Question

Given that Ka for HCN is 6.2 × 10-10 at 25 °C, what is the value of Kb for CN– at 25 °C? Given that Kb for (CH3)3N is 6.3 × 10-5 at 25 °C, what is the value of Ka for (CH3)3NH at 25 °C?

Answer 1

• H2O -------> H+ + OH-

Equilibrium constant K is written as-

K = [H+][OH-]/[H2O]

Or K × [H2O] = [H+] [ OH-]

Or Kw = Ka × kB

= [10^-7] [10^-7] at 25°C temperature.

= 10^-14

Hence, Kw = 10^-14 at 25°C.

Where Kw = dissociation constant of water (H2O)

Ka = dissociation constant of acid (H+)

Kb = dissociation constant of base (OH-)

• Now you can easily calculate Ka and Kb as-

1) Ka = 6.2 × 10^-10 (given)

Kb = ?

We know Kw = Ka × Kb

Or. Kb = Kw/Ka (We know Kw = 1×10^-14 at 25°C)

Hence, Kb = 1× 10^-14/6.2 × 10^-10

= 1.6 × 10^-5.

Hence,Kb of CN- at 25°C = 1.6 × 10^-5.

2) Kb = 6.3 × 10^-5 (given)

Ka = ?

Because, Ka = Kw/kb

Hence,. Ka = 1 × 10^-14/6.3 × 10^-5

= 1.58 × 10^-10.

Hence Ka of(CH3)3NH at 25°C = 1.58 × 10^-10.