Question

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 :
\[\left( a \right)5 \times {10^{ - 8}}\]
$\left( b \right)3 \times {10^{ - 20}}$
$\left( c \right)6 \times {10^{ - 21}}$
$\left( d \right)5 \times {10^{ - 19}}$

Answer 1

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 .

Complete answer:
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 ${k_2}$
Now , add the add both the chemical equation
${H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }}.....\left( 3 \right)$
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 $\left( b \right)3 \times {10^{ - 20}}$

Additional information:
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.

Note:
Always construct the table for concentration of each reactant and products .Equilibrium constant is independent of initial and final values of reactants and products and also independent of presence of catalyst and inert material present . It only depends upon the temperature .