Question

$7.3$ grams of $HCl$ is dissolved in $20$ lit solution. $50$ ml of this solution is taken in $250$ ml flask and water is added up to the mark. $40$ ml of this diluted $HCl$ solution exactly neutralizes $20$ ml of $Ba(OH{)}_2$ solution. $pH$ of given barium hydroxide solution is:

• A. $7.902$
• B. $11.601$
• C. $12.301$
• D. $2.699$

Answer 1

The correct option is B $11.601$

Number of moles of $HCl=\frac{7.3g}{36.5}=0.2$ moles

Thus molarity of the solutions= $\frac{n}{v}=\frac{0.2}{20}$

$M_{HCl}=0.01M={10}^{-2}M$

Now, $50ml$ of this solution is taken and diluted to $250ml$. Thus, new concentration of solution will be

$\Rightarrow M_1{V}_1=M_2{V}_2$

$\Rightarrow 0.01M\times 50ml=M_2\times 250ml$

$M_2=2\times {10}^{-3}M$

This is the concentration of diluted $HCl$ solution as $40ml$ of this solution dilute $20ml$ of $Ba(OH{)}_2$

Thus, concentration of $Ba(OH{)}_2$ is

$\Rightarrow 40\times 2\times {10}^{-3}=20\times M_3\times 2$ $nf=2$ for $Ba(OH{)}_2$

$\Rightarrow M_3=2\times {10}^{-3}M$

Thus, this is the concentration of $Ba(OH{)}_2$

$\underset{s}{Ba}(OH{)}_2\rightleftharpoons {\underset{s}{Ba}}^{2+}+\underset{2s}{2\stackrel{-}{O}H}$

Thus concentration of $\stackrel{-}{O}H$ is

$\Rightarrow {[}^{-}OH]=2\times (2\times {10}^{-3}M)$

$\Rightarrow \left[O{H}^{-}\right]=4\times {10}^{-3}M$

$pOH=-\log \left[OH\right]=2.397$

as $pH=14-pOH$

$pH=14-2.397$

$pH=11.6$

Thus, $pH$ of $Ba(OH{)}_2$ is $11.6$ .