Question

Ionic product of water is $1 \times {10^{ - 12}}$. pH of water will be:

Answer 1

Follow the steps given to calculate the pH of water:
Calculate the concentration of hydrogen ions: The dissociation of water occurs to a small constant because it is a weak electrolyte. Thus, water dissociates to hydrogen and hydroxide ions. The ionic product of water is the product of the concentrations of hydrogen and hydroxide ions.
Calculate the pH of water: pH of any solution is the negative logarithm of the hydrogen ion concentration. Thus, the pH can be calculated from the hydrogen ion concentration.

Complete step by step answer:
Step 1: Calculate the concentration of hydrogen ions as follows:
Water dissociates into hydrogen and hydroxide ions. The reaction is,
${{\text{H}}_2}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{O}}{{\text{H}}^ - }$
The initial and final concentrations of ${{\text{H}}_2}{\text{O}}$, ${{\text{H}}^{\text{ + }}}$ and ${\text{O}}{{\text{H}}^ - }$ are,
${{\text{H}}_2}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{O}}{{\text{H}}^ - }$
${\text{ }}C{\text{ }}0{\text{ }}0$
$C - x{\text{ }}x{\text{ }}x$
The ionic product of water is the product of the concentrations of ${{\text{H}}^{\text{ + }}}$ and ${\text{O}}{{\text{H}}^ - }$ ions. Thus,
${K_{\text{w}}} = \left[ {{{\text{H}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Where ${K_{\text{w}}}$ is the ionic product of water.
Substitute $1 \times {10^{ - 12}}$ for the ionic product of water, $x$ for the concentrations of ${{\text{H}}^{\text{ + }}}$ and ${\text{O}}{{\text{H}}^ - }$ ions. Thus,
$1 \times {10^{ - 12}} = x \times x$
${x^2} = 1 \times {10^{ - 12}}$
$x = \sqrt {1 \times {{10}^{ - 12}}}$
$x = 1 \times {10^{ - 6}}$
Thus, the concentration of ${{\text{H}}^{\text{ + }}}$ ions is $1 \times {10^{ - 6}}$.
Step 2: Calculate the pH of water as follows:
The pH of a solution is the negative logarithm of the ${{\text{H}}^{\text{ + }}}$ ion concentration. Thus, the equation for pH is,
${\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right]$
Substitute $1 \times {10^{ - 6}}$ for the ${{\text{H}}^{\text{ + }}}$ concentration of ${{\text{H}}^{\text{ + }}}$ ions. Thus,
${\text{pH}} = - \log \left( {1 \times {{10}^{ - 6}}} \right)$
${\text{pH}} = 6$
Thus, the pH of water is $6$.

Note:
The dissociation of water occurs to a small constant because it is a weak electrolyte. Thus, water dissociates to hydrogen and hydroxide ions. The ionic product of water is the product of the concentrations of hydrogen and hydroxide ions.
pH of any solution is the negative logarithm of the hydrogen ion concentration. Thus, the pH can be calculated from the hydrogen ion concentration.