Question

$25$ mL of a solution of ${Na}_{2}C{O}_3$ having a specific gravity of $1.25$ g/mL required $32.9$ mL of a solution of $HCl$ containing $109.5$ g of the acid per litre for complete neutralization. The volume (in mL) of $0.84N{H}_{2}S{O}_4$ that will be completely neutralized by $125$ g of ${Na}_{2}C{O}_3$ solution is :

Answer 1

The specific gravity of sodium carbonate is 1.25 g/mL.
Hence, the mass of 25 mL of solution of sodium carbonate is $1.25\times 25=31.25$ g.
The mass of HCl used for neutralization is $\frac{32.9}{1000}\times 109.5=3.60$ g.
The number of moles of HCl in 3.60 g is $\frac{3.60}{36.5}=0.0987$.
2 moles of HCl will neutralize 1 mole of sodium carbonate.
The number of moles of sodium carbonate neutralized by 0.0987 moles of HCl are $\frac{0.987}{2}=0.04935$
Thus, 31.25 g of sodium carbonate contains 0.04935 moles.
Hence, 125 g of sodium carbonate will contain $\frac{125}{31.25}\times 0.04935=0.1974$ moles.
They will neutrlaize 0.1974 moles of sulphuric acid which corresponds to $2\times 0.1974=0.3948$ g eq of sulphuric acid.
The volume of 0.84 N sulphuric acid required will be $\frac{0.3948}{0.84}=0.470$ L or 470 mL.