The $pH$ of solution formed by mixing 40ml of $0.1MHCl$ with 10ml of $0.45M$ of $NaOH$ is:

The correct option is B 12
Solution:- (B) $12$

As we know that,

Molarity of a solution $=\frac{\text{no. of moles of solute}}{{\text{volume of solution}}_{\left(\text{in L}\right)}}$

Given:-

Molarity of $HCl$ solution $=0.1M$

Volume of $HCl$ solution $=40mL=0.04L$

Therefore,

No. of moles of $HCl=0.04\times 0.1=0.004\text{mol}$

Again,

Molarity of $NaOH$ solution $=0.45M$

Volume of $NaOH$ solution $=10mL=0.01L$

Therefore,

No. of moles of $NaOH=0.01\times 0.45=0.0045\text{mol}$

Now, for the reaction-

$NaOH+HCl⟶NaCl+H_{2}O$

$NaOH$ is in excess.

Therefore,

Excess amount of $NaOH=0.0045-0.004=0.0005\text{mol}$

Total volume $=0.04+0.01=0.05L$

Now,

$\left[{OH}^{-}\right]=\frac{0.0005}{0.05}=0.01M={10}^{-2}M$

Therefore,

$pOH=-\log \left[{OH}^{-}\right]$

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

Now as we know that,

$pH+pOH=14$

Therefore,

$pH=14-pOH=14-2=12$

Hence the pH of the solution is $12$.