In a solution of pure water, the dissociation of water can be expressed by the following: h2o(l) + h2o(l) ⇌ h3o+(aq) + oh−(aq) the equilibrium constant for the ionization of water, kw, is called the ion-product of water. in pure water at 25 °c, kw has a value of 1.0 × 10−14. the dissociation of water gives one h3o+ ion and one oh− ion and thus their concentrations are equal. the concentration of each is 1.0 × 10−7 m. kw = [h3o+][oh−] kw = (1.0 × 10-7)(1.0 × 10-7) = 1.0 × 10-14 [h3o+][oh−] = 1.0 × 10-14 a solution has a [oh−] = 3.4 × 10−5 m at 25 °c. what is the [h3o+] of the solution? answer 2.9 × 10−9 m 2.9 × 10−15 m 3.4 × 109 m 2.9 × 10−10 m i don't know yet

[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.

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