Question

I have no idea how to approach this problem:

Calculate the concentration of all species in a 0.145 M solution of H2CO3.

[H2CO3],

[HCO−3],

[CO2−3],

[H3O+],

[OH−]

Express your answer using two significant figures.

Answer 1

carbonic acid   Ka1 = 4.3 x 10^-7 , Ka2 =4.8 x 10^-11

H2CO3 ---------------------> H+ + HCO3-

0.145                                  0           0    ---------------------> initial

0.145-x                               x             x --------------------> after dissociation

Ka1 = [H+][HCO3-]/[H2CO3]

4.3 x 10^-7 = (x) (x) / (0.145-x)

x^2 + 4.3 x 10^-7 x -6.235 x 10^-8 = 0

by solving this

x = 2.49 x 10^-4

[H+] = [HCO3-] = x

[H+] = 2.5 x 10^-4 M

[HCO3-] = 2.5 x 10^-4 M

[H2CO3] = 0.145-x = 0.145- 2.49 x 10^-4

[H2CO3] = 0.14 M

[OH-] = Kw / [H+]

          = 1.0 x 10^-14 / 2.5 x 10^-4

[OH-]   = 4.0 x 10^-11 M

second ionisation constant value always equal to A-2 species of dibasic acid . so here

Ka2 = CO3-2

4.8 x 10^-11 = CO3-2

[CO3-2] = 4.8 x 10^-11 M