Question

Use the concept of autoprotolysis to calculate the concentrations of all species in a 1.00 M solution of NaHCO₃ (sodium bicarbonate). Please show calculations and setups.

Species	Na+	HCO3-	H+	OH-	CO32-	CO2 (aq)
Molar Concentrations

K₁ = [H⁺][HCO₃^1-] / [CO₂] = 4.3 x 10^-7

K₂ = [H⁺] [CO₃^2-] / [HCO₃^1-] = 5.6 x 10^-11

Answer 1

Since NaHCO₃(aq) is completely dissociated into Na⁺(aq) and HCO₃^-(aq), [Na⁺] = [NaHCO₃] = 1.00 M (Answer)

Equation for dissociation of HCO₃^-(aq):

--HCO₃^-​​​​​​​(aq) <-----> CO₃^2-(aq) + H⁺​​​​​​​(aq): Ka2 = 5.6*10^-11

I: 1.00 M --------------- 0 M --------- 0 M

C: - X ------------------- +X ---------- +X

E:(1.00 - X) ------------ X ------------ X

Ka2 = 5.6*10^-11 = [CO₃^2-(aq)]*[H⁺​​​​​​​(aq)] / [HCO₃^-​​​​​​​(aq)]

=> 5.6*10^-11 = X*X / (1.00 - X)

Since X << 1.00, we can neglect X in denominator.

=>  5.6*10^-11 = X²

=> X = 7.48*10^-6 M

=> X = [CO₃^2-(aq)] = 7.48*10^-6 M (Answer)

Equation for hydrolysis of HCO₃^-​​​​​​​(aq):

--HCO₃^-​​​​​​​(aq) + H2O(l) <-----> CO₂(aq) + OH^-​​​​​​​(aq): Kb1 = Kw/Ka1 = 10^-14 / 4.3*10^-7 = 2.33*10^-8

I: 1.00 M ------------------------ 0 M --------- 0 M

C: - X ---------------------------- +X ---------- +X

E:(1.00 - X) ---------------------- X ------------ X

Kb1 = 2.33*10^-8 = [CO₂(aq)]*[OH^-​​​​​​​(aq)] / [HCO₃^-​​​​​​​(aq)]

=> 2.33*10^-8 = X*X / (1.00 - X)

Since X << 1.00, we can neglect X in denominator.

=>  2.33*10^-8 = X²

=> X = 1.525*10^-4 M

=> X = [CO₂(aq)] = 1.5*10^-4 M (Answer)  

Since OH^-(aq) is formed in higher amount than H⁺(aq), the solution in basic

Hence X = [OH^-​​​​​​​(aq)] = 1.5*10^-4 M (Answer)  

[H⁺​​​​​​​(aq)] = 10^-14 / [OH^-​​​​​​​(aq)] =  10^-14 / 1.5*10^-4 = 6.67*10^-11 M (Answer)

Since very negligible amount of HCO₃^-​​​​​​​(aq) is dissociated, [HCO₃^-​​​​​​​(aq)] = 1.00 M (Answer)