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Question: An aqueous solution contains \[0.10M\] \({H_2}S\) and \(0.20M\) \(HCl\) . If the equilibrium constan...

An aqueous solution contains 0.10M0.10M H2S{H_2}S and 0.20M0.20M HClHCl . If the equilibrium constant for HS−1H{S^{ - 1}} from H2S{H_2}S is 1.0×10−71.0 \times {10^{ - 7}} and that of S2−{S^{2 - }} from HS−H{S^ - } is 1.2×10−131.2 \times {10^{ - 13}}. Then the concentration of S2−{S^{2 - }} in aqueous solution is :
(a)5×10−8\left( a \right)5 \times {10^{ - 8}}
(b)3×10−20\left( b \right)3 \times {10^{ - 20}}
(c)6×10−21\left( c \right)6 \times {10^{ - 21}}
(d)5×10−19\left( d \right)5 \times {10^{ - 19}}

Explanation

Solution

First try to write the chemical equations , then adding the equations we obtain the final chemical equations where the term S2−{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
H2S⇌HS−+H+.......(1){H_2}S \rightleftharpoons H{S^ - } + {H^ + }.......\left( 1 \right)
HS−⇌S2−+H+......(2)H{S^ - } \rightleftharpoons {S^{2 - }} + {H^ + }......\left( 2 \right)
Here in question the value of equilibrium constant is given
For equation 11 Equilibrium constant given 1.0×10−71.0 \times {10^{ - 7}} let it be k1{k_1}
For equation 22 equilibrium constant given 1.2×10−131.2 \times {10^{ - 13}} let it be k2{k_2}
Now , add the add both the chemical equation
H2S⇌2H++S2−.....(3){H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }}.....\left( 3 \right)
Here the equilibrium constant will be k=[H+]2[S2−][H2S]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=k1×k2k = {k_1} \times {k_2}
Putting the values
[H+]2[S2−][H2S]=k1×k2\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
H2S⇌2H++S2−.....(3){H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }}.....\left( 3 \right)
⇒\Rightarrow (0.10−α)\left( {0.10 - \alpha } \right) (0.20+2α)\left( {0.20 + 2\alpha } \right) (α)\left( \alpha \right)
⇒\Rightarrow (0.20+2α)2(α)(0.10−α)=1.2×10−20\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 α\alpha will be very low so we can ignore the terms , and we can write
⇒\Rightarrow (0.20)2(α)(0.10)=1.2×10−20\dfrac{{{{\left( {0.20} \right)}^2}\left( \alpha \right)}}{{\left( {0.10} \right)}} = 1.2 \times {10^{ - 20}}
⇒\Rightarrow α=3.2×10−20\alpha = 3.2 \times {10^{ - 20}}
So the concentration of S2−{S^{2 - }} is (b)3×10−20\left( b \right)3 \times {10^{ - 20}}

Additional information:
The equation 11 and 22 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 .