Question

$0.023g$ of sodium metal is reacted with $100c{m}^3$ of water. The pH of the resulting solution is ______

• A. 11
• B. 10
• C. 12
• D. 9

Answer 1

Answer: (C)

$\underset{2mol}{2Na}+\underset{2mol}{2H_{2}O}⟶\underset{2mol}{2NaOH+H_2}$

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 $NaOH=1mol$

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

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

Concentration of

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

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

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

$=2$

$PH=14-2=12$