Berapa ph larutan 0,002m hclo2, jika ka = 1,0 x 10-2

Jawaban:

[H+] = √(KaxM)

[H+] = √(10^-2 x 2 x 10^-3)

[H+] = √(2×10^-5)

[H+] = 2^1/2 x 10^5/2

pH = – log [H+]

pH = 2,5 – log √2