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Question: Given that the dissociation constant for water is \( {k_w} = 1 \times {10^{ - 14}}mo{l^2}{L^{ - 1}} ...

Given that the dissociation constant for water is kw=1×10−14mol2L−1{k_w} = 1 \times {10^{ - 14}}mo{l^2}{L^{ - 1}} . The pH of a 0.001M0.001M KOH solution is

A. 10−11{10^{ - 11}}

B. 10−3{10^{ - 3}}

C. 3

D. 11

Explanation

Solution

The degree of dissociation of water (or) ionic product of water can be written as,

kw=[H+][OH−]{k_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] It is known that in KOH solution, it contains K+{K^ + } and OH−O{H^ - } ions.

It is also known that pH is the negative logarithm of the concentration of the H+{H^ + } ion. Thus, H+{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 [OH−]=10−3M\left[ {O{H^ - }} \right] = {10^{ - 3}}M , kw=1×10−14mol2L−1{k_w} = 1 \times {10^{ - 14}}mo{l^2}{L^{ - 1}}

It is known that kw=[H+][OH−]{k_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]

Thus, H+{H^ + } can be determined as,

⇒1×10−14mol2L−1=[H+][10−3M]\Rightarrow 1 \times {10^{ - 14}}mo{l^2}{L^{ - 1}} = \left[ {{H^ + }} \right]\left[ {{{10}^{ - 3}}M} \right]

⇒[H+]=1×10−14mol2L−110−3M\Rightarrow \left[ {{H^ + }} \right] = \dfrac{{1 \times {{10}^{ - 14}}mo{l^2}{L^{ - 1}}}}{{{{10}^{ - 3}}M}}

⇒[H+]=10−11M\Rightarrow \left[ {{H^ + }} \right] = {10^{ - 11}}M

Thus, the concentration of H+{H^ + } ion is obtained as 10−11M{10^{ - 11}}M and its pH is 11

The hydroxyl ion concentration [OH−]=10−3M\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+]= - \log [{H^ + }] .

It can be determined as,

pH=−log⁡[10−11]pH = - \log \left[ {{{10}^{ - 11}}} \right]

pH=−(−11)log⁡[10]pH = - ( - 11)\log [10]

pH=11pH = 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+{H^ + } ion. Thus, [H+]\left[ {{H^ + }} \right] has to be calculated with the help of the ionic product of the water formula.

Generally in pH value,

1−71 - 7 indicates that the compound is acidic

7 indicates that the compound is neutral

7−147 - 14 Indicates that the compound is basic.

It is known that KOH is basic and its pH must be under 7−147 - 14 . Hence, here we can rule out other options clearly.