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Answer on Inorganic Chemistry Question for George Ristic

Question #10099
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
Expert's answer
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 = pKa + log{[CO32-]/[HCO3-]}

The Ka value comes from the equilibrium: HCO3-(aq) + H2O(l) <=> H3O+(aq) + CO32-(aq)

Ka = [H3O+][CO32-]/[HCO3-] = 4.7 x 10-11
pKa = -log Ka = 10.3
log {[CO32-]/[HCO3-]} = log{[0.25]/[0.0343]} = log{7.288} = 0.8626
pH = 10.3 + 0.8626 = 11.16

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