Question

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

Answer 1

The value of
`K_b` is
`2.04* 10^(-5)`.

Step-by-step explanation:

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

The dissociation constant of the HCN (acid)=
`K_a=4.9* 10^(-10)`

Ionic product of water =
`K_w=1* 10^(-14)`

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

`K_w=K_a* K_b`

`K_b=(1* 10^(-14))/(4.9* 10^(-10))=2.04* 10^(-5)`

The value of
`K_b` is
`2.04* 10^(-5)`.

Answer 2

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