Question

1. Calculate the pH of the following solution using numerical methods.

a) 150 ml of 1.04 × 10-6 M CH3COOH (pKA = 4.7)

b) 150 ml of 1.04 × 10-6 M CH3COOH (pKA = 4.7) added with 500 ml of water

c) 250 ml of 2.71 × 10-5 M HOCl (pKA = 7.6)

d) 150 ml of 2.71 × 10-5 M HOCl (pKA = 7.6) added with 520 ml of water

Answer 1

1.

a) 150 ml of 1.04 × 10-6 M CH3COOH (pKA = 4.7)

pH of CH3COOH = 1/2(pka-logC)

   C = concentration of CH3COOH = 1.04*10^-6 M

   pKa = 4.7

pH = 1/2(4.7-log(1.04*10^-6)) = 5.34

b) 150 ml of 1.04 × 10-6 M CH3COOH (pKA = 4.7) added with 500 ml of water

net concentration of CH3COOH = 150*(1.04*10^-6)/650 = 2.4*10^-7 M

pH = 1/2(4.7-log(2.4*10^-7)) = 5.66

c) 250 ml of 2.71 × 10-5 M HOCl (pKA = 7.6)

   pH of HOCl = 1/2(pka-logC)

   C = concentration of HOCl = 2.71*10-5 M

   pKa = 7.6

pH = 1/2(7.6-log(2.71*10^-5)) = 6.083

d) 150 ml of 2.71 × 10-5 M HOCl (pKA = 7.6) added with 520 ml of water

    pH of HOCl = 1/2(pka-logC)

   C = concentration of HOCl = 150*(2.71*10-5)/670 = 6.07*10^-6 M

   pKa = 7.6

pH = 1/2(7.6-log(6.07*10^-6)) = 6.41