Question

What is the pH of a 0.153 M solution of sodium bicarbonate (NaHCO3)? (Ka1 = 4.45E-7, Ka2 = 4.69E-11)

Answer 1

HCO3^-    has Ka = 4.69 x 10^-11

                           HCO3^- (aq)    <----> H+ (aq) + CO3^2- (aq)

Initially                 0.153                              0            0

equilibrium           0.153-X                         X                 X

Ka = [H+] [ CO3^2-] / [ HCO3-]

4.69 x 10^-11 = X^2 / ( 0.153-X)

here we approximate 0.153 -X = 0.153 since we get very less value of X since Ka is very loss

4.69 x 10^-11 = X^2 / ( 0.153)

X = 2.7 x 10^- 6 = [H+]

pH = -log [H+] = - log ( 2.7 x 10^-6) = 5.6