Question

$x$ mole of a strong monoacidic base is dissolved in one litre of water. The $pH$of the solution will be
a.$ - \log x$
b.$14 - \log x$
c.$14 + \log x$
d.$ + \log (14 - x)$

Answer 1

Hint: As we know that a monoacidic base is a base that furnishes only the hydroxyl group in the water. It combines with one hydrogen ion and is called the monoacidic base. Some of the examples are $NaOH, KOH$, etc are monoacidic bases. We can find the $pH$ of the solution $ = - \log [concentration]$ or we can write it as $pH = - \log [{H^ + }]$.

Complete step by step answer:
Let us assume that the number of molecules base $ = x$. And we have been given that the volume of water is $1$ litre.
So we can say that concentration of monoacidic base $ = \dfrac{{\text{No. Of moles}}}{{\text{Volume of water}}}$.
It gives us concentration of monoacidic base $ = \dfrac{x}{1} = x$.
Therefore the $pH$of the solution by the formula is $ - \log [concentration]$ . By putting the value in the formula we have $pH = - \log x$.

Hence the correct option is (a) $ - \log x$.

Note:
We should know that the concept of $pH$ gives us an idea about the acidity and basicity of a given solution. The ionic product of water is given by ${K_w} = [{H^ + }][O{H^ - }]$. We should always try to avoid calculation mistakes and should know the formulas beforehand.