Question

Sulphuric acid reacts with sodium hydroxide as follows.H2SO4 + 2NaOH ⇒ Na2SO4 + 2H2Ohen 1L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is

• (A) 0.1 mol L^-1
• (B) 7.10 g
• (C) 0.025 mol L^-1
• (D) 3.55 g

Answer 1

The Correct Option is B, C

Explanation:
$1\mathrm{L}$ of $0.1\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4$ contains $0.1\mathrm{mol}{\mathrm{H}}_2{\mathrm{SO}}_4$1 L of $0.1\mathrm{M}$. NaOH contains $=0.1\mathrm{mol}\mathrm{NaOH}$ According to the given equation, 1 mol of ${\mathrm{H}}_2{\mathrm{SO}}_4$ reacts with $2\mathrm{mol}$ of $\mathrm{NaOH}$.Therefore, 0.1 mol of $\mathrm{NaOH}$ will react with $\left(\frac{0.1}{2}\right)=0.05\mathrm{mol}$ of ${\mathrm{H}}_2{\mathrm{SO}}_4$ and 0.05 mole ${\mathrm{H}}_2{\mathrm{SO}}_4$ will remain unreacted. Thus, $\mathrm{NaO}$is the limiting reagent. $\therefore 0.1$ mole of $\mathrm{NaOH}$ will produce 0.05 mole of ${\mathrm{Na}}_2{\mathrm{SO}}_4$Molar mass of ${\mathrm{Na}}_2{\mathrm{SO}}_4=2\times 23+1\times 32+4\times 16=142\mathrm{g}$Mass of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ formed $=142\times 0.05=7.10\mathrm{g}$Volume of solution after mixing $=2\mathrm{L}$ ${\mathrm{H}}_2{\mathrm{SO}}_4$ left unreacted $=0.05\mathrm{mol}$Molarity of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ formed $=\frac{0.05}{2}=0.025\mathrm{mol}{\mathrm{L}}^{-1}$