Question

$100$ ml of $\frac{N}{10}NaOH$ solution is mixed with $100$ ml of $\frac{N}{5}HCl$ solution and the whole volume is made to 1 liter, The pH of the resulting solution will be:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

As we know that,

$\text{No. of moles}=M\times V_{\left(\text{in L}\right)}$

Therefore,

Volume of $NaOH=100ml=0.1L$

$\text{Normality}=\frac{N}{10}$

$\text{Molarity}=\text{normality}=0.1M$

$\therefore$ moles of $NaOH=M\times V=0.1\times 0.1=0.01\text{mole}$

Volume of $HCl=100ml=0.1L$

$\text{Normality}=\frac{N}{5}=0.2N$

$\text{Molarity}=\text{normality}=0.2M$

$\therefore$ Moles of $HCl=0.2\times 0.1=0.02\text{mole}$

$NaOH+HCl⟶NaCl+H_{2}O$

Amount of $HCl$ required to neutralize 0.01 mol of $NaOH=0.01$

As $HCl$ is in excess, amount of HCl left after neutralization $=0.02–0.01=0.01\text{moles}$

Total volume of HCl solution $=1L$

Concentration of resulting solution $=\frac{\text{no. of moles left}}{{\text{volume of solution}}_{\left(\text{in L}\right)}}=\frac{0.01}{1}=0.01M={10}^{-2}M$

As we know that,

$pH=-\log \left[H^{+}\right]$

$pH=-\log \left({10}^{-2}\right)=2$

Hence, the pH of the resulting solution will be 2.