QJEEM18Ch9c1b: JEE Main 2018 Chemistry - Class 11 JEEM18Ch9c1b Solution: An aqueous solution contains 0.10 M H 2 S and 0.20 M HCl. If…

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 $-$} ] = ${{1.2 \times {{10}^{ - 20}} \times 0.1} \over {{{\left( {0.2} \right)}^2}}}$

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