Question

The ka for hcn is 4.9 ⋅ 10-10. what is the value of kb for cn-?

Answer 1

Kb=  Kw of  water/Ka  of  the acid

Kw  of water=   1  x  10^-14
Ka  of  acid=  4.9  x10^-10
Kb= ( 1  x10^-14) / ( 4.9  x10^-10) =   2.04  x10^-5

Answer 2

Answer:

The value of $K_b$ is $2.04\times 10^{-5}$.

Explanation:

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

The dissociation constant of the HCN (acid)= $K_a=4.9\times 10^{-10}$

Ionic product of water = $K_w=1\times 10^{-14}$

$K_b$ for $CN^-$(base) =$K_b=?$

$K_w=K_a\times K_b$

$K_b=\frac{1\times 10^{-14}}{4.9\times 10^{-10}}=2.04\times 10^{-5}$

The value of $K_b$ is $2.04\times 10^{-5}$.