Given:
Concentration of boric acid, H3BO3 = 0.050 M
pH = 5.28
To determine:
The Ka of H3BO3
Explanation
The dissociation of boric acid is represented as
H3BO3 ↔ H2BO3- + H+
Ka = [H2BO3-][H+]/[H3BO3]
pH = -log [H+]
[H+] = 10^-pH = 10^-5.28 = 5.25*10⁻⁶ M
At equilibrium we have:
[H2BO3-] = [H+] = 5.25*10⁻⁶M
[H3BO3]= 0.050 - 5.25*10⁻⁶≅ 0.050 M
Ka = [5.25*10⁻⁶]²/[0.050] = 5.51 * 10⁻⁴
Ans: Ka of H3BO3 = 5.51 * 10⁻⁴
