Question

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

Answer 1

Hint: 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.