Question

1.Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka= 1.2×10⁻⁵.

Find the percent dissociation of this solution.

2.Find the

pH of a 0.150 M solution of a weak monoprotic acid having Ka= 0.16.

Answer 1

Ka = CX^2

Ka = ionisation constant = 1.2*10^−5.

C = concentration = 0.150 M

x = degree of ionisation = ?

(1.2*10^-5) = 0.15*x^2

x = 0.00894

percentage of ionisation = x*100

                        = 0.00894*100

                        = 0.894%

2. pH of weak acid = 1/2(pka-logC)

   pka = -logka

       = -log0.16   = 0.796

   C = concentration of acid = 0.15 M

pH = 1/2(0.796-log0.15) = 0.81