Question

Calculate the pH of a solution prepared by dissolving 0.876 moles of carbonic acid (H2CO3) and 0.543 moles of sodium hydrogen carbonate (NaHCO3) in water sufficient to yield 1.00 L of solution. The Ka of carbonic acid is 4.5x10^-7. Explain answer please

Answer 1

no of moles of carbonic acid used = 0.876 moles

no of moles of sodium hydrogen carbonate used = 0.543 moles

reaction between carbonic acid and sodium hydrogen carbonate is given bellow

H₂CO₃ + 2NaHCO₃   Na₂CO₃ + 2H₂O + 2CO₂  

according to reaction 2 moles of sodium hydrogen carbonate reacts with 1 mole of carbonic acid

therefore 0.543 moles of sodium hydrogen carbonate will react with = 0.543/2 = 0.2715 mole H₂CO₃

but 0.876 moles of carbonic acid are taken which are excess than required

no of moles of carbonic acid remained unreacted = 0.876 - 0.543 = 0.333 moles

total volume of solution = 1liter

molarity = no of moles/volume in liter , substitute values to find molarity of carbonic acid

molarity of carbonic acid = 0.333/1 = 0.333M

carbonic acid dissociates as

H₂CO₃  + H₂O HCO₃^-+ H₃O⁺

K_a = [HCO₃^-][H₃O⁺] / [H₂CO₃]

but  [HCO₃^-] = [H₃O⁺] = x

K_a = [x][x] / [H₂CO₃] , given Ka of H₂CO₃ = 4.5 x 10^-7  

Substitute the value in equation

4.5 10^-7 = [x]²/ 0.333

[x]² = 4.5 10^-7 0.333 = 1.4985 x 10^-7

[x] = 3871.0463 10^-7

Concentration of H₃O⁺ = 3871.0463 10^-7 M

pH = - log[H⁺]

pH = - log (3871.0463 10^-7) = 3.41

ans pH of resulatant solution formed is 3.41