Question

Calculate the pH and the equilibrium concentrations of HCO3- and CO32- in a 0.0514 M carbonic acid solution, H2CO3 (aq). For H2CO3, Ka1 = 4.2×10-7 and Ka2 = 4.8×10-11 pH = [HCO3-] = M [CO32-] = M

Answer 1

H2CO3 (aq)   <---------------> H+ (aq)   +   HCO3- (aq)

0.0514                                     0                    0

0.0514 - x                                 x                      x

Ka1 = x^2 / 0.0514 - x

4.2 x 10^-7 = x^2 / 0.0514 - x

x = 1.47 x 10^-4

[H+] = 1.47 x 10^-4 M

pH = -log (1.47 x 10^-4)

pH = 3.83

[HCO3-] = 1.47 x 10^-4 M

[CO32-] = 4.80 x 10^-11 M