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Dejanras
Answer is: pH of monoprotic weak acid is 3.28.
Chemical reaction: HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c(monoprotic acid) = 0.23 M.
Ka = 1.2·10⁻⁶.
[A⁻] = [H⁺] = x.
[HA] = 0.23 M - x.
Ka = [A⁻]·[H⁺] / [HA]. 
1.2·10⁻⁶ = x² / (0.23 M - x).
Solve quadratic equation: [H⁺] = 0.000524 M.
pH = -log[H⁺].
pH = -log(0.000524 M).
pH = 3.28.
dsdrajlin

The pH of a 0.23 M  monoprotic acid, whose Ka is 1.2 × 10⁻⁶, is 3.3.

Let's consider the dissociation reaction for a generic monoprotic weak acid.

HA(aq) ⇄ H⁺(aq) + A⁻(aq)

Given that the acid dissociation constant (Ka) is 1.2 × 10⁻⁶ and the concentration of the acid (Ca) is 0.23 M, we can calculate the concentration of H⁺ using the following expression.

[tex][H^{+} ] = \sqrt{Ca \times Ka } = \sqrt{0.23 \times (1.2 \times 10^{-6} ) } = 5.3 \times 10^{-4} M[/tex]

The pH of the solution is:

[tex]pH = -log [H^{+} ] = -log (5.3 \times 10^{-4} )= 3.3[/tex]

The pH of a 0.23 M  monoprotic acid, whose Ka is 1.2 × 10⁻⁶, is 3.3.

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