Question

Find the pH of a solution containing 0.025 M chloric acid and 0.35 M hypochlorous acid.

I don’t know if you need the Ka value to solve this so I included it anyway.

Ka of HClO2 = 1.2x10^-2
Ka of HOCl = 3.5x10^-8

Answer 1

1.91

Explanation:

The chlorous acid is 340 000 times stronger than the hypochlorous acid. It will suppress the ionization of HOCl to such an extent that we can ignore the HOCl in our calculations.

The equation for the equilibrium of HClO₂ is

HClO₂ + H₂O ⇌ H₃O⁺ + ClO₂⁻; Kₐ= 1.2 × 10⁻²

For ease of typing, let’s rewrite the equation as

HA + H₂O ⇌ H₃O⁺ + A⁻

Set up an ICE table:

HA + H₂O ⇌ H₃O⁺ + A⁻

I: 0.025 0 0

C: -x +x +x

E: 0.025-x x x

Set up the Kₐ expression

Kₐ = {[H₃O⁺][A⁻]}/[HA] = 1.2 × 10⁻² Substitute values

(x × x)/(0.0250-x) = 1.2 × 10⁻² Combine like terms

x²/(0.025-x) = 1.2 × 10⁻²

Check that x ≪ 0.025

0.025/(1.2 × 10⁻²) = 2.1

The ratio is less than 400. We must solve a quadratic equation

Set up the quadratic equation.

x²/(0.025-x) = 1.2 × 10⁻²

x² = (0.025 - x)(1.2 × 10⁻²)

x² = 3.00 × 10⁻⁴ - 1.2 × 10⁻²x

x² + 1.2×10⁻²x – 3.00×10⁻⁴ = 0

Solve the quadratic equation

The solutions to the quadratic equation are

`x = (-b\pm√(b^2-4ac))/(2a)`

a = 1

b = 1.1 × 10⁻²

c = -1.65 × 10⁻³

Substituting the values into the equation, we get

x = 0.0123 and x = -0.02435

b =We reject the negative value, so

x = [H₃O⁺] = 0.0123 mol·L⁻¹

Calculate the pH

pH = -log[H₃O⁺]

pH = -log(0.0123)

pH = 1.91