Answer
pH = 11.12
Work Step by Step
1. $KCN$ is a salt formed by $K^+$ and $CN^-$:
$K^+$ has insignificant acidity (conjugate acid of KOH, which is a strong base), so we don't need to consider.
$CN^-$ is a weak base.
2. We don't have the $K_b$ for $CN^-$, but we can use the $K_a$ for HCN to find it:
$HCN (K_a) = 4.9 \times 10^{-10}$
$K_a * K_b = K_w = 10^{-14}$
$ 4.9\times 10^{- 10} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 4.9\times 10^{- 10}}$
$K_b = 2.041\times 10^{- 5}$
- Note that, $KCN$ is a ionic salt, so:
$[KCN] = [CN^-]initial = 0.0852M$
3. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HCN] = x$
-$[CN^-] = [CN^-]_{initial} - x = 0.0852 - x$
For approximation, we consider: $[CN^-] = 0.0852M$
4. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HCN]}{ [CN^-]}$
$Kb = 2.041 \times 10^{- 5}= \frac{x * x}{ 0.0852}$
$Kb = 2.041 \times 10^{- 5}= \frac{x^2}{ 0.0852}$
$ 1.739 \times 10^{- 6} = x^2$
$x = 1.319 \times 10^{- 3}$
Percent ionization: $\frac{ 1.319 \times 10^{- 3}}{ 0.0852} \times 100\% = 1.548\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HCN] = x = 1.319 \times 10^{- 3}M $
$[CN^-] \approx 0.0852M$
5. Calculate the pH value.
$pOH = -log[OH^-]$
$pOH = -log( 1.319 \times 10^{- 3})$
$pOH = 2.88$
$pH + pOH = 14$
$pH + 2.88 = 14$
$pH = 11.12$
