Question

Calculate the weight of $N{a}_{2}C{O}_3$ of 85% purity required to prepare to neutralize 45.6 mL of 0.235 N $H_{2}S{O}_4$ ?

• A. 2.05 g
• B. 1.44 g
• C. 0.25 g
• D. 0.67 g

Answer 1

The correct option is D 0.67 g

Meq. of $N{a}_{2}C{O}_3=$ Meq. of $H_{2}S{O}_4$
Meq. of $H_{2}S{O}_4=45.6\times 0.235$
$\therefore \frac{W_{N{a}_{2}C{O}_3}}{E_{N{a}_{2}C{O}_3}}\times 1000=45.6\times 0.235$
$\Rightarrow \frac{W_{N{a}_{2}C{O}_3}}{106/2}\times 1000=45.6\times 0.235$
$\therefore W_{N{a}_{2}C{O}_3}=0.5679$

For 85 g of pure $N{a}_{2}C{O}_3,$ weighed sample = 100 g
For 0.5679 g of pure $N{a}_{2}C{O}_3,$ weighed sample $=\frac{100}{85}\times 0.5679$
$=0.6681$ g