CAIE A-Level Chemistry 25.1.6 Buffer Concentrations and pH

Practise calculating buffer composition after neutralisation and using Ka or pKa to obtain the final pH.

[Maximum number: 4]

A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents.

Fig. 6.1 shows the chromatogram obtained by the student using water as a solvent.

Fig. 6.1

In a third experiment, the pH of the mixture of metal ions is kept constant using a buffer solution.

The student prepares the buffer solution by mixing $20.0 \mathrm{~cm}^{3}$ of $0.150 \mathrm{moldm}^{-3} \mathrm{KOH}(\mathrm{aq})$ and $50.0 \mathrm{~cm}^{3}$ of $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}(\mathrm{aq})$ .
$\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}$ is a weak carboxylic acid that has $\mathrm{pK}_{\mathrm{a}}=5.40$ .

Fig. 6.2

Calculate the pH of the buffer solution. Show all your working.

\mathrm{pH}=

M1 initially: moles of $\mathrm{KOH}=0.150 \times 20.0 \div 1000 \quad=3.0 \times 10^{-3}$

AND moles of acid HA $=0.100 \times 50.0 \div 1000=5.0 \times 10^{-3}$

M2 at equilibrium: moles of salt KA $\quad=3.0 \times 10^{-3}$ ecf

AND moles of acid $\mathrm{HA}=5.0 \times 10^{-3}$ - moles of NaOH $=2.0 \times 10^{-3}$ ecf

$\begin{aligned} \text { M3 } K_{a}=10^{-5.40}=3.98 \times 10^{-6} 3.98 \times 10^{-6}=\left[\mathrm{H}^{+}\right]\left[3.0 \times 10^{-3}\right] \div\left[2.0 \times 10^{-3}\right] {\left[\mathrm{H}^{+}\right]=2.65 \times 10^{-6} \text { eff } } \end{aligned}$

M4 $\mathrm{pH}=-\log \left(2.65 \times 10^{-6}\right)=\mathbf{5 . 5 8} / \mathbf{5 . 6}$ ecf on a calculated and identified $\left[\mathrm{H}^{+}\right]$

ALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4

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25.1.6—Relevant concentrations and the pH of buffer question 1