JEE Main 2018 — Ionic Equilibrium Question with Solution

HCl $$\to$$ H⁺ + Cl^$$-$$

H⁺ concentration is = 0.2 M.

H₂S $$\rightleftharpoons$$ H⁺ + HS^$$-$$; K₁ = 1.0 $$\times$$ 10^$$-$$7

HS^$$-$$ $$\rightleftharpoons$$ H⁺ + S^2$$-$$; K₂ = 1.2 $$\times$$ 10^$$-$$13

H₂S $$\rightleftharpoons$$ S^2$$-$$ + 2H⁺

K = K₁ $$\times$$ K₂ = 1.0 $$\times$$ 10^$$-$$7 $$\times$$ 1.2 $$\times$$ 1.0^$$-$$13 = 1.2 $$\times$$ 10^$$-$$20

as K₁ and K₂ both are very low for this reaction so dissociation of H₂S and HS^$$-$$ will be very low so, the produced H⁺ from this reaction will also be very low.

So, we can say the concentration of H⁺ will be almost same as H⁺ in HCl.

$$\therefore$$ [ H⁺ ] = 0.2 M.

From the reaction, H₂S $$\rightleftharpoons$$ 2H⁺ + S^2$$-$$

We get [ H⁺ ]² [ S^2$$-$$] = K $$\times$$ [ H₂ S ]

$$\Rightarrow$$ [ S^2$$-$$ ] = $$\frac{1.2\times {10}^{-20}\times 0.1}{{\left(0.2\right)}^2}$$

$$\Rightarrow$$ [ S^2$$-$$ ] = 3 $$\times$$ 10^$$-$$20 M