25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was

Some Basic Concepts of Chemistry PYQ — JEE Main 2003 · Shift 0 · Q54 with Solution

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Correct answer: D

Write the balanced chemical equation

Barium hydroxide reacts with hydrochloric acid as: $\mathrm{Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O}$

So, 1 mole of $\mathrm{Ba(OH)_2}$ reacts with 2 moles of $\mathrm{HCl}$ .
Given data
- Volume of $\mathrm{Ba(OH)_2}$ solution $= 25\,\text{mL}$
- Molarity of $\mathrm{HCl} = 0.1\,\text{M}$
- Volume of $\mathrm{HCl}$ used $= 35\,\text{mL}$
Use titration relation

First calculate moles of $\mathrm{HCl}$ used: $n_{\mathrm{HCl}} = M \times V = 0.1 \times \frac{35}{1000} = 3.5 \times 10^{-3}\,\text{mol}$
Find moles of $\mathrm{Ba(OH)_2}$

From the stoichiometric ratio: $\mathrm{Ba(OH)_2 : HCl = 1:2}$ Therefore, $n_{\mathrm{Ba(OH)_2}} = \frac{3.5 \times 10^{-3}}{2} = 1.75 \times 10^{-3}\,\text{mol}$
Calculate molarity of $\mathrm{Ba(OH)_2}$

Volume of $\mathrm{Ba(OH)_2}$ solution: $25\,\text{mL} = 0.025\,\text{L}$

Hence, $M_{\mathrm{Ba(OH)_2}} = \frac{1.75 \times 10^{-3}}{0.025} = 0.07\,\text{M}$
Check options
- A: $0.14$ ❌
- B: $0.28$ ❌
- C: $0.35$ ❌
- D: $0.07$ ✅

Therefore, the molarity of barium hydroxide solution is: $\boxed{0.07\,\text{M}}$