Given:

pH = 2.34

Concentration of HCNO = 0.1M

Ionization of HCNO,

HCNO↔ H⁺ + CNO^-

As pH = -log[H⁺]

pH = 2.34 (given)

∴ log[H⁺] = -2.34 = -3.66

By taking antilog of both the side, we get

[H⁺] = antilog -3.66

⇒[H⁺] = 4.57 × 10^-3 M

At equilibrium, [H⁺] = [CNO^-] =4.57 × 10^-3 M

∴ Using the formula,

K_a =

⇒K_a =

⇒K_a =

⇒K_a = 2.09 × 10^-4

To calculate degree of ionization (α) applying formula directly, we have α = √

⇒α = √

⇒α = 0.0457

Thus, the ionization constant of the acid is 2.09 × 10^-4 and the degree of ionization is 0.00457