Question

2 g of $NaOH$ are dissolved in water to make 1 litre solution. What is $pH$ of the solution?

Answer 1

As we know that,

$\text{no. of moles}=\frac{\text{Wt.}}{\text{Mol. wt.}}$

Mol. wt. of $NaOH=40g$

Wt. of $NaOH=2g\left(\text{Given}\right)$

Therefore,

No. of mles $=\frac{2}{40}=0.05\text{mol}$

As we know that molarity of a solution is defined as the no. of moles of solute present in $1L$ of solution.

Since $0.05$ mole of $NaOH$ are pesent in $1L$ of of solution.

Therefore,

Molarity of solution $\left(M\right)=0.05$

$\Rightarrow \left[{OH}^{-}\right]=0.05$

As we know that,

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

$\Rightarrow pOH=-\log (5\times {10}^{-2})$

$\Rightarrow pOH=1.3$

Therefore,

$pH+pOH=1$

$\Rightarrow pH=14-pOH=14-1.3=12.7$

Hence pH of the solution is $12.7$.