# USNCO question - Finding pH of salt solution from pKa of its parent acid

The question from 2017 USNCO is as below:

What is the $$\mathrm{pH}$$ of a $$\pu{0.25 M}$$ solution of $$\ce{NaCN}$$? (The $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{HCN}$$ is $$9.21$$.)

(A) $$4.91$$
(B) $$8.61$$
(C) $$11.30$$
(D) $$13.40$$

My understandings/reasoning:

• salt hydrolysis: solution of salt derived from strong base and weak acid is basic;
• $$\ce{NaCN}$$ is derived from the base $$\ce{NaOH}$$ (strong) and acid $$\ce{HCN}$$ (weak according to given $$\mathrm{p}K_\mathrm{a}$$);
• Solution is basic, so (A) is incorrect.

However, I am unable to proceed from here.

Sodium cyanide is a salt formed by a strong base and a weak acid, so the reduced ionic equation of hydrolysis is

$$\ce{CN- + H2O <=>[K_\mathrm{h}] HCN + OH-}$$

\begin{align} K_\mathrm{h} &= \left.\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]} \qquad \right|\cdot \frac{[\ce{H+}]}{[\ce{H+}]}\\ &= \frac{[\ce{HCN}][\ce{OH-}][\ce{H+}]}{[\ce{CN-}][\ce{H+}]}\\ &=\frac{K_\mathrm{w}}{K_\mathrm{a}} \label{eq:1}\tag{1} \end{align}

Sodium cyanide is highly soluble salt, so one can neglect auto-protolysis of water, and also express hydrolysis constant (assuming $$[\ce{OH-}] = [\ce{HCN}]$$ and $$C$$ – concentration of sodium cyanide) as follows:

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{C-[\ce{OH-}]} \label{eq:2}\tag{2}$$

Equating \eqref{eq:1} and \eqref{eq:2}:

$$\frac{10^{-14}}{10^{-9.21}} = \frac{[\ce{OH-}]^2}{0.25-[\ce{OH-}]}\, ,\label{eq:3}\tag{3}$$

and solving the quadratic equation \eqref{eq:3} for $$[\ce{OH-}]$$, one can find that $$[\ce{OH-}] \approx \pu{2e-3 M}$$, $$\mathrm{pOH} = 2.70$$, hence

$$\mathrm{pH} = 14.00 - \mathrm{pOH} = 14.00 - 2.70 = 11.30$$

• Thank you. I noticed that you used water's dissociation constant at 25°C. However, many questions specify the temperature, while this doesn't, leading me to believe that a temperature-independent solution would be preferred. Commented Jan 2, 2018 at 8:12
• @GeorgeTian There is none unless the problem itself explicitly tells you the temperature or $K_\mathrm{w}$. In the absence of such info, it's safe to assume that $K_\mathrm{w} = 10^{-14}$ and move along. Commented Jan 2, 2018 at 8:17