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# An aqueous solution contains $0.10M$ ${H_2}S$ and $0.20M$ $HCl$ . If the equilibrium constant for $H{S^{ - 1}}$ from ${H_2}S$ is $1.0 \times {10^{ - 7}}$ and that of ${S^{2 - }}$ from $H{S^ - }$ is $1.2 \times {10^{ - 13}}$. Then the concentration of ${S^{2 - }}$ in aqueous solution is :$\left( a \right)5 \times {10^{ - 8}}$$\left( b \right)3 \times {10^{ - 20}}$$\left( c \right)6 \times {10^{ - 21}}$$\left( d \right)5 \times {10^{ - 19}}$

Last updated date: 11th Sep 2024
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Hint: First try to write the chemical equations , then adding the equations we obtain the final chemical equations where the term ${S^{2 - }}$ should present as we need to find the concentration of this term . putting the values we can obtain the final concentration .

According to the question the reactions which is occurring here are
${H_2}S \rightleftharpoons H{S^ - } + {H^ + }.......\left( 1 \right)$
$H{S^ - } \rightleftharpoons {S^{2 - }} + {H^ + }......\left( 2 \right)$
Here in question the value of equilibrium constant is given
For equation $1$ Equilibrium constant given $1.0 \times {10^{ - 7}}$ let it be ${k_1}$
For equation $2$ equilibrium constant given $1.2 \times {10^{ - 13}}$ let it be ${k_2}$
${H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }}.....\left( 3 \right)$
Here the equilibrium constant will be $k = \dfrac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{S^{2 - }}} \right]}}{{\left[ {{H_2}S} \right]}}$
When we add any two chemical equations then their equilibrium constant becomes a product .
So , $k = {k_1} \times {k_2}$
Putting the values
$\dfrac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{S^{2 - }}} \right]}}{{\left[ {{H_2}S} \right]}} = {k_1} \times {k_2}$
We need to put the concentration for this look at the equation
${H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }}.....\left( 3 \right)$
$\Rightarrow $$\left( {0.10 - \alpha } \right) \left( {0.20 + 2\alpha } \right) \left( \alpha \right) \Rightarrow \dfrac{{{{\left( {0.20 + 2\alpha } \right)}^2}\left( \alpha \right)}}{{\left( {0.10 - \alpha } \right)}} = 1.2 \times {10^{ - 20}} Now the value of \alpha will be very low so we can ignore the terms , and we can write \Rightarrow$$\dfrac{{{{\left( {0.20} \right)}^2}\left( \alpha \right)}}{{\left( {0.10} \right)}} = 1.2 \times {10^{ - 20}}$
$\Rightarrow$$\alpha = 3.2 \times {10^{ - 20}}$
So the concentration of ${S^{2 - }}$ is $\left( b \right)3 \times {10^{ - 20}}$

The equation $1$ and $2$ both are reversible reactions which means they can proceed in both forward and backward directions .The double arrow shows the reversible reaction . equilibrium is the point at which the rate of forward reaction is equals to reverse reaction . The equilibrium constant is calculated after the reaction proceeds and then measures each molecule's concentrations.