$0.023 \mathrm{~g}$ of sodium metal is reacted with $100 \mathrm{~cm}^{3}$ of water. The $\mathrm{pH}$ of the resulting solution is

• A

  10
• B

  11
• C

  9
• Correct answer
  D

  12

Solution

Given, $\frac{0.023}{23}$ mol $\frac{100}{22400} \mathrm{~mol}$ $=1 \times 10^{-3} \mathrm{~mol}=4.46 \times 10^{-3} \mathrm{~mol}$ Thus, Na is the limiting reagent and decide the amount of $\mathrm{NaOH}$ formed. $\because 1$ mole Na give $\mathrm{NaOH}=1 \mathrm{~mol}$ $\therefore 1 \times 10^{-3}$ mole Na will give $\mathrm{NaOH}$ $=1 \times 10^{-3} \mathrm{~mol}$ Concentration of $$ \begin{aligned} {\left[\mathrm{OH}^{-}\right] } &=\frac{1 \times 10^{-3} \times 1000}{100}=1 \times 10^{-2} \\ \mathrm{pOH} &=-\log \left[\mathrm{OH}^{-}\right] \\ &=-\log \left(1 \times 10^{-2}\right) \\ &=2 \\ \mathrm{pH} &=14-2=12 \end{aligned} $$

Hence, the correct answer is option: 'D'.