Question

1. consider a solution of hcn. list all of the aqueous species in this solution.

if the solution has a ph of 5.3, determine the concentration of all of the species you listed above.

Explanation:

Step1: Identify species in HCN solution

HCN is a weak acid. In water, it dissociates as $HCN(aq)
ightleftharpoons H^{+}(aq)+CN^{-}(aq)$ and water also auto - ionizes $H_2O(l)
ightleftharpoons H^{+}(aq)+OH^{-}(aq)$. So the aqueous species are $HCN$, $H^{+}$, $CN^{-}$, $OH^{-}$.

Step2: Calculate $[H^{+}]$ from pH

The formula for pH is $pH = -\log[H^{+}]$. Given $pH = 5.3$, then $[H^{+}]=10^{-pH}=10^{- 5.3}\ M\approx5.01\times10^{-6}\ M$.

Step3: Use $K_a$ of HCN

The $K_a$ of HCN is $6.2\times10^{-10}$. Let $[HCN]=x$, $[H^{+}]=y$, $[CN^{-}]=z$. From the dissociation of HCN, $K_a=\frac{[H^{+}][CN^{-}]}{[HCN]}$. Since $[H^{+}]$ from water auto - ionization is negligible compared to $[H^{+}]$ from HCN dissociation when dealing with weak acids in non - extremely dilute solutions, and at equilibrium $[H^{+}]\approx[CN^{-}]$ (from HCN dissociation). So $6.2\times10^{-10}=\frac{y\times z}{x}$. Also, from the charge balance $[H^{+}]=[CN^{-}]+[OH^{-}]$.

Step4: Calculate $[OH^{-}]$

We know that $K_w=[H^{+}][OH^{-}]=1.0\times10^{-14}$. Since $[H^{+}]=5.01\times10^{-6}\ M$, then $[OH^{-}]=\frac{K_w}{[H^{+}]}=\frac{1.0\times10^{-14}}{5.01\times10^{-6}} = 1.996\times10^{-9}\ M$.

Step5: Calculate $[CN^{-}]$

Since $K_a$ of HCN is very small, $[H^{+}]\approx[CN^{-}]=5.01\times10^{-6}\ M$. And $[HCN]$ can be approximated as the initial concentration of HCN. If we assume the initial HCN concentration is $C$, at equilibrium $[HCN]=C - [H^{+}]\approx C$ (because $[H^{+}]$ is very small due to the small $K_a$). But if we assume we started with pure HCN solution, and using $K_a=\frac{[H^{+}][CN^{-}]}{[HCN]}$, we can say $[HCN]=\frac{[H^{+}][CN^{-}]}{K_a}=\frac{(5.01\times10^{-6})\times(5.01\times10^{-6})}{6.2\times10^{-10}}\approx4.05\times10^{-2}\ M$.

Answer:

$[HCN]\approx4.05\times10^{-2}\ M$, $[H^{+}]\approx5.01\times10^{-6}\ M$, $[CN^{-}]\approx5.01\times10^{-6}\ M$, $[OH^{-}]\approx1.996\times10^{-9}\ M$