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