Edexcel IAL Chemistry 14.9 Weak acid pH from Ka or pKa

Start by converting pKa when needed, use [H+]^2 = Ka[HA] only when the assumptions are justified and finish by checking that the pH remains acidic.

Exam points

Calculate [H+] and pH from Ka and the initial weak-acid concentration.
Convert pKa to Ka before rearranging the weak-acid expression for pH or required mass.
State assumptions such as negligible water ionisation and [HA]initial approximately [HA]equilibrium.

[Maximum number: 3]

Propanoic acid is a weak acid.

Calculate the pH of $0.500 \mathrm{~mol} \mathrm{dm}^{-3}$ propanoic acid at $25^{\circ} \mathrm{C}$ .

K_{\mathrm{a}}(\text { propanoic acid })=1.30 \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3} \text { at } 25^{\circ} \mathrm{C} .

expression for acid dissociation constant

(1)

- calculation of hydrogen ion concentration

(1)

- evaluation of pH

(1)

Example of calculation:

(3)

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

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

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14.9—The pH of a weak acid from Ka or pKa values, making relevant assumptions Students will not be expected to solve quadratic question 1