Question

To $1 \,L$ of $1.0\, M$ impure $H _{2}\, SO _{4}$ sample, $1.0 \,M\, NaOH$ solution was added and a plot was obtained as follows:

The $\%$ purity of $H _{2} SO _{4}$ 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 $NaOH =1.0 M \times 1.5 \,L =1.5\, mol$
Moles of $H _{2} SO _{4}=1.0\, M \times 1\, L =1.0\, mol$
$\underset{\text{2 mol}}{2 NaOH} + \underset{1 mol}{H _{2} SO _{4}} \longrightarrow Na _{2} SO _{4}+2 H _{2} O$
Moles of $H _{2} SO _{4}$ reacted with $NaOH$
$=\frac{1}{2} mol $ of $ NaOH =\frac{1}{2} \times 1.5=0.75\, mol \,H _{2} SO _{4}$
$\%$ purity of $H _{2} SO _{4}=\frac{0.75 mol \times 100}{1.0 \,mol }=75 \%$
(ii) For slope: $\frac{x}{a}+\frac{y}{b}=1$
Slope $\left(\frac{y}{x}\right)=\frac{-b}{a}=\frac{1.5}{1.5}=-1$