Question

How much water from $5$ litre of ${10}^{-3}M$ $HCl$ should be evaporated to change its pH by $2$ units?

• A. $3$ litre
• B. $0.5$ litre
• C. $2.54$ litre
• D. $4.95$ litre

Answer 1

Answer: (A)

$3$ litre

Current concentration is $\left[H^{+}\right]={10}^{-3}M$

no. of moles of $HCl$ is ${10}^{-3}\times 5=5\times {10}^{-3}$

Desired concentration is when $pH=2.$ $pH=-log\left[H^{+}\right]$

$\Rightarrow 2=-log\left[H^{+}\right]$

$\Rightarrow \left[H^{+}\right]={10}^{-2}$ is desired concentration.

The concentration should increase by 10-fold.

so $n=\frac{C_{new}}{V_{new}}=\frac{{10}^{-2}}{V_{new}}$

$\Rightarrow 5\times {10}^{-3}=\frac{{10}^{-2}}{V_{new}}$

$V_{new}=\frac{{10}^{-2}}{5\times {10}^{-3}}=2liter$

So $5-2=3$ liter of water should be evaporated to change concentration from ${10}^{-3}$ to $pH=2$