Answer:
7.5x10⁻¹³M = [H⁺]
Explanation:
When a solution of Na₂CO₃ is dissolved in water, the equilibrium produced is:
Na₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁺(aq) + OH⁻(aq) + 2Na⁺
Where Kb is defined from equilibrium concentrations of reactants, thus:
Kb = [HCO₃⁺][OH⁻] / [Na₂CO₃] (1)
It is possible to obtain Kb value from Ka2 and Kw thus:
Kb = Kw / Ka2
Kb = 1x10⁻¹⁴ / 5.6x10⁻¹¹
Kb = 1.8x10⁻⁴
Replacing in (1):
1.8x10⁻⁴ = [HCO₃⁺][OH⁻] / [Na₂CO₃]
The equilibrium concentrations are:
[Na₂CO₃] = 1.0M - X
[HCO₃⁺] = X
[OH⁻] = X
Thus:
1.8x10⁻⁴ = [X][X] / [1-X]
1.8x10⁻⁴ - 1.8x10⁻⁴X = X²
X² + 1.8x10⁻⁴X - 1.8x10⁻⁴ = 0
Solving for X:
X = -0.0135 → False answer, there is no negative concentrations
X = 0.0133
As [OH⁻] = X;
[OH⁻] = 0.0133
From Kw:
Kw = [OH⁻] [H⁺]
1x10⁻¹⁴ = 0.0133[H⁺]
7.5x10⁻¹³M = [H⁺]
