Edexcel A-Level Chemistry A2 14 Acid Base Equilibria Questions

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.

\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

An answer that makes reference to the following point:

- $\mathrm{pH}=-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$

A description in words could take a number of forms

Accept $\mathrm{pH}=-\mathrm{lg}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$

Accept $\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]$

Accept $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

Accept $\mathrm{pH}=-\mathrm{lg}\left[\mathrm{H}^{+}\right]$

Award pH is the negative $\log$ to the base 10 of oxonium / hydronium / hydroxonium / hydrogen ion concentration

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

a weak acid is dissociated (into its ions) to a small extent
and
a strong acid is (almost) completely dissociated

Allow propanoic acid is partially ionised and hydrochloric acid is fully ionised

calculation of pH

$$\begin{aligned} \mathrm{pH}=-\log _{10}(0.500)=0.30103 / 0.301 / 0.30 / 0.3 \text { Ignore SF } \end{aligned}$$

correct expression

Allow use of skeletal structure or molecular formula
$$K_{\mathrm{a}}=\frac{\left[\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{NO}_{3}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_{4} \mathrm{H}_{12} \mathrm{NO}_{3}^{+}\right]}$$

Do not accept round brackets

NB If no answer is given here, check in the answer to 22(b)(iii)

expression for acid dissociation constant

(1)

- calculation of hydrogen ion concentration

(1)

- evaluation of pH

(1)

Example of calculation:

$$K_{\mathrm{a}}=\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$$

Allow $\quad K_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$

Allow $\quad K_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right][\text {salt }]}{[\text { acid }]}$

Allow $1.3 \times 10^{-5}=\frac{\left[\mathrm{H}^{+}\right]^{2}}{0.5}$$$\left[\mathrm{H}^{+}\right]=2.5495 \times 10^{-3}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$$$$\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=2.59 / 2.6$$

Correct answer with some working scores 3 Ignore SF except 1SF

B

pOH values: 14 − 12.43 = 1.57 and 14 − 12.00 = 2.00. Therefore [OH−] values are 10⁻¹·⁵⁷ = 0.026915 mol dm⁻³ and 10⁻²·⁰⁰ = 0.0100 mol dm⁻³. Using c1v1 = c2v2: v2 = (0.026915 × 50.0)/0.0100 = 134.58 cm³. Volume of water to add = 134.58 − 50.0 = 84.58 cm³, so 84.6 cm³ / 85 cm³.

D

calculation of concentration of acid (1)

- calculation of $\left[\mathrm{H}^{+}\right]$ value

(1)

- expression or substitution into expression

(1)

- evaluation of $K_{a}$

(1)

Example of a calculation:
$0.0150 \div 94.5=0.0001587 / 1.59 \times 10^{-4}$$0.0001587 \div 1.5=0.00010582 / \mathbf{1 . 0 5 8} \times \mathbf{1 0}^{-\mathbf{4}}\left(\mathrm{mol} \mathrm{dm}^{-\mathbf{3}}\right)$$\left[\mathrm{H}^{+}\right]=10^{-3.42}=\mathbf{3 . 8 0 2} \times \mathbf{1 0}^{-\mathbf{4}}\left(\mathrm{mol} \mathrm{dm}^{-3}\right)$
( $\left[\mathrm{H}^{+}\right]^{2}=1.445 \times 10^{-7}$ )
$K_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]^{2}}{[\text { acid }]}$ or $K_{\mathrm{a}}=\frac{\left[1.445 \times 10^{-7}\right]}{\left[1.058 \times 10^{-4}\right]}$$K_{\mathrm{a}}=0.0013655 / 1.37 \times 10^{-3}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$
TE from M3 if answer < 1
Allow $1.36 \times 10^{-3}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$ for 4 marks
Ignore SF except 1 SF
Correct answer with no working scores 4
Penalise incorrect units for M4

Expert

An answer that makes reference to three of the following points:

- start pH between 2 and 4 and finish at 12-14

- vertical section of at least 3 pH units starting at or above 6 and finishing at or below 11 .

- S-shaped curve with equivalence( vertical section) at $25 \mathrm{~cm}^{3}$

(1)

(1)

(1)

Example of sketch

An answer that makes reference to two of the following points:

- phenolphthalein/ bromothymol blue / phenol red (1)

- because the indicator pH range $/ \mathrm{p} K_{\text {ind }} \pm 1$ lies (completely) (1) within the vertical section of the graph

TE from incorrect vertical section provided indicator chosen from Data Booklet

Allow a pH range for their vertical section

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

calculation of $\left[\mathrm{H}^{+}\right]$

(1)

- rearrangement of $K_{\mathrm{a}}$ equation

(1)

(1)

- substitution of values

and

calculation of $\left[\mathrm{RCOO}^{-}\right]=$moles required

(1)

OR Use of Henderson-Hasselbalch

- quotation of equation

(1)

- substitution of values

(1)

- calculation of $\left[\mathrm{RCOO}^{-}\right]=$moles required

(1)

Example of calculation
$$\left[\mathrm{H}^{+}\right]=10^{-3.71}=1.9498 \times 10^{-4}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$$$$\left[\mathrm{RCOO}^{-}\right]=\frac{K_{\mathrm{a}} \mathrm{x}[\mathrm{RCOOH}]}{\left[\mathrm{H}^{+}\right]}$$$$\begin{aligned} {\left[\mathrm{RCOO}^{-}\right] } =\frac{1.55 \times 1.38 \times 10^{-4}}{1.9498 \times 10^{-4}} =1.0970 / 1.10 / 1.1\left(\mathrm{~mol} / \mathrm{mol} \mathrm{dm}^{-3}\right) \end{aligned}$$
& $=1.0970 / 1.10 / 1.1\left(\mathrm{~mol} / \mathrm{mol} \mathrm{dm}^{-}\right)$$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\underset{[\mathrm{RCOOH}]}{\log \left[\mathrm{RCOO}^{-}\right]}$ & $\begin{array}{l}\text { Ignore round brackets } \text { or no brackets }\end{array}$

&
$$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\underset{[\mathrm{RCOOH}]}{\log \left[\mathrm{RCOO}^{-}\right]} \quad \begin{aligned} \text { Ignore round brackets } \text { or no brackets } \end{aligned}$$$3.71=3.86+\log \frac{\left[\mathrm{RCOO}^{-}\right]}{1.55}$$\left[\mathrm{RCOO}^{-}\right]=1.0970 / 1.10 / 1.1\left(\mathrm{~mol} / \mathrm{mol} \mathrm{dm}^{-3}\right)$
Henderson-Hasselbalch equation with acid and salt
concentration reversed gives 2.19 mol scores (2)
Correct answer with some working scores

An explanation that makes reference to the following points:

- (when half the acid has been neutralised)

(1)
$$\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]=\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]$$

(1)

- evaluation of pH

Allow in words Allow pH at half neutralisation $=\mathrm{p} K_{\mathrm{a}}$ propanoic acid Allow $\left[\mathrm{H}^{+}\right]$ at half neutralisation =K a propanoic acid
$$\mathrm{pH}=-\log _{10} 1.30 \times 10^{-5}$$
=4.8861 = 4.89/4.9

Ignore SF except 1SF

Correct answer scores 2

An answer that makes reference to one of the following points:

- pH is important for enzyme function

- constant pH is important for living organisms

- pH regulates growth of bacteria and fungi

- pH is linked to oxygen availability in water

Allow enzymes may be denatured at high or low pH

Allow enzymes may be denatured at the wrong pH

NB The word denatured is not on the specification so not essential

Allow to have the correct pH for metabolic processes / reactions

Allow to maintain the pH of blood Allow sudden pH change can be dangerous to organs of the body

Ignore "resists change to pH "

Expert