Question

The pH of $HCl$ is $3$. Then the pH of $NaOH$ solution having same molar
concentration is:
(A) $3$
(B) $6$
(C) $9$
(D) $11$

Answer 1

The pH is equal to negative logarithm to the base ten of x, where, x, is the hydrogen ion concentration. Its unit is in moles per litre/s. In the given question we have to find concentration of $HCl$ to find the pH of $NaOH$ and conclude whether it is the same or not as one is acid which will lose hydrogen atom and other is base which will give away hydroxyl ion.

Step by step answer:
The range of pH scale is in between $0$ to $14$ including $0$and $14$.The pH value is divided in three parts first, from $0$ to $7$ that indicates the acidic nature of solution. Second, from $7$ to $14$ that indicates the basic nature of the solution.
The intermediate value that is $7$ indicates neutral nature of solution such as pure water or neutral substance which will not give hydrogen ion nor hydroxyl ion. The pH value of Ammonia or Sodium Hydroxide is in the range $7$-$14$, which reflects their basic nature.
Whereas acidic substances like vinegar$C{H_3}COOH$, stomach acid ($HCl$), battery acid, etc, show pH range from $0$-$7$.
The given pH OF HCl is $3$.

pH{\rm{ }} of {\rm{ }} HCl {\rm{ }} = {\rm{ }}3\\\ \left[ {{H^ + }} \right] = {\rm{ }}{10^{ - 3}} \end{array}$$ It's given that the molar concentration of $$HCl$$ and $$NaOH$$ is the same. So, moles of concentration of $$O{H^ - }$$ion from$$NaOH$$ is $$\begin{array}{c} \left[ {O{H^ - }} \right]{\rm{ }} = {\rm{ }}{10^{ - 3}}\\\ pOH{\rm{ }} = {\rm{ }} - log\left[ {{{10}^{ - 3}}} \right]{\rm{ }}\\\ pOH = {\rm{ }}3\\\ pH{\rm{ }} = {\rm{ }}14{\rm{ }} - {\rm{ }}3{\rm{ }} = {\rm{ }}11 \end{array}$$ **Hence, the correct option is (D).** **Note:** In such types of questions usually we can get confused with the value of pH of acid should be same as that of the value of pH of base, as in the question, it is given that molar concentration is same of both the acid and base that means the value of $$pOH$$=$$pH$$. Hence we can directly use the given formula: $$pOH{\rm{ }} = {\rm{ }}14{\rm{ }} - {\rm{ }}pH$$.