JEE Main 2018 — Ionic Equilibrium Question with Solution

Let initially concentration of Ba⁺² = x m.

After adding 50 ml Na₂SO₄ in Ba⁺² solution final volume becomes 500 ml.

$$\therefore$$ Initial volume of Ba⁺² solution

= (500 $$-$$ 50) ml = 450 ml

As at the begining of precipitation, ionic product = solubility product.

$$\Rightarrow$$ [Ba²⁺] [SO$${}_4^{-2}$$] = K_sp of BaSO₄

$$\Rightarrow$$ [Ba²⁺] $$\left(\frac{50\times 1}{500}\right)$$ = 1 $$\times$$ 10^$$-$$10

$$\Rightarrow$$ [Ba²⁺] = 10^$$-$$9 M.

So, the concentration of Ba⁺² in final solution is 10^$$-$$9 M.

$$\therefore$$ concentration of Ba⁺² in original solution,

M₁ V₁ = M₂ V₂

$$\Rightarrow$$ x $$\times$$ 450 = 10^$$-$$9 $$\times$$ 500

$$\Rightarrow$$ x = 1.1 $$\times$$ 10^$$-$$9 M