Answer : The pH of a 0.05 M solution of hydrocyanic acid (HCN) is, 5.305
Solution :
The balanced equilibrium reaction will be,
[tex]HCN\rightleftharpoons H^++CN^-[/tex]
The expression for dissociation constant will be,
[tex]k_a=\frac{[H^+][CN^-]}{[HCN]}[/tex]
Let the concentration of [tex][H^+][/tex] and [tex]CN^-[/tex] will be, 'x'
[tex]4.9\times 10^{-10}=\frac{(x)\times (x)}{0.05}[/tex]
[tex]x=4.95\times 10^{-6}M[/tex]
That means the concentration of [tex][H^+][/tex] and [tex]CN^-[/tex] are [tex]4.95\times 10^{-6}M[/tex]
Now we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (4.95\times 10^{-6})[/tex]
[tex]pH=5.305[/tex]
Therefore, the pH of a 0.05 M solution of hydrocyanic acid (HCN) is, 5.305
