Question

The $pH$of ${10^{ - 3}}M$ monoacidic base, if it is $1\% $ ionized is:
A. $5$
B. $8$
C. $3$
D. $9$

Answer 1

Hint: We know that a monoacidic base is a base which furnishes only one hydroxyl group in water. Similarly diacidic bases will furnish two hydroxyl groups in water. The concept of $pH$ gives us an idea about the acidity and basicity of a given solution.

Complete step by step solution:
As we know that monoacidic base furnishes only one hydroxyl group in water. Let the monoacidic base be $BOH$. It is given that in the solution it is $1\% $ ionized. Let the degree of ionization be $\alpha $. Then $\alpha = 1\% = {10^{ - 2}}$. We have given the value of concentration $(C)$ of the $BOH$ solution, that is $C = {10^{ - 3}}M$. So ,
$BOH \rightleftarrows {B^{ + 1}} + O{H^{ - 1}}$
In the above reaction , the concentration of $O{H^{ - 1}}$ will be $[O{H^{ - 1}}] = C\alpha = {10^{ - 3}} \times {10^{ - 2}} = {10^{ - 5}}$.
We know that,
 $
  {K_w} = [{H^ + }][O{H^ - }] = {10^{ - 14}} \\
   \Rightarrow [{H^ + }] = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 5}}}} = {10^{ - 9}} \\
 $
Here , ${K_w} = $ ionic product of water.
So now as we have got the value of the concentration of hydrogen ion, that is $[{H^ + }] = {10^{ - 9}}$.
Now we can directly apply the formula of $pH$, to get the value of $pH$of the given solution.
$
  pH = - \log [{H^ + }] \\
   \Rightarrow pH = - \log [{10^{ - 9}}] \\
   \Rightarrow pH = 9 \\
 $
So from the above explanation and calculation it is clear to us that the correct answer of the given question is option : $9$

Hence, the correct answer is option D.

Additional information:
The range of $pH$ of an acidic solution is less than $7$ and the $pH$of a basic solution is greater than $7$ and less than $14$. Some of the examples of strong acids are $HCl,{H_2}S{O_4},HN{O_3}$ and some examples of strong bases are $NaOH,KOH,LiOH$. Strong bases and acids dissociate completely in water . They are also known as strong electrolytes.

Note: Always remember that $pH$ of a solution is given by $pH = - \log [{H^ + }]$. The ionic product of water is given by ${K_w} = [{H^ + }][O{H^ - }] = {10^{ - 14}}$. Monoacidic bases furnish only one hydroxyl group and diacidic bases furnish two hydroxyl groups. Always try to avoid silly mistakes and calculation errors while solving the question.