Question

Hydrofluoric acid, HF, has a K_a of 6.8 ×10⁻⁴. What are [H₃O⁺], [F⁻], and [OH⁻] in 0.880 M HF?

Answer 1

Answer

[H3O⁺] = 0.02412 M

[F^-] = 0.02412 M

[OH^-] = 4.15 ×10^-13 M

Explanation

Dissociation equilibrium of HF is

HF(aq) + H2O(aq) <--------> H3O⁺(aq) + F^-(aq)

K_a = [H3O⁺][F^-] /[HF] = 6.8 ×10^-4

Initial concentration

[HF] = 0.880

[H3O⁺] = 0

[F^-​​​​​​] = 0

Change in concentration

[HF] = - x

[H3O⁺] = +x

[F^-] = + x

Equilibrium concentration

[HF] = 0.880 - x

[H3O⁺] = x

[F^-] = x

so,

x²/(0.880 - x) = 6.8 ×10^-4

solving for x

x = 0.02412

[HA] = 0.880 - 0.02412= 0.8559 M

[F^-] = 0.02412 M

[H3O⁺] = 0.02412 M

[H3O⁺] [OH^-] = K_w

K_w is ionic product of water , 1.00 × 10^-14

0.02412 M × [OH^-] = 1.00 ×10^-14M²

[OH^-] = 4.15 ×10^-13M