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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

Respuesta :

Edufirst

Answer:

  • [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.

Answer:

1. pH = -log[H^+]

2. Acidic: pH < 7.00, neutral: pH = 7.00, basic: pH > 7.00

3. pH + pOH = 14.00 at room temperature

Explanation:

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