we used to rinse the pH probe. 

Results

You should have obtained a pH value of less than 7 for all the acids that you measured using your pH probe. Vice versa, you should have obtained a pH value of more than 7 for all the bases that you have measured using your pH probe. For neutral substances, such as distilled water, they should have a pH of 7. 

The key objective of this practical, apart from learning how to measure pH of substances, is for you to identify that: 

  • Acids have a pH of less than 7.
  • Bases have a pH of more than 7.
  • Neutral substances have a pH equal to 7.

Learning Objective #2 - Calculate the pH, pOH, [H+] and [OH-] for a range of solutions

Now that we have measured the pH values for a range of acids and bases, we did not learn how to actually calculate the pH value mathematically. This is because we used an instrument (pH probe or indicator pH colour chart) to give us the pH value. 

The formula to calculate the pH value is:

pH = -log10[H+]

  • The ‘p’ in pH just means that the ‘H’ (i.e the concentration of hydrogen ions) is being multiplied by -log10.

Sometimes you may see the mathematical formula in calculating pH written as:

pH = -log10[H3O+]

This is because the acid that you are dealing is normally diluted with water or is not a pure liquid, i.e. the acid is dissolved in water. This means that the acid solution is in aqueous state. Therefore, the hydrogen ions will react with water to form hydronium ions, H3O+.

H+(aq) + H2O(l) -> H3O+(aq).

This means that pH = log10[H+] does not take into account of the reaction between hydrogen ions with water. Either way, both of the formulae are accepted in HSC chemistry. 

The point of the formula means that you can substitute the concentration value of H+ or H3O+ into the pH formula to obtain the pH value of the substance you are measuring.

Calculating pOH

The autoionisation of water constant (Kw) is equal to 10-14 at 25 degrees celsius.

Kw =[H3O+] x [OH] = 10-14

We have the autoionisation of water because research have found that the purest form of water conducts electricity. This would suggest the presence of ions in such pure water. These ions are hydrogen and hydroxide ion which are formed when water molecules (H2O) dissociate.

The reason towards why water dissociates is due to the interaction between water molecules. We have explored Bronsted-Lowry acid and base reactions in the previous week. 

The interaction between two water molecules would lead one molecule to act as an Bronsted-Lowry acid and the other water molecule to act as a Bronsted-Lowry base.

H2O(l) + H2O(l) <-> H3O+(aq) + OH(aq)

It is this reaction, as shown in the above equation, that leads to the autoionisation of water, which involves the interaction and dissociation between the two water molecules.

  • NOTE: Since the autoionisation of water can arrive to an equilibrium, if a closed system of water is allowed to reach equilibrium, there will be water molecules, hydrogen ions and hydroxide ions in the system.

This will therefore have an equilibrium constant. Since water are the only reactants, we call this the autoionisation of water constant, Kw

Kw = [H3O+] x [OH]

Recall that pure liquid acts as a solvent like water in this case, so it will not be included in the equilibrium constant expression. This is because water is on the reactant and on product side as the hydrogen and hydroxide ions are dissolved in water as aqueous ions. Therefore, water will therefore appear on the numerator and denominator of the Kw expression and will cancel each other out. So, we do not include water in our Kw expression as we have already explained in Week 3’s notes of Module 5 

At 25 degrees cel*, Kw = 1.0 x 10-14

Therefore, Kw = 1.0 x 10-14 = [H3O+] x [OH]

  • *Recall that equilibrium constant values, such as Kw, will change depending on temperature. 
  • If the pH of the solution is 7 at 25 degrees celsius, then [H+] and [OH] are both equal to 1.0 x 10-7
  • If the pH of the solution is less than 7 (acidic) at degrees celsius, the [H3O+] must be greater than 1.0 x 10-7.
  • If the pH of the solution is greater than 7 (basic) at 25 deg. celsius, the [OH-] must be greater than 1.0 x 10-7

Applying logarithmic laws in mathematics to Kw, it would mean that  pH + pOH = 14 

  • Where pH = -log10[H+]. 

Similarly, pOH, a measure of basicity, means the [OH] is being multipled with -log10.

  • This means that pOH = log10[OH].
What is the significance of all of this?
  • This means that if we know pH, we can determine the pOH using pH + pOH = 14.
  • This also means that if we know the [H+] or [OH] we can also calculate both pH and pOH. Or, you can calculate it through using using the ionisation of water constant. 
There will be homework questions at the end of this week’s notes for you to practice applying these formulas and relationships. 

Calculating the concentration of hydrogen ions, [H+]

There are two ways to calculate [H+].

The first way is to calculate the [H+] when given pH.

  • In this case, you can use pH = -log10[H+], so that 10-pH = [H+]

The second way is to calculate [H+] when given pOH or [OH].

  • You can use pOH = -log10[OH], so 10-pOH= [OH
  • You can then the autoionisation of water, [H+] x [OH] = 1.0 x 10-14, to solve for [H+]. 

OR

  • If given pOH or [OH], you can use pOH + pH = 14. Then, you can solve for pH and using 10-pH= [H+]
 
  • Side Note: You can also calculate [H+], [OH], and thus pH, using Ka and Kb which we will learn next week.

Calculating the concentration of hydroxide ions, [OH-]

Similar concept as calculating the concentration of hydrogen ions. 

There are two ways to calculate [OH]

The first way is to calculate the [OH] when given pOH.

  • In this case, you can use pOH = -log10[OH], so that 10-pOH = [OH]

The second way is to calculate the [OH] when given pH or [H+]

  • You can use pH = -log10[H+], so 10-pH = [H+]
  • You can then use the autoionisation of water constant, [H+] x [OH] = 1.0 x 10-14, to calculate [OH]

OR

  • If you are given pH or [H+], you can use pOH + pH = 14. Then, you can solve for pOH using 10-pOH = [OH]

Learning Objective #3 - Construct models to communicate the differences between strong, weak, concentrated and dilute acids and bases

Difference between strong acids and weak acids 

The difference between strong acids and weak acids are their acid molecules’ degree of ionisation or dissociation. 

All of the acid molecules of a strong acid dissociate completely into its constituent ions in water. This is called complete ionisation or 100% degree of ionisation or dissociation. 

Comparatively, weak acid molecules do not all dissociate completely into its constituent ions in water. Some of the acid molecules completely dissociate whereas some do not, i.e. partial dissociation

Those acid molecules that do not dissociate in water remain as acid molecules dissolved in water rather than ions (H+ ions and its corresponding anion e.g. Cl- in HCl). 

It is the difference in the degree of ionisation of acid molecules in water that allows us to classify an acid into a strong acid or weak acid. 

HSC-Chemistry-Strong-and-Weak-Acids

Difference between strong bases and weak bases

On the same token as the classification of strong and weak acids, we can also classify the strength of a base depending on its degree of ionisation of its base molecules in water.

Strong bases have 100% degree of ionisation where all of its base molecules are ionised in water. 

For weak bases, not all of its base molecules are ionised in water but only a portion do, i.e. partial ionisation.

Those base molecules that do not dissociate in water remain as base molecules dissolved in solution rather than ions (e.g. OH,aka conjugate base, and corresponding cation such as Na+ in NaOH).

Therefore, it is the degree of ionisation of base molecules in water whereby we can say if a base is strong or weak. 

HSC-Chemistry-Strong-Base-and-Weak-Base

Difference between a concentrated acid vs a dilute acid

To classify or define a strong or weak acid, we need to examine the amount (moles) of acid molecules that are dissolved in water. 

The higher the concentration of acid molecules dissolved in water, the more concentrated it is. 

Hence, compared to a dilute acid solution, a concentrated acid solution have more moles of acid molecules dissolved in the same volume of solvent (water). 

When comparing two acids in terms of concentration and their volumes are different, we need to look at which acid solution have greater concentration of dissolved acid molecules using c = n/V and compare their molarity. The higher the molarity, the more concentrated the acid is. 

  • NOTE: We DO NOT look at the degree of ionisation (or H+ ions) of acid molecules when we are talking about concentrated verses dilute acid solutions.
HSC-Chemistry-Concentrated-and-Dilute-Acid

Difference between a concentrated base vs a dilute base

To classify or define a strong or weak base, we need to examine the amount (moles) of base molecules that are dissolved in water. The higher concentration of base molecules dissolved in water, the more concentrated it is. 

Hence, compared to a dilute base solution, a concentrated base solution have more moles of base molecules dissolved in the same volume of solvent (water). 

When comparing two bases in terms of concentration and their volumes are different, we need to look at which base solution have greater concentration of dissolved base molecules (molarity) using c = n/V. 

  • That is, we compare the molarity. The base solution with greater molarity would be more concentrated or have higher concentration.

NOTE: We DO NOT look at the degree of ionisation (or OH- ions) of base molecules when we are talking about concentrated verses dilute base solutions.

HSC-Chemistry-Concentrated-and-Dilute-Base

Molarity value of a concentrated acid/base solution

You may be wondering what is the numerical definition of an concentrated acid or concentrated base in terms of molarity. 

Well, the majority of chemist agree on that a solution that has a molarity of 3M or higher is considered concentrated. Some, however, consider a solution to be concentrated if it has a molarity of 6M or higher. 

Therefore, it is best practice to describe a concentrated or dilute solution is by stating their molarity. 

  • For example: It is much less vague to say prepare 2M of NaOH rather than saying prepare a concentrated NaOH solution. 

We can however use the term ‘concentrated’ to compare between acid solutions (or between base solutions). The acid solution with highest concentration of dissolved acid molecules would be the most concentrated acid. Vice versa, the base solution with the highest concentration of dissolved base molecules would be the most concentrated base.

Insoluble strong bases 

Recall from HSC Preliminary Course, there are soluble and insoluble bases. Soluble bases are called alkalis. 

NaOH is an example of an alkali because sodium hydroxide is soluble in water. We know this from the solubility rules that we explored in Module 5.

Mg(OH)2 is an example of an insoluble base. This is because magnesium hydroxide is not very soluble in water, i.e. sparingly soluble compound. So, is magnesium hydroxide a strong or weak base? 

The answer is both! It is a strong, insoluble base! However, it is a weak, soluble base (weak alkali). 

Magnesium hydroxide is a strong, insoluble base because for all the magnesium hydroxide is that dissolved in water, all of the magnesium hydroxide is 100% ionised into magnesium and hydroxide ions.

This is because the number of moles of hydroxide ions that was formed in solution is twice the number of magnesium hydroxide molecules that was dissolved in water. 

  • Mg(OH)2(s) (dissolved) -> Mg2+ (aq) + 2OH (aq)
For example, you know that 0.05 moles of magnesium hydroxide is dissolved in water. You would expect 0.1 mol of hydroxide ions to be formed using mole ratio of 1:2.
 
This is because magnesium hydroxide is a strong, insoluble base. All of the magnesium hydroxide that is dissolved will completely dissociate.  
 

Magnesium hydroxide is a weak, soluble base because not all of the magnesium hydroxide that you pour into the water will dissolve. Hence, it is a weak alkali where the number of moles of hydroxide ions formed in solution is NOT twice the number of magnesium hydroxide that you poured into the water.

Mg(OH)2(s) (poured into beaker) -> Mg2+(aq) + 2OH(aq)

For example, you poured 2 mol of Mg(OH)2 into a beaker of water and only 0.1 mol of OH ions were formed. This is because only 0.05 mol of Magnesium hydroxide was dissolved in water. The 0.05 mol of magnesium hydroxide then completely dissociated into hydroxide and magnesium ions, giving you 0.1 mol of OH ions which you calculated when you, perhaps, measured the pH of the solution. This is why magnesium hydroxide is a weak, soluble base (weak alkali).

Learning Objective #4 - Conduct an investigation to demonstrate the use of pH to indicate the differences between the strength of acids and bases

This learning objective is similar to the practical investigation we explored in Learning Objective #1 in this week’s notes. 

However, the only difference is that the objective of this learning to determine the strength of acids and bases using pH. 

In the investigation this week in Learning Objective #1, we were only concerned about the pH of an acid and base and that’s it. We did not relate the pH value to the strength of acids and bases. 

In this investigation, we will relate the pH of different acids to determine their relative strengths and compare the pH of different bases to determine their relative strengths. So, let’ see how we can do that.

Procedure (Same as Learning Objective #1)

Step 1: Attach the pH probe and adaptor to the pH meter. 

  • NOTE: On the pH meter, there is a electronic display screen that shows the pH value which you read off from and determine the pH of the substance.

Step 2: Attach the adaptor to the electrical power supply

Step 3: Wash the lower part of the probe with distilled water.*

Step 4: Pour 25ml of 0.1M HCl, HNO3, acetic acid (CH3COOH), H3PO4, KOH, NaOH, ammonia (NH3) and NaF into separate test tubes and label as test tubes 1-8 respectively along with the name of solution it contains. 

Step 5: Submerge the pH probe into test tube 1 and record the pH shown on the pH meter. 

Step 6: Wash the pH thoroughly with distilled water.*

Step 7: Repeat steps 5-6 to measure the pH of a different substance in another test tube. 

*NOTE: You can also insert the pH probe in a buffer solution with a pH of 7 (i.e. neutral buffer) in a separate beaker to reset the pH reference point to 7 rather than rinsing the probe with distilled water. Note that distilled water have a pH of 7 which we used to rinse the pH probe. 

Expected Results

In the condition of equal acid molecules concentrations, compared to strong acids, weak acids will have a higher pH. 

This is because strong acids have more moles of hydrogen ions in solution due to complete dissociation as we have explored in the previous learning objective. This would mean that strong acids have higher concentration of hydrogen ions in solution than weak acids provided the strong and weak acids have equal concentration to start with.

In the condition of equal base molecule concentrations, compared to strong bases, weak bases will have a lower pH. 

This is because strong bases have more moles of hydroxide ions in solution due to 100% ionisation into hydroxide ions and its corresponding cation. This would mean that strong bases have higher concentration of hydroxide ions in solution than weak bases provided than the strong and weak bases have equal concentration.

Acids experiment results:

  • HCl and HNO3 are strong acids.
  • CH3COOH and H3PO4 are weak acids. You will expect these to have higher pH (less acidic) than HCl and HNO3 since the concentration of the strong and weak acids are the same. 

Bases experiment results:

  • KOH, NaOH are strong bases
  • NH3 and NaF are weak bases. You will expect these weak bases to have lower pH (less basic) than KOH and NaOH since the concentration of the strong and weak bases are the same.

Learning Objective #5 - Calculate the pH of the resultant solution when solutions of acids and/or bases are diluted

Q: Calculate the pH of the resulting solution when you mix/dilute 500mL of 0.15M HCl with 300mL of 0.14 NaOH?

HSC-Chemistry-Question-pH-of-resultant-solution-2
HSC-Chemistry-Question-pH-of-resultant-solution

Learning Objective #6 - Write ionic equations to present the dissociation of acids and bases in water, conjugate acid/bases in solution and amphiprotic nature of some salts, for example:

- Sodium hydrogen carbonate

- Potassium dihydrogen phosphate

Amphiprotic salts

The term ‘Amphiprotic‘ means capable of acting both as a proton donor and proton acceptor. The word ‘Amphi’ means ‘both’ or ‘two’

If you recall from last week’s notes, we said that a Bronsted-Lowry acid is a proton donor and a Bronsted-Lowry base is a proton acceptor. 

Therefore, if a substance is amphiprotic, it is capable of both acting as a Bronsted-Lowry acid or Bronsted-Lowry base. Depending on the substance which the amphiprotic substance is reacting with, the substance will determine whether the amphiprotic substance will act as an acid or a base. 

  • Although the following method is not assessable in HSC Chemistry, you can determine if an amphiprotic substance will act as an acid or base by comparing the equilibrium constant of the two possible reactions to determine which reaction is more favoured.

Generally speaking, amphiprotic substance with Group 3 metals will tend to act as a base and so can dissolve in acidic solutions. Recall that, from last week’s notes, we explored how metal oxides will dissolve in water to produce basic solutions.

NOTE: For a substance to be amphiprotic, it MUST have at least one hydrogen atom. This is because if it does not have an hydrogen atom, it cannot act as a Bronsted-Lowry acid. Hence, it cannot be amphiprotic.

Amphiprotic Substance Example: Sodium hydrogen carbonate

Sodium hydrogen carbonate acting as a Bronsted-Lowry Acid (proton donor):

Full Ionic Equation: NaHCO3(s) + H2O(l) <-> H3O+ (aq) + CO32-(aq) + Na+ (aq)

  • When sodium hydrogen carbonate was dissolved in water, it reacted with water and donated a proton (H+)
  • Note: Na+ is a spectator ion so we can remove it from both sides of the equation to get net ionic equation.

Sodium hydrogen carbonate acting as a Bronsted-Lowry Base (proton acceptor):

Full Ionic Equation: NaHCO3(s) + H2O(l) <-> 2H+(aq) + CO32-(aq) + OH(aq) + Na+(aq)

  • When sodium hydrogen carbonate was dissolved in water, it reacted with water and accepted a proton (H+)
  • Note: Na+ is a spectator ion so we can remove it from both sides of the equation to get a net ionic equation.

Amphiprotic Substance Example: Potassium dihydrogen phosphate 

Potassium dihydrogen phosphate acting as a Bronsted-Lowry Acid (proton donor):

KH2PO4(aq) + H2O(l) <-> HPO42-(aq) + H3O+(aq) + K+(aq)

Full Ionic Equation: K+ (aq) + H2PO4 (aq) + H2O(l) <-> HPO42-(aq) + H3O+(aq) + K+(aq)

  • When potassium dihydrogen phosphate was dissolved in water, it reacted with water and donated a proton (H+).
  • Note: K+ is a spectator ion so we can remove it from both sides of the equation to get net ionic equation.

Potassium dihydrogen phosphate acting as a Bronsted-Lowry Base (proton acceptor):

KH2PO4(aq) + H2O(l) <-> H3PO4(aq) + OH (aq) + K+ (aq)

Full Ionic equation: K+ (aq) + H2PO4– (aq) + H2O(l) <-> 3H+(aq) + PO43-(aq) + OH– (aq) + K+ (aq)

  • When potassium dihydrogen phosphate was dissolved in water, it reacted with water and accepted a proton (H+)
  • Note: K+ is a spectator ion so we can remove it from both sides of the equation to get net ionic equation.

How to determine if a substance is acidic or basic?

When determining if a substance is acidic or base, here is the general rule of thumb. 

We first look at the two scenarios where your substance is acting as an acid or base. We then do not discard the one that is least likely. 

We determine this be excluding the reaction where it forms potential products that you have not seen before. 

Let’s do an example to see what we mean by this!

Suppose you want to determine if ammonium chloride is acidic or basic. Well, one method to determine this is that you can dissolve it in water. 

  • If the resulting solution is acidic, then ammonium chloride is an acid. 
  • Vice versa, if the resulting solution is basic, then ammonium chloride is a base. 
  • Lastly, if the resulting solution is neutral, then the substance is neutral.

Suppose that ammonia acts as a Bronsted-Lowry Base:

NH4+ (aq) + H2O(l) <-> NH5+(aq) + OH(aq)   [Equation 1]

Suppose that ammonia acts as a Bronsted-Lowry Acid:

NH4+ (aq) + H2O(l) <-> H3O+(aq) + NH3(aq)   [Equation 2]

We have never seen NH5+ ion in our HSC course yet. Therefore, in the exam, we will NOT write or consider Equation 1.

We have seen H3O+ and NH3 in our HSC course. Therefore, we will write and consider Equation 2.

In equation 2, since hydronium ion is a strong acid and ammonia (NH3) is a weak base, the resulting solution is acidic. Therefore, ammonium chloride is an acidic substance or is an acid. 

  • How we know that ammonia is a weak base? Well, in the next section, we will have a condensed list of acids and bases based on their strength. You will get more exposure to strong/weak acids and bases throughout HSC Chemistry and you can add to the list. 

In next week’s note, you will learn that how the chloride ion is a conjugate base of the strong acid HCl. The conjugate base of strong acids is very weakly basic and cannot react with water (to accept a H+) and thus has no effect on the pH. 

Acid and Base strength 

Strength of acid in decreasing order (i.e. strongest to weakest):

#1 – H3O+ 

#2- HClO2

#3- HF

#4 – H2CO3

#5 – CH3COOH

#6 – NH4+

#7 – HPO42-

#8 – H2O

Strength of base in decreasing order (i.e. strongest to weakest):

#1 – OH

#2 – PO43-

#3 – NH3

#4 – CH3COO

#5 – HCO3

#6 – F

#7 – ClO2

#8 – H2O

NOTE: From the chart, clearly H+ (or H3O+) is the strongest acid and OH is the strongest base. So a solution contains only H+ and OH ions and they’re both present in equal moles, the pH will be 7. 

  • If you have a solution that contains #1 acid (H+) and #1 base (OH), it means that the solution’s pH = 7 if they’re equal in moles and there is NO other ion or substance in solution that affects the pH.
  • Similarly, if a solution contains #2 acid (HClO2) and #2 base (PO43-), it means the solution’s pH = 7 if they’re equal in moles and there is NO other ion or substance in solution that affects the pH.
  • If you have a solution #1 acid (H+) and #2 base (PO43-), it would mean that the solution’s pH is less than 7 if they’re equal in moles and there is NO other ion in solution or substance that affects the pH. This is because #1 is always stronger than #2 as the list is in decreasing order of acidity/basicity.

Dissociation of strong and weak acid in water

HSC-Chemistry-Dissociation-Of-Acids-In-Water

Dissociation of strong and weak base in water

HSC-Chemistry-Dissociation-Of-Bases-In-Water

Net Ionic Equations between Strong Base - Weak Acid and Strong Acid - Weak Bases.

When writing net ionic equation involving strong base and weak acid, you don’t split the weak acid molecule. This is because undissociated weak acid molecules dominant the ions from the dissociated acid molecules (i.e. H+ and the corresponding anion).

For example, the reaction between sodium hydroxide (strong base) and acetic acid (weak acid)

Chemical equation: KOH(aq) + CH3COOH(aq) -> KCH3COO(aq) + H2O(l)

Full Ionic Equation: K+(aq) + OH-(aq) + CH3COOH(aq) -> K+(aq) + CH3COO-(aq) + H2O(l)

Net Ionic Equation: OH-(aq) + CH3COOH(aq) -> CH3COO-(aq) + H2O(l)

The same logic applies for reaction between weak bases with strong acid, don’t split the weak base molecule. This is because most of the weak base molecules are in the undissociated state. So, you write equation based on the most dominant state.

Week 6 Homework Set

Homework Question #1: Explain the difference between a named strong acid and a named weak acid.

Homework Question #2: Explain the difference between a named weak acid and a named weak base

Homework Question #3: Suppose you have three solutions, those being 0.1M of citric acid, acetic acid and HCl. Their respective pH were 2.1, 2.9 and 1.0. Account for the pH difference between the three solution. 

  • HINT: Citric acid is triprotic meaning that each citric acid molecule has the capacity to donate 3 hydrogen ions in an acid-base reaction.

Homework Question #4: Explain the difference between a concentrated acid solution and a dilute acid solution

Homework Question #5: Explain the difference between a concentrated base solution and a dilute base solution

Homework Question #6: Suppose that you mix 0.1L of 8M sodium hydroxide to a beaker containing 300mL of HCl at a concentration of 0.00005M, determine the pH of the resulting solution

Homework Question #7: Define what is meant when a substance is ‘amphiprotic’ and provide a named example.

Week 6 Curveball Questions

Curveball Question #1: Explain why magnesium hydroxide is a strong base despite it not having high solubility in water.

Curveball Question #2: Determine whether or not dissolving 20g of ammonium chloride in distilled water will produce a solution that is acidic, basic or neutral. Explain why using appropriate chemical equations.

Curveball Question #3: Determine whether or not a solution of 30g of sodium fluoride in distilled water will produce a solution that is acidic, basic or neutral. Explain why using appropriate chemical equation 

Week 6 Extension Questions

Extra questions for this Inquiry Questions uploaded regularly here!

Solutions to Week 6 Questions

Solution to Week 6 Homework Question #1: Strong acids, such as hydrochloric acid, ionise completely in water. Comparatively, weak acids such as acetic acid partially ionise in water. This means that for a monoprotic weak acid, the concentration of the acid will NOT be equal to the concentration of the hydrogen ions as there is only partial ionisation. 

Solution to Week 6 Homework Question #2: Weak acids such as acetic acid partially dissociate into hydrogen ions and corresponding conjugate base ions in water.

For example: CH3COOH(aq) <-> CH3COO-(aq) + H+(aq)

Weak bases, such as ammonium hydroxide, partially dissociate into hydroxide ions and corresponding cation ions in water. 

For example: NH4OH(aq) <-> NH4+(aq) + OH-(aq)

Solution to Week 6 Homework Question #3: HCl is a strong acid and therefore all acid molecules (HCl molecules) will ionise completely in water. Comparatively, acetic and citric acid are weak acids. Therefore, acetic and citric acid molecules will partially ionise in water. 

Since the concentration of all three acids are equal, the concentration of hydrogen ions in HCl will be the highest as it’s a strong acid. The reason why citric acid has a lower pH (i.e. more acidic) than acetic acid is because citric acid molecules are triprotic. That is, each citric acid molecule has the capacity to donate 3 hydrogen ions. However, this dissociation is not complete as citric acid is a weak acid. 

Comparatively, acetic is also a weak acid and all of its acid molecules will only partially dissociate to form ions in water. However, acetic acid molecules are monoprotic rather than triprotic, i.e. each acid molecule has the capacity to dissociate to form one hydrogen ion in water. This means that the concentration of hydrogen ions will be less than in the case of citric acid.

Therefore, due to the higher concentration of hydrogen ions in the 0.1M citric acid solution than in the 0.1M acetic solution, the citric acid molecules have a lower pH or is more acidic. 

Solution to Week 6 Homework Question #4: A concentrated acid solution has greater moles of acid molecules than a dilute acid solution. Due to this, if the concentrated and dilute acid solution has comprised of the same acid AND both solutions have the same volume, the concentrated acid solution will have a greater molarity. That is, the concentrated acid solution is more acidic.

Solution to Week 6 Homework Question #5: A concentrated base solution has greater moles of base molecules than a dilute base solution. Due to this, if the concentrated and dilute base solution is comprised of the same base AND both solutions have the same volume, the concentrated base solution will have a greater molarity. That is, the concentrated base solution more basic.

Solution to Week 6 Homework Question #6:

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

n(NaOH) = cV = 8 x 0.1 = 0.8 moles

n(HCl) = cV = 0.00005 x 0.3 = 1.5 x 10^-5 moles

Limiting reagent is HCl. 

n(NaOH)excess after neutralisation = 0.8 – (1.5 x 10^-5) = 0.8 moles (0.7999.. stored in calculator)

[ NaOH ]excess = n (NaOH) excess / V(acid + base) = 0.8 / 0.4 = 2.0M

[OH-] = 8M as the mole ratio between NaOH : OH- is 1:1.

pOH = -log[OH-] = -log(2) = -0.3

Yes , it’s possible to have negative pH (concentrated, strong acid solutions) and pH that exceeds 14 (concentrated, base solutions)

pOH = 14 – pH 

pH = 14 – (-0.3) = 14.3 

Solution to Week 6 Homework Question #7: Amphiprotic refers to a substance that is capable of donating and accepting a proton depending on the system’s environment (e.g. solvent) that it is located in.

For instance, sodium hydrogen carbonate is an amphiprotic substance. 

Sodium hydrogen carbonate is capable of acting as a Bronsted-Lowry Acid (proton donor):

NaHCO3(s) + H2O(l) <-> H3O+(aq) +CO32-(aq) + Na+(aq)

Sodium hydrogen carbonate is capable of acting as a Bronsted-Lowry Base (proton acceptor):

NaHCO3(s) + H2O(l) <-> 2H+(aq) + CO32-(aq) + OH–(aq) + Na+(aq)

Solution to Week 6 Curveball Question #1: Magnesium hydroxide is a strong, sparingly soluble base because, for all of the magnesium hydroxide that is dissolved in water, they are completely ionised to form magnesium and hydroxide ions. However, magnesium hydroxide is a weak, soluble base (weak alkali). 

Solution to Week 6 Curveball Question #2: 

NH4+(aq) + H2O(l) -> NH3(aq) + H3O+(aq)

Chloride ions are too weak of a conjugate base to interact and react with water and so they will not affect the resulting solution’s pH.

Therefore, the resulting solution’s pH is governed by the reaction between ammonium ions and water molecules. From here, water is neutral, ammonia is a weak base and hydronium ions are strong acids. Therefore, the resulting solution’s pH is acidic due to the presence of hydronium ions. 

Solution to Week 6 Curveball Question #3: 

F-(s) + H2O(l) -> HF(aq) 

Sodium ions are too weak of an acid to interact and react with water through hydrolysis. Hence, they will not affect the resulting solution’s pH.

So, the resulting solution’s pH is determined by neutral water molecules and hydrogen fluoride that is formed. Hydrogen fluoride is a weak acid as shown in the following equation. 

HF(aq) <-> H+(aq) + F-(aq)

Therefore, the resulting solution will be acidic. 

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