Chapter 16, Problem 16.62QP

Interpretation Introduction

Interpretation:

The ${\text{K}}_w$ of water at a particular temperature has to be calculated

Concept Information:

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

Relationship between $\text{pH}$ and $\text{pOH}$

$\text{pOH}$ is similar to $\text{pH}$.  The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

The equation of equilibrium for autoionization of water is,

${\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}$

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Taking negative logarithm on both sides, we get

$\begin{array}{l}-\log ({\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])= -log(1}\times {\text{10}}^{\text{-14}})\\ (-\log {\text{[H}}^{\text{+}}{\text{])+(-log[OH}}^{\text{-}}\text{])= 14})\end{array}$

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The ${\text{K}}_w$ of water at a particular temperature