Question

Calculate the pH of a 1.50 M solution of boric acid, H3BO3. Include the ionization equilibrium of H3BO3 as part of your answer.

Answer 1

H3BO3+H2O=H†+[B(OH)4]–

pKa=9.24; Ka=5,8*10^-10 M

Because Kacid<Kawater

[H+]=√(Ka*CHA+Kw)

[H+]=√(5.8*10^-10*1,5M+10^-14)

[H+]=2,94*10^-5

pH=-lg[H+]

pH=4,53