Chapter 16, Problem 1PE

Interpretation Introduction

Interpretation:

The concentration of the hydroxide ion in a HCl solution to be calculated.

Concept introduction:

Concentration is expressed as the amount of a solute in the solvent in the solution.

The product of the concentrations of the $\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ ions is constant at a constant temperature and is called the ionic product of water.

Answer to Problem 1PE

Solution: $8.33\times {10}^{-16}\text{ M}$

Explanation of Solution

Given information:

The molarity of the hydronium ion is $12\text{ M}$.

The moles of HCl in the solution is $12\text{ mol/L}$.

Calculate the concentration of the hydroxide ion.

The equation for the auto ionization of water is as:

${\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The equilibrium expression for a reaction is written as:

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]={\text{K}}_w$.

Here, $\left[{\text{H}}^{+}\right]$ is the concentration of the hydronium ion, $\left[{\text{OH}}^{-}\right]$ is the concentration of the hydroxide ion and ${\text{K}}_w$ is the ionic product of water.

Substitute the value of $\left[{\text{H}}^{+}\right]$ as $12\text{M}$ and the value of ${\text{K}}_w$ as ${10}^{-14}$ in the above expression,

$\begin{array}{c}12\text{ M}\times \left[{\text{OH}}^{-}\right]={10}^{-14}\\ \left[{\text{OH}}^{-}\right]=\left(\frac{{10}^{-14}}{12}\right)\\ =8.33\times {10}^{-16}\text{ M}\end{array}$

Conclusion

The concentration of the hydroxide ion in the HCl solution is $8.33\times {10}^{-16}\text{ M}$.