Question

Calculate the pH of the following dilute solutions:

(a) 3.42x10^-8 M hydroiodic acid, HI.

(b) 5.77x10^-9 M KOH.

Answer 1

(a) 3.42x10^-8 M hydroiodic acid, HI.

[H+]_HI = 3.42x10^-8

[H+]_H2O = 1.0 x10^-7

[H+]net = [H+]_HI +[H+]_H2O = 3.42x10^-8 +1.0 x10^-7

[H+]net = 1.34 x10^-7

pH= -log[H+]= -log(1.34 x10^-7) =6.87

      pH=    6.87

(b) 5.77x10^-9 M KOH.

[OH-]_KOH = 5.77x10^-9

[OH-]_H2O = 1.0 x10^-7

[OH-]_net = [OH-]_KOH + [OH-]_H2O

                          =5.77x10^-9 +1.0 x10^-7

                    = 1.0577 x 10^-7

pOH= -log [OH-]

        = -log(1.0577 x 10^-7)

          = 6.97

pH= 14 -pOH

    pH= 7.03