Chapter 8, Problem 8.15P

Interpretation Introduction

Interpretation:

The $\text{pH}$ value for $\text{3}\text{.0}\times {\text{10}}^{\text{-5}}\text{ M}$$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ has to be calculated.

Concept Introduction:

$\text{pH}$: The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.

It is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Ionic-product constant for water: It is the hydronium ion concentration times the $O{H}^{-}$ concentration present in the solution.

$\begin{array}{l}{H}_2{O}_{\left(l\right)}+H_2{O}_{\left(l\right)}{H}_3{O}^{+}{}_{\left(aq\right)}+O{H}^{-}{}_{\left(aq\right)}\\ \\ K=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[H_{2}O\right]}\\ \\ \left[H_3{O}^{+}\right]=\left[O{H}^{-}\right]=1.00\times {10}^{-7}M\\ \\ \therefore K_W=K\left[H_{2}O\right]\left[H_{2}O\right]=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ \\ K_W=1.00\times {10}^{-14}M\end{array}$