Chapter 16.3, Problem 1PPA

Interpretation Introduction

Interpretation:

The concentration of hydronium ${\text{[H}}_3{\text{O}}^{+}\text{]}$ ions in the given solution at $25°\text{C}$ is to be calculated.

Concept introduction:

At a particular temperature, the product of concentrations of water is called the ionic product of water and it is denoted by $K_w$.

Water is a universal solvent and it is a weak electrolyte. Based on the self-ionization of two moles of water, molecules combine to form hydronium ions ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ and hydroxide ions ${\text{[OH}}^{\text{-}}\text{]}$.

${\text{H}}_2\text{O}+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}$

At $25°\text{C}$, the concentrations of hydronium and hydroxide ions, respectively, in water is,

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=1.0\times {10}^{-7}M\\ {\text{[OH}}^{\text{-}}\text{]}=1.0\times {10}^{-7}M\end{array}$

The concentrations of hydronium and hydroxide ions are equal in a neutral solution.

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

By using the equilibrium expression, $K_w$ is as follows:

$K_w={\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}=1.0\times {10}^{-14}$

Rearrange the above equation for hydronium ions as:

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{{\text{[OH}}^{\text{-}}\text{]}}$ …… (1)