A1.0 m h2s solution has a ph = 3.75 at equilibrium. what is the value of ka?

Ka = 3.16 x 10⁻⁸.

Explanation:

3.75 = -log[H⁺].

log[H⁺] = -3.75.

∴ [H⁺] = 1.78 x 10⁻⁴ M.

where, Ka is the dissociation constant of the acid,

C is the concentration of the acid (C = 1.0 M).

∴ 1.78 x 10⁻⁴ M = √(Ka)(1.0 M),

by squaring the both sides:

∴ (1.78 x 10⁻⁴ M)² = (Ka)(1.0),

∴ Ka = 3.16 x 10⁻⁸.

10 years is the answer