Question

Why does $p{\rm H} + p{\rm O}{\rm H} = 14$

Answer 1

The ionic product controls the relative concentration of ${{\rm H}^ + }$ ions and ${\rm O}{{\rm H}^ - }$ ions in aqueous solutions. Relative concentration of these ions defines the net $pH$ of the solution. In a pure water sample, the concentration of ${{\rm H}^ + }$ ions and ${\rm O}{{\rm H}^ - }$ ions are equal.

Complete answer:
Water is an example of weak electrolyte and it undergo self-dissociation upto a small extent-
${H_2}{O_{\left( l \right)}} + {H_2}{O_{\left( l \right)}} \rightleftharpoons {H_3}{O^ + }_{\left( {aq} \right)} + O{H^ - }_{\left( {aq} \right)}$
Dissociation constant for water hydrolysis process is expressed as-
${\rm K} =$ $\dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{{{\left[ {{H_2}O} \right]}^2}}}$
Where ${\rm K} =$ dissociation constant of water
$\left[ {{H_3}{O^ + }} \right]$ $=$ Concentration of hydronium ion
$\left[ {O{H^ - }} \right]$ $=$ Concentration of hydroxide ion
${\left[ {{H_2}O} \right]^2}$ $=$ Concentration of water
As we already know water undergoes dissociation to a very small extent therefore concentration of undissociated water is considered as constant value.
${\rm K} \times$ ${\left[ {{H_2}O} \right]^2}$ $=$ $\left[ {{H_3}{O^ + }} \right]$ $\times$ $\left[ {O{H^ - }} \right]$
On taking the value of concentration of water as constant, the equation will become-
${{\rm K}_w}$ $=$ $\left[ {{H_3}{O^ + }} \right]$ $\times$ $\left[ {O{H^ - }} \right]$
Where ${{\rm K}_w}$ $=$ it is a constant which is known as an ionic product of water.
value of $- p{{\rm K}_w}$ at ${25^ \circ }C$ is $14$
On converting the above-mentioned equation into $\log$ , equation is written as-
$\log {{\rm K}_w}$ $=$ $\log \left[ {{H_3}{O^ + }} \right]$ $\times$ $\log \left[ {O{H^ - }} \right]$
Now convert the equation into $p{\rm H}$ form- As we know negative logarithm of hydrogen ion concentration is expressed as $p{\rm H}$ and negative logarithm of hydroxide ion concentration is expressed as $p{\rm O}{\rm H}$ .
$- p{{\rm K}_w}$ $=$ $- p{\rm H} - p{\rm O}{\rm H}$
Convert the above equation into positive term, now the equation will modify into
${{\rm K}_w}$ $=$ $p{\rm H} + p{\rm O}{\rm H}$
As we already know value of $- p{{\rm K}_w}$ at ${25^ \circ }C$ is $14$
$\Rightarrow$ Therefore, $p{\rm H} + p{\rm O}{\rm H} = 14$ .

Note:
Value of ${{\rm K}_w}$ is constant at a particular temperature but it may change with change in temperature because the rate of dissociation of the water also changes. There is a $p{\rm H}$ scale which is used to express the acidity and basicity of the solution.