The ionization constant of dimethylamine is 5.4 × 10^–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

Given:

K_b = 5.4 × 10^–4.

c=0.02M

To calculate the degree of ionization, we apply the formula

α=

Where K_b is the ionization constant

c is the concentration

∴ α =√

⇒α =0.1643

Thus, the degree of ionization is 0.1643

Ionization of 0.1 M NaOH;

Since, (CH₃)₂ NH⁺₂]=x

[OH^– ]=x+0.1 ≈ 0.1 M

Using the formula,

K_b =

⇒K_b =

As K_b = 5.4 × 10^–4.(given)

[(CH₃)₂ NH] = 0.02 M (given)

∴ 5.4 × 10^–4 =

⇒x = 0.054

Thus, It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.