Question

Plz give detailed solutions:

1. Calculate the pH of a 0.504 M aqueous solution of phenol (a weak acid) (C₆H₅OH, K_a = 1.0×10^-10).

2. Calculate the pH of a 0.0308 M aqueous solution of hydrofluoric acid (HF, K_a = 7.2×10^-4).

3. Calculate the pH of a 0.0703 M aqueous solution of hydrofluoric acid (HF, K_a = 7.2×10^-4).

Answer 1

1. K_a = 1.0×10^-10

    PKa = -logKa

             = -log1*10^-10

             = 10

PH = 1/2Pka -1/2logc

     =1/2*10-1/2log0.504

    = 5 +0.14875   = 5.14875

2. Ka = 7.2*10^-4

   Pka = -logka

            = -log7.2*10^-4

          = 3.1426

PH = 1/2Pka-1/2logc

        = 1/2*3.1426-1/2log0.0308

        = 1.5713+0.7557     = 2.327

3.

   Ka = 7.2*10^-4

   Pka = -logka

            = -log7.2*10^-4

          = 3.1426

PH = 1/2Pka-1/2logc

          = 1/2*3.1426-1/2log0.0703

         = 1.5713 +0.5765   = 2.1478 >>>>answer