Question

a} The pH of a 0.25 M solution of a weak base is 9.23. what is the K_a value of its conjugate acid?

b) The pH of a .35 M solution of a weak acid is 4.72. What is the K_b value of its conjugate base?

c) Calculate the pH of a .52 M aqueous solution of a weak acid with a value of K_a = 2.1X10^-8

If you can explain each question step by step please!

Answer 1

a) pH = 9.23

pOH = 14 - 9.23 = 4.77

for weak base

pOH = 1/2 [pKb - logC]

4.77 = 1/2 [pKb - log 0.25]

9.54 = pKb + 0.60

pKb = 8.94

Kb = 10^-pKb = 10^-8.94   = 1.15 x 10^-9

Ka = 1.0 x 10^-14 / 1.15 x 10^-9

ka = 8.69 x 10^-6

b) for weak acids

pH = 1/2 [pKa - log C]

4.72 = 1/2 [pKa - log0.35]

9.44 = pKa + 0.46

pKa = 8.98

Ka = 10^-pKa   = 10^-8.98  = 1.05 x 10^-9

Kb = 1.0 x 10^-14  / 1.05 x 10^-9

Kb = 9.52 x 10^-6

c) for weak acid

pH = 1/2 [pKa - log C]

pKa = -log Ka = - log [2.1 x 10^-8] = 7.68

pH = 1/2 [7.68 - log 0.52]

pH = 3.98