Question

Calculate the hydronium ion concentration and pH for a 0.050 M solution of sodium formate, NaHCO2. Kb for reaction of HCO2- ion with water is 5.6 × 10-11.

[H3O+] =_____ M

pH =______

Answer 1

for simplicity lets write weak base HCO2- as A-

A-        + H2O   ----->     AH   +   OH-
0.0500                        0         0
0.0500-x                      x         x

Kb = [AH][OH-]/[B-]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.60E-11)*0.0500) = 1.67E-6
so,
[OH-] = 1.67E-6 M

[H+] = 10E-14 / [OH-]
= 10E-14 / (1.67E-6)
=5.99*10E-9 M

pH = -log [H+]
= -log (5.99*10E-9)
= 8.22

Answer:
[H3O+] = 5.99*10E-9 M
pH = 8.22