Question

The $pH$ of ${10^{ - 4}}M\,KOH$ solution will be:
A. $4$
B. $11$
C. $10.5$
D. $10$

Answer 1

Hint: pH is the measurement of acidic or basic property of a solution.
pH of a solution is the negative logarithm of \[{H^ + }\] ion concentration of the solution.
\[pH + pOH = 14\]

Complete step by step answer:
pH is a measure of acidic or basic characteristic of a solution. $pH$ of a solution is defined as the negative logarithm of \[{H^ + }\] ion concentration of the solution.
The aqueous solution of \[KOH\] gives rise to two ions i.e., \[{K^ + }\] and \[O{H^ - }\].
\[KOH \rightleftharpoons {K^ + } + O{H^ - }\]
Since, the concentration of KOH is \[{10^{ - 4}}M\], the concentration of \[{K^ + }\] is \[{10^{ - 4}}M\]and concentration of \[O{H^ - }\] is also \[{10^{ - 4}}M\].
We know that the product of concentration of \[{H^ + }\] and \[O{H^ - }\] is always equal to \[{10^{ - 14}}\] at room temperature.
Or, \[[{H^ + }][O{H^ - }] = {10^{ - 14}}\]
\[[{H^ + }]\] is the concentration of \[{H^ + }\] ion and \[[O{H^ - }]\] is the concentration of \[O{H^ - }\] ion.
Now, \[[{H^ + }] = \dfrac{{{{10}^{ - 14}}}}{{[O{H^ - }]}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 4}}}} = {10^{ - 10}}\]
\[pH = - \log [{H^ + }] = - \log ({10^{ - 10}}) = 10\]
Alternatively, we know that, \[pH + pOH = 14\]
Here, concentration of \[O{H^ - }\] ion =\[[O{H^ - }]\]=\[{10^{ - 4}}M\]
So, \[pOH = - \log [O{H^ - }] = - \log ({10^{ - 4}}) = 4\]
\[ \Rightarrow pH = 14 - pOH = 14 - 4 = 10\]
pH of \[KOH\] is \[10\] which implies that \[KOH\] has basic property.
Hence, the correct answer is (D).

Note:
\[pH\] stands for. The pH value is measured from $0$ to $14$. pH value of pure water at normal temperature is$7$. The pH value less than $7$ denotes the acidic properties of the solution and pH greater than $7$ denotes the basic properties of the solution.