Question

Find out ka for 10-2M HCN acid having poh is 10

Answer 1

Answer:The dissociation constant of $10^{-2} M$ HCN is $K_a=1.01\times 10^{-6}$.

Explanation:

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

At e'qm  c-x           x    x

$K_a=\frac{[H^+][CN^-]}{[HCN]}=\frac{x\times x}{(c-x)}$

$pH+pOH=14$

pOH=10, So pH = 4

$pH=4=-\log[H^+]$

$[H^+]=x=10^{-4} M$

$K_a=\frac{[H^+][CN^-]}{[HCN]}=\frac{x\times x}{(c-x)}$

$K_a=\frac{10^{-4}\times 10^{-4}}{10^{-2}-10^{-4}}=1.01\times 10^{-6}$

The dissociation constant of $10^{-2} M$ HCN is $K_a=1.01\times 10^{-6}$.

Answer 2

Answer: Dissociation constant for $HCN$ is $10^{-6}$.

Explanation:

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

At t=0     c              0   0

At e'qm   c-x           x    x

$K_a=\frac{[H^+][CN^-]}{[HCN]}=\frac{x\times x}{(c-x)}$

pH+pOH=14

pOH=10, So $pH=14-10=4$

$pH=-log[H^+]$

$4=-log[H^+]$

$[H^+]=10^{-4}M$

Thus $x=10^{-4}M$

$c=10^{-2}M$

$c-x=10^{-2}-10^{-4}=9.9\times 10^{-3}$

Putting in the values we get:

$K_a=\frac{10^{-4}[10^{-4}}{9.9\times 10^{-3}}=10^{-6}$

Thus dissociation constant for $HCN$ is $10^{-6}$.