Question

The pH of NH4OH solution is 10.72 in 0.015M solution. Calculate it's dissociation constant.

Answer 1

1. Write the equation for the dissociation of a generic monoprotic acid:

HA + H₂O ⇌ H₃O⁺ + A⁻

2. Write the dissociation constant expression.

Ka=[H₃O⁺][A⁻][HA]

3. Determine the equilibrium concentrations.

pH = -log[H₃O⁺] = 10.72

[H₃O⁺] = 10^{-10.72}mol/L = 1.91 × 10^{-11} mol/L

From the equation, [A⁻] = [H₃O⁺] = 1.91 × 10^{-11}mol/L

[HA] = (0.015 - 1.91 × 10⁻11) mol/L = 0.015 mol/L

4. Substitute these values in the expression and that is it!

Ka=[H₃O⁺][A⁻]/[HA]= (1.91 × 10⁻11 1.91 × 10⁻11 )/0.031 = 2.4 × 10⁻⁸