Question

For carbonic acid, Ka2 = 4.7 x 10^- 11 . What is [HCO3− ] in a 0.10 M solution of sodium carbonate? Report the answer to two sig figs.

Answer 1

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/4.7*10^-11

Kb = 2.128*10^-4

CO32- dissociates as

CO32- + H2O -----> HCO3- + OH-

0.1 0 0

0.1-x x x

Kb = [HCO3-][OH-]/[CO32-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.128*10^-4)*0.1) = 4.613*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

2.128*10^-4 = x^2/(0.1-x)

2.128*10^-5 - 2.128*10^-4 *x = x^2

x^2 + 2.128*10^-4 *x-2.128*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 2.128*10^-4

c = -2.128*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 8.515*10^-5

roots are :

x = 4.507*10^-3 and x = -4.72*10^-3

since x can't be negative, the possible value of x is

x = 4.507*10^-3

we have:

[HCO3-] = x = 4.507*10^-3

Answer: 4.5*10^-3