Question

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–

Answer 1

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Step-by-step explanation:

The following equilibrium goes on in aqueous solutions:

`\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^(+)\;(aq) + \text{OH}^(-)\;(aq)`.

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

`K_w = [\text{H}^(+)]\cdot[\text{OH}^(-)]`.

Note that water isn't part of this constant.

The value of
`K_w` at 25 °C is
`10^(-14)`. How to memorize this value?

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

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

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