JEE MAIN - Chemistry (2018 (Offline) - No. 19)
An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS– from H2S is 1.0 $$\times$$ 10–7 and that of S2- from HS– ions is 1.2 $$\times$$ 10–13 then the concentration of S2- ions in aqueous solution is :
5 $$\times$$ 10–19
5 $$\times$$ 10–8
3 $$\times$$ 10–20
6 $$\times$$ 10–21
Explanation
HCl $$ \to $$ H+ + Cl$$-$$
H+ concentration is = 0.2 M.
H2S $$\rightleftharpoons$$ H+ + HS$$-$$; K1 = 1.0 $$ \times $$ 10$$-$$7
HS$$-$$ $$\rightleftharpoons$$ H+ + S2$$-$$; K2 = 1.2 $$ \times $$ 10$$-$$13
H2S $$\rightleftharpoons$$ S2$$-$$ + 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+ + S2$$-$$
We get [ H+ ]2 [ S2$$-$$] = K $$ \times $$ [ H2 S ]
$$ \Rightarrow \,\,\,$$ [ S2$$-$$ ] = $$ {{1.2 \times {{10}^{ - 20}} \times 0.1} \over {{{\left( {0.2} \right)}^2}}}$$
$$ \Rightarrow \,\,\,$$ [ S2$$-$$ ] = 3 $$ \times $$ 10$$-$$20 M
H+ concentration is = 0.2 M.
H2S $$\rightleftharpoons$$ H+ + HS$$-$$; K1 = 1.0 $$ \times $$ 10$$-$$7
HS$$-$$ $$\rightleftharpoons$$ H+ + S2$$-$$; K2 = 1.2 $$ \times $$ 10$$-$$13
H2S $$\rightleftharpoons$$ S2$$-$$ + 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+ + S2$$-$$
We get [ H+ ]2 [ S2$$-$$] = K $$ \times $$ [ H2 S ]
$$ \Rightarrow \,\,\,$$ [ S2$$-$$ ] = $$ {{1.2 \times {{10}^{ - 20}} \times 0.1} \over {{{\left( {0.2} \right)}^2}}}$$
$$ \Rightarrow \,\,\,$$ [ S2$$-$$ ] = 3 $$ \times $$ 10$$-$$20 M
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