Edexcel IAL Chemistry Topic 14 Acid–base equilibria

Write the acid equilibrium first, identify the justified approximation and carry concentration units and logarithms carefully before interpreting acid strength or pH.

Exam points

Write proton-transfer equations and label Brønsted–Lowry conjugate acid–base pairs.
Calculate pH, hydrogen or hydroxide concentration using strong-acid, Ka or Kw relationships.
Compare acid strength through dissociation, Ka and pKa rather than concentration alone.

Covered syllabus points

14.1—A Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry base is a proton acceptor and that acid-base reactions involve 14.2—Identify Brønsted–Lowry conjugate acid-base pairs 14.3—Define the term ‘pH’14.4—PH from hydrogen ion concentration 14.5—Hydrogen-ion concentration in mol dm−3 from pH using [H+] = 10^(−pH)14.6—The difference between a strong acid and a weak acid in terms of the degree of dissociation 14.7—The pH of a strong acid 14.8—Deduce the expression for the acid dissociation constant, Ka, for a weak acid 14.9—The pH of a weak acid from Ka or pKa values, making relevant assumptions Students will not be expected to solve quadratic 14.10—Define the ionic product of water, Kw 14.11—The pH of a strong base from its concentration 14.12—Define the terms ‘pKa’ and ‘pKw’14.13—Analyse data from the following experiments: i measuring the pH of a variety of substances 14.14—Ka for a weak acid from experimental data given the pH of a solution containing a known mass of acid 14.15—Draw and interpret titration curves 14.16—Select a suitable indicator for a titration 14.17—What is meant by the term ‘buffer solution’14.18—The action of a buffer solution 14.19—The pH of a buffer solution given appropriate data 14.20—The concentrations of solutions required to prepare a buffer solution of a given pH 14.21—How to use a weak acid-strong base or strong acid-weak base titration curve to: i demonstrate buffer action ii determine Ka 14.22—The importance of buffer solutions in biological environments: i buffers in cells and in blood (H2CO3/HCO-3) ii in foods 14.23—CORE PRACTICAL 11 Finding the Ka value for a weak acid

[Maximum number: 6]

The equation for the formation of ammonia in the Haber Process is shown

1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+1 \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})

(a)

Ammonia from the Haber Process reacts with acids.

With phosphoric acid, $\mathrm{H}_{3} \mathrm{PO}_{4}$ , a number of products are formed in solution. One of these is the fertiliser diammonium hydrogenphosphate.

[ 6 ]

(i)

Write an equation for the production of this fertiliser.

State symbols are not required.

[ 2 ]

formula of diammonium hydrogenphosphate

(1)

- balanced equation

(1)

$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$ $2 \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$

Allow multiples

Allow ions for the product

Allow for M2
$\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow\left(\mathrm{NH}_{4}\right) \mathrm{H}_{2} \mathrm{PO}_{4}$

Allow ions for the product

No other TE

Ignore state symbols even if incorrect

(ii)

Write an ionic equation to show that ammonium ions are acidic in aqueous solution.
State symbols are not required.

[ 1 ]

$\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{NH}_{3}+\mathrm{H}^{+}$

OR

- $\mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+}$

Allow → instead of ⇌

Do not award reactions reversed

Allow
$\begin{aligned} & \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} & \text { Allow } \rightleftharpoons \text { instead of } \rightarrow \end{aligned}$

Ignore state symbols even if incorrect

(iii)

A solution containing both ammonia and ammonium ions acts as a buffer. Explain, using a relevant ionic equation, the effect of adding a small amount of acid to this buffer.

[ 3 ]

An answer that makes reference to the following points:

- the mixture contains a large amount/ (large) reservoir of both ammonium ions and ammonia (1) / of $\mathrm{NH}_{4}^{+}$ and $\mathrm{NH}_{3}$

Either

- added $\mathrm{H}^{+}$ reacts with ammonia to form ammonium ions $/ \mathrm{H}^{+}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+}$

Or

- added $\mathrm{H}^{+}$ combines with $\mathrm{OH}^{-}$ ions in water to form water $/ \mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}$

And ammonia reacts with water to produce $\mathrm{OH}^{-} \quad$ (1) ions $/ \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}$

- ratio of ammonium ions to ammonia hardly (1) changes

Do not award incorrect formulae such as $\mathrm{NH}_{3}{ }^{-}$ in M1 and M2 but allow TE in M3 Ignore comments about acid / base in relation to $\mathrm{NH}_{4}^{+}$ / $\mathrm{NH}_{3}$ unless defined

Allow → instead of ⇌ Allow $\mathrm{H}_{3} \mathrm{O}^{+}$

Allow → instead of ⇌

This marking point must include at least one ionic equation

Allow remains constant

Allow pH is unchanged / changes very little because added $\mathrm{H}^{+}$ removed and change in concentration of $\mathrm{NH}_{3}$ and $\mathrm{NH}_{4}^{+}$ is small

[Maximum number: 17]

This question is about acids and bases.

(a)

Some $K_{\mathrm{a}}$ and $\mathrm{p} K_{\mathrm{a}}$ values for several weak acids are shown.

Table for Question (a) — Edexcel A-Level Chemistry A2

[ 9 ]

(i)

The structure of $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$ is shown.

Figure for Question (i) — Edexcel A-Level Chemistry A2

Write the $K_{\mathrm{a}}$ expression for this weak acid.

[ 1 ]

An answer that makes reference to the following point:

- $(\mathrm{Ka}=) \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right]}{\left[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\right]}$

Figure for Answer to Question (i) — Edexcel A-Level Chemistry A2

Do not award $\mathrm{C}_{6}$ for $\mathrm{C}_{5}$

Do not award non-square brackets

Allow [ $\mathrm{H}_{3} \mathrm{O}^{+}$ ] for [ $\mathrm{H}^{+}$ ]

Allow use of

for $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}$

Do not award $\mathrm{C}_{5} \mathrm{H}_{4} \mathrm{NH}$ for $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}$

Do not award charged $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}$ , eg $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}^{-}$

Allow use of

Allow $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}^{+}$ for $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$

Do not award omission of charge from $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$

(ii)

When methanoic acid is added to propanoic acid, an equilibrium is set up containing two acid-base pairs.

Complete the equilibrium, labelling the acid-base pairs as A1, B1 and A2, B2.

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{HCOOH} \rightleftharpoons \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .

*(iv) A student measured the pH of separate $0.050 \mathrm{~mol} \mathrm{dm}^{-3}$ solutions of $\mathrm{CH}_{2} \mathrm{ClCOOH}$ and of $\mathrm{CHCl}_{2} \mathrm{COOH}$ , using a pH meter.

The student also calculated the pH of these solutions, making two assumptions:
- $\left[\mathrm{H}^{+}\right]=\left[\mathrm{A}^{-}\right]$
- $[\mathrm{HA}]_{\text {equilibrium }}=[\mathrm{HA}]_{\text {initial }}$

The measured and calculated pH values are shown.

Table for Question (ii) — Edexcel A-Level Chemistry A2

Discuss the differences between the student's measured and calculated pH values.

In your answer, you should
- show how the student calculated their pH values

\begin{aligned} & {\left[K_{\mathrm{a}}\left(\mathrm{CH}_{2} \mathrm{ClCOOH}\right)=1.4 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3} ;\right.} \\ & \left.K_{\mathrm{a}}\left(\mathrm{CHCl}_{2} \mathrm{COOH}\right)=4.5 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3}\right] \end{aligned}

- explain, with reference to the assumptions made, why there is a difference between the calculated and measured pH values
- suggest why the measured pH values are higher than the calculated pH values.

[ 2 ]

An answer that makes reference to the following points:

- $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}_{2}{ }^{+}$

and

$\mathrm{HCOO}^{-}$

- HCOOH A1 and $\mathrm{HCOO}^{(-)} \mathrm{B} 1$ and
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \mathrm{~B} 2 \text { and } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}_{2}{ }^{(+)} \text {A2 }$

(1)

Allow $\mathrm{COOH}^{-}$

19(a)(iv)

This question assesses a student's ability to show a coherent and logically structured answer with linkages and fully-sustained reasoning.

Marks are awarded for indicative content and for how the answer is structured and shows lines of reasoning.

The following table shows how the marks should be awarded for indicative content.

Number of indicative marking points

seen in answer

Number of marks awarded for

indicative marking points

6

5-4

3-2

1

0

\end{tabular}

The following table shows how the marks should be awarded for structure and lines of reasoning.

Figure for Answer to Question (ii) — Edexcel A-Level Chemistry A2

\end{tabular}

The mark for indicative content should be added to the mark for lines of reasoning. For example, an answer with five indicative marking points that is partially structured with some linkages and lines of reasoning scores 4 marks (3 marks for indicative content and 1 mark for partial structure and some linkages and lines of reasoning).

If there are no linkages between points, the same five indicative marking points would yield an overall score of 3 marks (3 marks for indicative content and no marks for linkages).

If there is any incorrect chemistry, deduct mark(s) from the reasoning. If no reasoning mark(s) awarded, do not deduct mark(s).

Comment: Look for the indicative marking points first, then consider the mark for the structure of the answer and sustained line of reasoning.

Indicative points:

- IP1: use of $\left[\mathrm{H}^{+}\right]=\sqrt{\left(K_{\mathrm{a}} \times[\mathrm{HA}]\right)}$

- IP2: use of $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

- IP3: indication that $[\mathrm{HA}]_{\text {equilibrium }}$ is lower than $[\mathrm{HA}]_{\text {initial }}$

- IP4: (because) dissociation (of both acids) is significant

- IP5: (calculated pH values lower than measured pH values because) [HA] is overestimated in the calculations

- IP6: (difference greatest for) $\mathrm{CHCl}_{2} \mathrm{COOH}$ (as is) stronger acid or two Cl atoms in $\mathrm{CHCl}_{2} \mathrm{COOH}$ are more electron withdrawing than one / stabilise anion more / weaken O-H bond more

If calculations shown, pH values are 2.0775 and 1.3239

Ignore $\left[\mathrm{H}^{+}\right]=\left[\mathrm{A}^{-}\right]$ assumption is not valid Ignore $\left[\mathrm{H}^{+}\right]>\left[\mathrm{A}^{-}\right]$

Do not award $\left[\mathrm{A}^{-}\right]>\left[\mathrm{H}^{+}\right]$

Allow dissociation is not negligible Allow dissociation occurs

Do not award dissociation is negligible Do not award dissociation is complete

Do not award $\left[\mathrm{H}^{+}\right]$ overestimated in calculation due to dissociation of water

Allow more dissociated for stronger Ignore strong acid for stronger acid Ignore $\mathrm{CHCl}_{2} \mathrm{COOH}$ has larger $K_{\mathrm{a}}$ / smaller pKa

(iii)

A student measured the pH of separate 0.050 mol dm−3 solutions of $\ce{CH2ClCOOH}$ and $\ce{CHCl2COOH}$ , using a pH meter. The student also calculated the pH of these solutions, making two assumptions: [ $\ce{H+}$ ] = [A−] and [HA]equilibrium = [HA]initial. The measured and calculated pH values are shown.
0.050 mol dm−3 $\ce{CH2ClCOOH}$ : measured pH 2.11, calculated pH 2.08.
0.050 mol dm−3 $\ce{CHCl2COOH}$ : measured pH 1.52, calculated pH 1.32.
Discuss the differences between the student's measured and calculated pH values.
In your answer, you should show how the student calculated their pH values, explain why there is a difference between the calculated and measured pH values, and suggest why the measured pH values are higher than the calculated pH values.

[ 6 ]

Indicative points include: use [ $\ce{H+}$ ] = sqrt(Ka × [HA]); use pH = −log[ $\ce{H+}$ ]; [HA]equilibrium is lower than [HA]initial because dissociation is significant; calculated pH values are lower because [HA] is overestimated; the difference is greatest for $\ce{CHCl2COOH}$ because it is the stronger acid / has two electron-withdrawing Cl atoms. Award structure and sustained reasoning as in the markscheme.

(b)

4-Hydroxybenzoic acid, $\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH}$ , is a diprotic acid, which dissociates in two stages. Each stage has a different $\mathrm{p} K_{\mathrm{a}}$ value.

Stage $1 \mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COO}^{-}$

\mathrm{p} K_{\mathrm{a} 1}=4.5

Stage $2 \mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COO}^{-} \rightleftharpoons \mathrm{H}^{+}+{ }^{-} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COO}^{-}$

\mathrm{p} K_{\mathrm{a} 2}=9.3

In a titration, $75.0 \mathrm{~cm}^{3}$ of $0.025 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaOH}$ was added to $25.0 \mathrm{~cm}^{3}$ of $0.025 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH}$ .

Complete the titration curve.

(3)

[ 3 ]

An answer that makes reference to the following points:

- vertical section at $25 \mathrm{~cm}^{3}$

and

with height in the range of 1 to 4 pH units

(1)

- vertical section at $50 \mathrm{~cm}^{3}$

and

with height greater than 0.4 pH units

1)

- (buffered section with) pH 9.3 at $37.5 \mathrm{~cm}^{3}$

(1)

Example of completed titration curve:

Allow slight slope in range of $24 \mathrm{~cm}^{3}$ to $26 \mathrm{~cm}^{3}$

(1) Allow any pH between 9.0 and 9.6

Allow reading of pH from midpoint volume between two vertical sections as TE on M1/M2

(c)

A student was given $50.0 \mathrm{~cm}^{3}$ of a solution of sodium hydroxide.

The pH of this solution was 12.43.
The student was asked to adjust the pH to 12.00 , by dilution with deionised water. The student did not have access to a pH meter.

Calculate the volume of deionised water, in $\mathbf{c m}^{3}$ , the student should add to the original solution.

[ 5 ]

[Maximum number: 10]

This question is about sulfuric acid and its salts.

(a)

Sulfur trioxide is used to produce sulfuric acid.

[ 3 ]

(i)

The pH of a $0.10 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of sulfuric acid at $25^{\circ} \mathrm{C}$ is 0.97 .

Calculate the concentration of hydrogen ions, in $\mathrm{mol} \mathrm{dm}^{-3}$ , in this solution.

[ 1 ]

calculation of $\left[\mathrm{H}^{+}(\mathrm{aq})\right] /\left[\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\right]$

Example of calculation:
$\begin{aligned} & {\left[\mathrm{H}^{+}(\mathrm{aq})\right] /\left[\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\right]=10^{-0.97}} & \quad=0.10715 / 0.1072 / 0.107 / 0.11\left(\mathrm{~mol} \mathrm{dm}^{-3}\right) \end{aligned}$

Ignore SF except 1 SF Ignore incorrect units

Correct answer with no working scores (1)

Do not award 0.1 / 0.10 / 0.214

(1)

(ii)

In an aqueous solution of sulfuric acid, the following equilibria exist.

\begin{array}{ll} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) & K_{\mathrm{a}} \text { very large } \\ \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) & K_{\mathrm{a}}=0.012 \mathrm{moldm}^{-3} \end{array}

Explain, in terms of these equilibria, why the concentration of hydrogen ions in a $0.10 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of sulfuric acid is not $0.20 \mathrm{~mol} \mathrm{dm}^{-3}$ . No calculation is required.

[ 2 ]

An explanation that makes reference to the following points:

First equilibrium

- the first ionisation of sulfuric acid is complete or the equilibrium position of the first equation lies very far to the right

(1)

Second equilibrium

- so $\left[\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\right]$ (from the second equilibrium) is very small

(1)

Allow $\left[\mathrm{H}^{+}(\mathrm{aq})\right]$ for $\left[\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\right]$ Ignore missing state symbols

Allow high $\left[\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\right]$ from first equilibrium Allow acid fully dissociates in first equilibrium

Ignore just acid fully dissociates

Allow second equilibrium shifts to the left Allow second dissociation is suppressed / further dissociation is prevented

(2)

(b)

A buffer solution is made from $\mathrm{HSO}_{4}^{-}$ and $\mathrm{SO}_{4}^{2-}$ ions.

[ 7 ]

(i)

Write two ionic equations involving $\mathrm{HSO}_{4}^{-}$ and $\mathrm{SO}_{4}^{2-}$ ions to show how this

[ 2 ]

$\mathrm{HSO}_{4}{ }^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{SO}_{4}{ }^{2-}+\mathrm{H}_{2} \mathrm{O}$

- $\mathrm{SO}_{4}{ }^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{HSO}_{4}{ }^{-}$

or
$\mathrm{SO}_{4}{ }^{2-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightarrow \mathrm{HSO}_{4}{ }^{-}+\mathrm{H}_{2} \mathrm{O}$

Allow equations in either order

Allow ⇌ provided equations written in directions shown

Ignore state symbols even if incorrect

Penalise non-ionic equations once only e.g. using HCl and NaOH

Allow $\mathrm{HSO}_{4}{ }^{-} \rightarrow \mathrm{SO}_{4}{ }^{2-}+\mathrm{H}^{+}$

and $\mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}$

for M1

(2)

(ii)

A buffer solution is formed by mixing
$25.0 \mathrm{~cm}^{3}$ of a solution that is $0.150 \mathrm{~mol} \mathrm{dm}^{-3}$ with respect to $\mathrm{SO}_{4}^{2-}$ ions with $75.0 \mathrm{~cm}^{3}$ of a solution that is $0.100 \mathrm{~mol} \mathrm{dm}^{-3}$ with respect to $\mathrm{HSO}_{4}^{-}$ ions.

Calculate the pH of this buffer solution.
$\left[K_{\mathrm{a}}\right.$ for $\mathrm{HSO}_{4}^{-}$ ions $=0.012 \mathrm{~mol} \mathrm{dm}^{-3}$ ]

Figure for Question (ii) — Edexcel A-Level Chemistry A2

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[ 5 ]

All question bank results loaded

Covered syllabus points

Topic 14: Acid-base Equilibria question 1

Question (a)

Question (i)

Question (ii)

Question (iii)

Topic 14: Acid-base Equilibria question 2

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question (c)

Topic 14: Acid-base Equilibria question 3

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)