Answer:
pH of solution B is 5
Step-by-step explanation:
A weak acid, HA, is in equilibrium with water as follows:
HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)
Where Ka (10^-pKa = 1x10⁻⁹) is:
Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
Where concentrations of this species are equilibrium concentrations
As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:
[HA] = 0.1M - X
[A⁻] = X
[H₃O⁺] = X
Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.
Replacing in Ka expression:
1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
1x10⁻⁹ = [X] [X] / [0.1 - X]
1x10⁻¹⁰ - 1x10⁻⁹X = X²
1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0
Solving for X:
X = -0.00001 → False solution, there is no negative concentrations.
X = 1x10⁻⁵ → Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 1x10⁻⁵M
And pH = -log[H₃O⁺]
pH = 5
pH of solution B is 5