Question

The pH of the water is measured as 8.5 and the total concentration of carbonate species is found to be 0.00180M. Find the concentrations of H2CO3, HCO3-and CO32- by both graphically and using the equations

Answer 1

At equilibrium, what is the relationship between [H₂CO₃] and [CO₃^2-]?
They have to be equal, since for every H₂CO₃ molecule produced, you will produce a CO₃^2- ion according to the reaction.

Ka₁ refers to this equation: H₂CO₃ H⁺ + HCO₃^-

The Ka₁ expression would be Ka₁ = [H⁺] [HCO₃^-]/[H₂CO₃]

Ka₂ refers to this equation: HCO₃^- H⁺ + CO₃^2-

The Ka₂ expression would be Ka₂ = [H⁺] [CO₃^2-]/ [HCO₃^-]

Solving this for [HCO₃^-] = [H⁺] [CO₃^2-]/Ka₂

Substitute this expression into the Ka₁ equation for [HCO₃^-] and you get

Ka₁ = [H⁺] ([H⁺] [CO₃^2-]/Ka₂) / [H₂CO₃]

Rearranging yields Ka₁ x Ka₂ = [H⁺]² [CO₃^2-] / [H₂CO₃]

But remember we said that [CO₃^2-] must equal [H₂CO₃] so [CO3^2-] / [H₂CO₃] must equal 1, leaving us with

Ka₁ x Ka₂ = [H⁺]²

Taking the square root of both sides will give [H+]
[H+] = sq. root of (Ka₁ x Ka₂)

pH = - log [H⁺]

8.5 = - log [H⁺]

[H⁺] = 3.16 10^-9

Ka₁ x Ka₂ = [H⁺]² = 9.98 10^-18

Since all the species are in equilibrium

[H₂CO₃] = [HCO₃^-] = [CO₃^2-] = 0.0018/3 = 6 10^-4 M