Question

An aqueous solution of $NaOH$ having density $1.1$ $kg/{dm}^3$ contains $0.02$ mole fraction of $NaOH$. The pH of NaOH solution is :

• A. 13.9939
• B. 0.0061
• C. 12.0218
• D. 14

Answer 1

The correct option is A 13.9939

Given data,

$d_{NaOH}=1.1kg/d{m}^3=1.1g/mL$

Mole fraction $x=0.02$

We know that,

$x_{NaOH}=\frac{n_{NaOH}}{n_{NaOH}+n_{H_{2}O}}=0.02$

Let we consider $100mL$ water

moles of $H_{2}O{n}_{H_{2}O}=\frac{1000}{18}=55.5$

$\frac{n_{NaOH}+n_{H_{2}O}}{n_{NaOH}}=\frac{1}{0.02}=50$

$1+\frac{55.5}{n_{NaOH}}=50$

$n_{NaOH}=\frac{55.5}{49}=1.132$ moles

mass of $NaOH=1.132\times 40=45.28g$

Volume of solution $=1.1\times 1045.28=1149.81mL$

Concentration of $O{H}^{-}=\frac{1.132}{1.1498}=0.985$

$pOH=-\log \left(0.985\right)=0.0067$

$P_H+P_{OH}=14$

$pH=13.99$

Option A is correct.