Question

The $ \left[ Ag+ \right]~ $ in a solution prepared by shaking $ A{{g}_{2}}S~ $ with saturated $ {{H}_{2}}S~(0.01M) $ in a solution of $ 0.15M~{{H}_{3}}{{O}^{+}} $ is $ X\times {{10}^{-15}}M $ , value of $ X $ is $ {{K}_{sp}}~of~A{{g}_{2}}S=1.6\times {{10}^{-49}}~and~{{K}_{1}},~{{K}_{2}}~for~{{H}_{2}}S~are~1.0\times {{10}^{-7}}~and~1.0\times {{10}^{-14}}~ $ respectively.

Answer 1

Hint :We know that the situation in which both reactants and products present in concentration do not change within the period of time in which a chemical reaction occurs, it means the solution is in chemical equilibrium. When the rate of both forward and backward reaction is the same, this means that the chemical reaction has achieved chemical equilibrium.

Complete Step By Step Answer:
There is no net change in the concentrations of the reactants and products as the rates of the reactions are equal. This state is referred to as a dynamic equilibrium and the rate constant is known as equilibrium constant.
When we define equilibrium constant, we can say that as the product of the molar concentration of products which is raised during a reaction is equal to tits stoichiometric coefficient and then divided by products of the molar concentrations of the reacts, each raised to power equal to its stoichiometric coefficients is constant at constant temperature.
It can be defined in terms of partial pressure of reactants and products. However, if it is expressed in terms of the partial pressure, it is denoted by $ {{K}_{p.}} $ the law of mass action influences the equilibrium. It states that the rate of a chemical reaction is directly proportional to product of the concentrations of the reactants with power to their respective stoichiometric coefficients.
For the dissociation $ $ $ {{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}} $
 $ {{K}_{1}}\times {{K}_{2}}=1.0\times {{10}^{-7}}\times 1.0\times {{10}^{-14}}=1.0\times {{10}^{-21}} $
Equilibrium concentrations: $ (M)0.01 $ $ 0.15 $ $ x $
The equilibrium constant expression is: $ K=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{[{{H}_{2}}S]} $ ………………………………. $ eqn(i) $
Now we substitute the given values for following in $ eqn(i) $
 $ 1.0\times {{10}^{-21}}=\dfrac{x{{(0.15)}^{2}}}{0.01} $
By cross multiplying we get the values of $ x $
 $ \Rightarrow x=4.44\times {{10}^{-22}} $
And now the concentration for $ {{[A{{g}^{+}}]}^{2}} $ is given by;
 $ {{[A{{g}^{+}}]}^{2}}=\dfrac{Ksp}{[{{S}^{2}}]}=\dfrac{1.6\times {{10}^{-49}}}{4.44\times {{10}^{-22}}}=3.6\times {{10}^{-27}} $
 $ [A{{g}^{+}}]=6\times {{10}^{-14}}=60\times {{10}^{-15}} $
Thus the value of $ x $ can be given as
 $ \Rightarrow X=60 $

Note :
Note that Le Châtelier's principle is an important phenomenon which predicts the behaviour of an equilibrium system when changes to its reaction conditions occur. It states that “If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to partially reverse the change.”