[H₃O⁺] = 2.90 × 10⁻¹⁰ M
Explanation:
1) Ionization equilibrium equation: given
H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
2) Ionization equilibrium constant, at 25°C, Kw: given
Kw = 1.0 × 10⁻¹⁴
3) Stoichiometric mole ratio:
As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:
[H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M
4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C and you need to calculate what the [H₃O⁺(aq)] is.
Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:
Kw = 1.0 × 10⁻¹⁴ M², and Kw = [H3O⁺] [OH⁻]
Then you can substitute the known values and solve for the unknown:
1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M ⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M
As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.