Question

To $1L$ of $1.0M$ impure $H_{2}S{O}_4$ sample, $1.0MNaOH$ solution was added and a plot was obtained as follows:

The $\%$ purity of $H_{2}S{O}_4$ and the slope of curve, respectively, are:

• A. $75\%-1/2$
• B. $75\%,-1$
• C. $50\%,-1/3$
• D. $50\%,-1/2$

Answer 1

Answer: (B)

(i) Moles of $NaOH=1.0M\times 1.5L=1.5mol$
Moles of $H_{2}S{O}_4=1.0M\times 1L=1.0mol$
$\underset{\text{2 mol}}{2NaOH}+\underset{1mol}{H_{2}S{O}_4}⟶N{a}_{2}S{O}_4+2H_{2}O$
Moles of $H_{2}S{O}_4$ reacted with $NaOH$
$=\frac{1}{2}mol$ of $NaOH=\frac{1}{2}\times 1.5=0.75mol{H}_{2}S{O}_4$
$\%$ purity of $H_{2}S{O}_4=\frac{0.75mol\times 100}{1.0mol}=75\%$
(ii) For slope: $\frac{x}{a}+\frac{y}{b}=1$
Slope $\left(\frac{y}{x}\right)=\frac{-b}{a}=\frac{1.5}{1.5}=-1$