Question

On ignition, Rochelle salt, $NaK{C}_4{H}_4{O}_6.4H_{2}O$ (molar mass $=282$ g/mol) is converted into $NaKC{O}_3$ (molar mass $=122$ g/mol). $0.9546$ g sample of the Rochelle salt on ignition gives $NaKC{O}_3$, which is titrated with $41.72$ mL $H_{2}S{O}_4$. From the following data, find the percentage purity of the Rochelle salt. The solution after neutralisation required $1.91$ mL of $0.1297NNaOH$. The $H_{2}S{O}_4$ used for the neutralisation requires its $10.27$ mL against $10.35$ mL of $0.1297NNaOH$.

• A. $72\%$
• B. $89\%$
• C. $67\%$
• D. $77\%$

Answer 1

The correct option is D $77\%$

Meq. of $H_{2}S{O}_4=$ Meq. of $NaOH$
$N\times 10.27=10.35\times 0.1297$ $⟹N_{H_{2}S{O}_4}=0.1307$
Meq. of $H_{2}S{O}_4$ used for $NaKC{O}_3=$ Meq. of $H_{2}S{O}_4$ added$-$Meq.of $H_{2}S{O}_4$ used by $NaOH$
$=(0.1307\times 41.72)-1.91\times 0.1297=5.2050$
Also, for change
$NaK{C}_4{H}_4{O}_6\cdot 4H_{2}O⟶NaKC{O}_3+C{O}_2\uparrow +H_{2}O\uparrow$
Now, meq. of $NaKC{O}_3$ using valence factor, $2$ during its neutralization with $H_{2}S{O}_4=5.2050$
Mass of $NaKC{O}_3=\frac{5.2050\times 122}{2\times 1000}=0.3175$ g
For $1:1$ mole ratio of conversion, mass of Rochelle salt $=\frac{282}{122}\times 0.3175$ $=0.7339$ g
Percentage purity of Rochelle salt $=\frac{0.7339}{0.9546}\times 100$ $=76.87\%$