Question

A 0.180 m solution of a weak acid (ha) has a ph of 2.91. calculate the acid ionization constant (ka) for the acid.

Answer 1

Answer is: the acid ionization constant (Ka) for the acid is 8.46·10⁻⁶.
Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
pH = 2,91.
[H⁺] = 10∧-2,91 = 0,00123 M.
[H⁺] = [A⁻] = 0,00123 M.
[HA] = 0,18 M - 0,00123 M.
[HA] = 0,1787 M.
Ka = [H⁺] · [A⁻] / [HA].
Ka = (0,00123 M)² / 0,1787 M.
Ka = 8,46·10⁻⁶ M.