Question

The $pH$ of a solution containing $0.1N$ $NaOH$ is:
(A) $1$
(B) $10^{ - 1}$
(C) $13$
(D) $10^{ - 13}$

Answer 1

Here concentration of $NaOH$ is given. It means we are indirectly given the concentration of $O{H^ - }$ ions. $pOH$ can be easily calculated when the concentration of $O{H^ - }$ ion is known by using the formula, $pOH = - \log [O{H^ - }]$
After getting the value of $pOH$ , $pH$ can be calculated by using the formula:
$pH + pOH = 14$ .

Complete step by step solution:
In this question, we are given the normality of $NaOH$ .
Normality of $NaOH$ $= 0.1N$
As, $pOH = - \log [O{H^ - }]$
Where, $[O{H^ - }]$ is the concentration of $O{H^ - }$ ions (molarity).
Here, the concentration is given in normality. But we can easily convert normality into molarity us using the formula:
Normality $=$ molarity $\times$ $\dfrac{{{\text{molar mass}}}}{{{\text{equivalent mass}}}}$
Where, equivalent mass $=$ $\dfrac{{{\text{molar mass}}}}{{{\text{acidity or basicity}}}}$
As we know $NaOH$ is a base. For bases, the above formula is reduced to:
Normality $=$ molarity $\times$ acidity
$NaOH$ dissociates to give one $N{a^ + }$ and one $O{H^ - }$ ion.
$\therefore$ acidity $= 1$
Substituting the known values in the above relation between normality and molarity for bases,
Normality $=$ molarity $\times$ acidity
$\Rightarrow$ $0.1 = 1 \times$ molarity
$\Rightarrow$ molarity $=$ $\dfrac{{0.1}}{1}$
$\Rightarrow$ molarity $= 0.1M$
Now, we got the molarity of $NaOH$ . Means we got the concentration of $O{H^ - }$ in the solution.
$[O{H^ - }]$ $=$ $0.1M$
Substituting the value of $[O{H^ - }]$ , we can easily find $pOH$ .
$pOH = - \log [O{H^ - }]$
$\Rightarrow pOH = - \log [0.1]$
$\Rightarrow pOH = 1$
Finally, substitute the value of $pOH$ in the formula $pH + pOH = 14$ to get the required $pH$ .
$pH + pOH = 14$
$\Rightarrow 1 + pOH = 14$
$\Rightarrow pH = 14 - 1$
$\Rightarrow pH = 13$
Hence, the $pH$ of a solution containing $0.1N$ $NaOH$ is $13$ .

Additional Information
Molarity is the amount of solute (in moles) per unit volume of solution whereas normality is the equivalent concentration of solute per unit volume of solution. So, we got here the difference between molarity and normality.

Note:
In this case, we got the equal value of normality and molarity. But this is not true for every case. We can simply use the relation (normality= molarity $\times$ acidity or basicity) to get the desired value of molarity if the concentration is given in normality. Once we get the concentration in molarity, we can easily find the value of $pH$ and $pOH$ by using the suitable formula.