What is the Ka of a 0.653 m solution of hydrocyanic acid with a ph of 5.47

1.758 x 10⁻¹¹.

Explanation:

∵ pH = - log[H⁺].

∴ 5.47 = - log[H⁺].

log[H⁺] = - 5.47.

∴ [H⁺] = 3.388 x 10⁻⁶ M.

∵ [H⁺] = √ (Ka. C)

∴ [H⁺]² = Ka. C

∴ Ka = [H⁺]²/C = (3.388 x 10⁻⁶) ² (0.653) = 1.758 x 10⁻¹¹.