Calculate the pH of the conjugate base of 0.100 moldm-3 with pKa of the acid 3.76

Answer 1

If the pKa of the acid is 3.76, then the pKb of the conjugate base will be 14 - 3.76 = 10.24

If the pKb = 10.24, then Kb = 1x10^-10.24 = 5.75x10^-11

Let HA be the weak acid

A- is the conjugate base

A- + H2O ==> HA + OH-

Kb = [HA][OH-] / [A-]

5.75x10^-11 = (x)(x) / 0.100-x and assuming x is small relative to 0.100 M, we can ignore it in denominator

5.75x10^-11 = x² / 0.1

x² = 5.75x10^-12

x = [OH-] = 2.40x10^-6 (very small relative to 0.100 M so above assumption was valid)

pOH = -log [OH-] = -log 2.40x10^-6

pOH = 5.62

pH = 14 - 5.62

pH = 8.38