Answer to Question #10099 in Inorganic Chemistry for George Ristic

Question

Volume of solution is 350 cubic centimeters. It contains 0,12 moles of sodium bicarbonate and 0.08 moles of sodium carbonate. How to calculate pH of such solution?
Acidity constants of carbonic acid are Ka1=4,3x10^-7 and Ka2=5,6x10^-11

Answer 1

350/100 = 3,5 L
C of sodium bicarbonate is 0.12/3.5 = 0.0343
C of sodium carbonate is 0.88/3.5 = 0.25
pH = pK_a + log{[CO₃^2-]/[HCO₃^-]}

The K_a value comes from the equilibrium: HCO₃^-(aq) + H₂O(l) <=> H₃O⁺(aq) + CO₃^2-(aq)

K_a = [H₃O⁺][CO₃^2-]/[HCO₃^-] = 4.7 x 10^-11
pK_a = -log K_a = 10.3
log {[CO₃^2-]/[HCO₃^-]} = log{[0.25]/[0.0343]} = log{7.288} = 0.8626
pH = 10.3 + 0.8626 = 11.16