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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) $13.40$

The answer sheet states (C).

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.

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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

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

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    $\begingroup$ 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. $\endgroup$ – George Tian Jan 2 '18 at 8:12
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    $\begingroup$ @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. $\endgroup$ – andselisk Jan 2 '18 at 8:17

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