Question

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

Answer 1

The answer is actually 1.76x10^-11

Step-by-step explanation:

Fill in the box 1.76 and then -11

Answer 2

1.758 x 10⁻¹¹.

Step-by-step 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.