Calculate pH of 0.02 M - HA solution. Ka for HA = 2 * 10^-12. (log 2 = 0.3, log 3 = 0.48, log 5 = 0. — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Ka is extremely small so water's contribution is significant. [H+]^2 = Ka*[HA] + Kw = 4*10^-14 + 10^-14 = 5*10^-14. [H+] = √(5)*10^-7. pH = 7 - 0.5*log5 = 7 - 0.35 = 6.65.