Question

Calculate the pH of a solution that is prepared by dissolving 0.23 mol of hydrofluoric acid (HF) and 0.57 mol of hypochlorous acid (HClO) in water and diluting to 3.60 L. Also, calculate the equilibrium concentrations of HF, F2, HClO, and ClO2. (Hint: The pH will be determined by the stronger acid of this pair.)

Answer 1

The equilibrium concentrations of HF = 0.058 , F2 = 0.006M , HClO =0.16M , and ClO2 = 7.7 × 10⁻⁷M.

Step-by-step explanation:

The Ka values for HClO₃ and HF are given as 2.9 × 10⁻⁸ and 6.6 × 10⁻⁴ respectively. The molar concentration for HF = 0.23/ 3.60L = 0.064 M and 0.57/ 3.60 = 0.16 M.

When HF is reacted with water, it ionizes to form H₃O⁺ and F⁻. The concentration of H₃O⁺ and F⁻ can be calculated below:

HF(aq) <------------------------> H30^+ + F^-.

Ka = [H^+] [F^-]/[HF] .

6.6× 10^-4 = [x][x]/ ( 0.064- x).

x = 0.0060 M.

The concentration of H₃O⁺ and F⁻ = 0.0060 M respectively.

The pH = - log [ H₃O⁺ ] = -log [0.0060] = 2.22.

When HClO is reacted with water, it ionizes to form H₃O⁺ and F⁻. The concentration of H₃O⁺ and ClO⁻ can be calculated below:

HClO(aq) <------------------------> H30^+ + ClO^-.

Ka = [H^+] [ClO^-]/[HClO] .

6.6× 10^-4 = [0.006 + x] [x]/ ( 0.16 - x).

x = 7.7 × 10^-7M.

[ClO^-] = 7.7 × 10^-7 M.

[HClO] = 0.16 - 7.7 × 10^-7 = 0.16M.

[F^-] = 0.006 M.

[HF] = 0.064 - 0.006 = 0.058 M.