VerifiedHint: The degree of dissociation of water (or) ionic product of water can be written as,
$ {k_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] $ It is known that in KOH solution, it contains $ {K^ + } $ and $ O{H^ - } $ ions.
It is also known that pH is the negative logarithm of the concentration of the $ {H^ + } $ ion. Thus, $ {H^ + } $ concentration can be determined by using the ionic product of water formula since the hydroxyl ion concentration of KOH is known.
Complete answer:
It is given that $ \left[ {O{H^ - }} \right] = {10^{ - 3}}M $ , $ {k_w} = 1 \times {10^{ - 14}}mo{l^2}{L^{ - 1}} $
It is known that $ {k_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] $
Thus, $ {H^ + } $ can be determined as,
$ \Rightarrow 1 \times {10^{ - 14}}mo{l^2}{L^{ - 1}} = \left[ {{H^ + }} \right]\left[ {{{10}^{ - 3}}M} \right] $
$ \Rightarrow \left[ {{H^ + }} \right] = \dfrac{{1 \times {{10}^{ - 14}}mo{l^2}{L^{ - 1}}}}{{{{10}^{ - 3}}M}} $
$ \Rightarrow \left[ {{H^ + }} \right] = {10^{ - 11}}M $
Thus, the concentration of $ {H^ + } $ ion is obtained as $ {10^{ - 11}}M $ and its pH is 11
The hydroxyl ion concentration $ \left[ {O{H^ - }} \right] = {10^{ - 3}}M $ and its pH can be calculated by using the dissociation of water. We can find out the pH by the formula pH $ = - \log [{H^ + }] $ .
It can be determined as,
$ pH = - \log \left[ {{{10}^{ - 11}}} \right] $
$ pH = - ( - 11)\log [10] $
$ pH = 11 $
Hence, the correct option is D.
Note: But this answer is incorrect since KOH contains hydroxyl ions and pH is the negative logarithm of $ {H^ + } $ ion. Thus, $ \left[ {{H^ + }} \right] $ has to be calculated with the help of the ionic product of the water formula.
Generally in pH value,
$ 1 - 7 $ indicates that the compound is acidic
7 indicates that the compound is neutral
$ 7 - 14 $ Indicates that the compound is basic.
It is known that KOH is basic and its pH must be under $ 7 - 14 $ . Hence, here we can rule out other options clearly.
