Question

$V_{1}mL$ of $NaOH$ of normality $x$ and $V_{2}mL$ of $Ba(OH{)}_2$ of normality $y$ are together sufficient to neutralize exactly $100mL$ of $0.1NHCl$. If $V_1:V_2=1:4$ and if $x:y=4:1$, what fraction of the acid is neutralised by $Ba(OH{)}_2$?

• A. $0.5$
• B. $0.33$
• C. $0.67$
• D. $0.25$

Answer 1

The correct option is A $0.5$

$\begin{array}{c}given;\\ NaOH\to V_{1}mlofnormalityx\\ Ba{\left(OH\right)}_2\to V_{2}mlofmormalityy\\ and,\\ \frac{V_1}{V_2}=\frac{1}{4}\Rightarrow V_1=\frac{V_2}{4};\frac{x}{y}=\frac{4}{1}\Rightarrow x=4y\\ Now,\\ MolarityofNaOH=N_1{V}_1=x{V}_1\\ MolarityofBa{\left(OH\right)}_2=N_2{V}_2=y{V}_2\\ accordingtothequestion\\ xVl1+y{V}_2=100\times 0.1\left[MolarityofHCl\right]\\ x{V}_1+y{V}_2=10---\left(1\right)\\ Now,\\ resplacingthegivenvaluesinequation\left(1\right)weget\\ \left(4y\right)\frac{V_2}{4}+y{V}_2=10\\ 2V_{2}y=10\\ \therefore V_{2}y=5\\ Now,\\ thefractionofacidneutralisedbyBa{\left(OH\right)}_2=\frac{totalmolesofBa{\left(OH\right)}_2}{totalmolesofHcl}\\ =\frac{5}{10}=0.5\\ Hence,theoptionAisthecorrectanswer\end{array}$