If the pH of a weak acid solution is 2.500 and the solution has a concentration of 0.100M, what is the Ka of the weak acid HA?

The Kₐ of the weak acid is 1.033×10⁻⁴

Explanation:

The dissociation of a weak acid in aqueous solution is limited to about 5 to 10%

The acid dissociation reaction is given as follows;

HA (aq) + H₂O (l) → H₃O⁺(aq) + A⁻ (aq)

Given that the pH = 2.5, we have

pH = -log₁₀[H₃O⁺] = 2.5

∴ [H₃O⁺] = 10^(-2.5) = 0.0031623

Kₐ = [H₃O⁺][A⁻]/[HA] = (0.0031623^2)/(0.1 - 0.0031623) = 1.033×10⁻⁴

The acid dissociation constant, Kₐ for weak acid is very low as obtained

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.