contestada

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?


1.0 × 10–5 M OH–

1.0 × 10–14 M OH–

1.0 × 105 M OH–

1.0 × 10–9 M OH–

Respuesta :

jacob193

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

[tex]\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)[/tex].

The equilibrium constant for this reaction is called the self-ionization constant of water:

[tex]K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}][/tex].

Note that water isn't part of this constant.

The value of [tex]K_w[/tex] at 25 °C is [tex]10^{-14}[/tex]. How to memorize this value?

  • The pH of pure water at 25 °C is 7.
  • [tex][\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}[/tex]
  • However, [tex][\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}[/tex] for pure water.
  • As a result, [tex]K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}[/tex] at 25 °C.

Back to this question. [tex][\text{H}^{+}][/tex] is given. 25 °C implies that [tex]K_w = 10^{-14}[/tex]. As a result,

[tex]\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}[/tex].

Otras preguntas