Question

Determine the pH of a 0.05 M solution of hydrocyanic acid (HCN). Hydrocyanic acid has an acid-dissociation constant of 4.9 × 10-10.

Answer 1

Answer : The pH of a 0.05 M solution of hydrocyanic acid (HCN) is, 5.305

Solution :

The balanced equilibrium reaction will be,

$HCN\rightleftharpoons H^++CN^-$

The expression for dissociation constant will be,

$k_a=\frac{[H^+][CN^-]}{[HCN]}$

Let the concentration of $[H^+]$ and $CN^-$ will be, 'x'

$4.9\times 10^{-10}=\frac{(x)\times (x)}{0.05}$

$x=4.95\times 10^{-6}M$

That means the concentration of $[H^+]$ and $CN^-$ are $4.95\times 10^{-6}M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (4.95\times 10^{-6})$

$pH=5.305$

Therefore, the pH of a 0.05 M solution of hydrocyanic acid (HCN) is, 5.305

Answer 2

The answer is 5.3. Just did it in e2020