Question

$0.023\, g$ of sodium metal is reacted with $100 \,cm^3$ of water. The pH of the resulting solution is ______

• A. 11
• B. 10
• C. 12
• D. 9

Answer 1

Answer: (C)

$\underset{2 mol}{2 Na} +\underset{2 mol }{2 H _{2} O } \longrightarrow \underset{2 mol }{2 NaOH + H _{2}}$

Given, $\frac{0.023}{23}$ mol $\frac{100}{22400}$ mol

$=1 \times 10^{-3} mol =4.46 \times 10^{-3}$ mol

Thus, Na is the limiting reagent and decide the amount of NaOH formed.

$\because 1$ mole Na give $NaOH =1 mol$

$\therefore 1 \times 10^{-3}$ mole Na will give $NaOH$

$=1 \times 10^{-3}$ mol

Concentration of

$\left[ OH ^{-}\right] =\frac{1 \times 10^{-3} \times 1000}{100}=1 \times 10^{-2}$

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

$=-\log \left(1 \times 10^{-2}\right)$

$=2$

$PH =14-2=12$