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[H+][OH-] = 10^-14
konsentrasi H+ dan OH- pada air murni adalah sama, maka bisa kita tuliskan:
[H+][H+] = 10^-14
[H+]^2 = 10^-14
[H+] = 10^-7 M
maka total [H+] = [H+] dari air + [H+] dari asam
total [H+] = 10^-7 + 10^-8
total [H+] = 1,1 x 10^-7
maka pH = -log[H+]
pH = -log(1,1 x 10^-7)
pH = 7 - log1,1