Question

A weak monobasic acid of 0.04% dissociated in 0.025 M solution. Calculate the pH of the solution.

Answer 1

Given: % α of monobasic acid = 0.04, c = 0.025

Now, \[\ce{HA + H2O <=> H3O+ + A-}\]

% α = α × 100

∴ α = `(% α)/100`

= `0.04/100`

= 4 × 10⁻⁴

Now, [H⁺] or [H₃O⁺] = α.C

= 4 × 10⁻⁴ × 0.025 M

= 4 × 10⁻⁵ M

Now, pH = − log₁₀ [H⁺]

∴ pH = − log₁₀ [1 × 10⁻⁵]

pH = 5