Hydrochloric acidis a strong acid, which can be considered as completelydissociated is aqueous solution. Thus in solution of 1.15×10⁻³M HCl you have 1.15×10⁻³M.
Chlorous acid is a weaker acid. So it ionize party tillconcentrationssatisfy following equilibrium equation.
Ka = [ClO₂⁻]∙[H⁺] / [HClO₂]
with
Ka = 10^-1.96 = 1.10×10⁻²
Let x be the concentration of clorite ion in the mixture, then concentrations are
[ClO₂⁻] = x
[H⁺] = 1.15×10⁻³ + x
[HClO₂] = 1.10×10⁻² - x
1.10×10⁻² = x∙(1.15×10⁻³ + x) / (1.10×10⁻² - x)
x² + 1.215×10⁻²∙x - 1.21×10⁻⁴ = 0
x = (1/2)∙1.215×10⁻² + √[(1/4)∙(1.215×10⁻²)² + 1.215×10⁻⁴]
= 1.864×10⁻² M
[H⁺] = 1.15×10⁻³ + 1.864×10⁻² = 1.979×10⁻² M
pH = - log₁₀[H⁺] = - log₁₀( 1.979×10⁻² ) = 1.70
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