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# Given that the dissociation constant for water is  ${k_w} = 1 \times {10^{ - 14}}mo{l^2}{L^{ - 1}}$ . The pH of a  $0.001M$  KOH solution isA.  ${10^{ - 11}}$ B.  ${10^{ - 3}}$ C. 3D. 11

Last updated date: 18th Jun 2024
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Hint: 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.

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.