Question

In the reversible reaction $Ca{{F}_{2\left( s \right)}}\rightleftharpoons C{{a}^{2+}}_{(aq)}+2{{F}^{-}}_{(aq)}$, the concentration of fluoride ions was made halved, then equilibrium concentration of $C{{a}^{2+}}$
(A) Increases by 2 times
(B) Decreases by 2 times
(C)Increases by 4 times
(D)Decreases by 4 times

Answer 1

Hint: The equilibrium constant of forward and backward reaction should have the equal value
The equilibrium constant for a reaction say $AB\rightleftharpoons {{A}^{+}}+{{B}^{-}}$
${{K}_{eq}}=\dfrac{\left[ {{A}^{+}} \right]\left[ {{B}^{+}} \right]}{\left[ AB \right]}$

Complete step by step answer:
So to solve the question let's analyze the data given in the question, it is asked that in the given reversible reaction, what happens to the concentration of $C{{a}^{2+}}$ if the concentration of ${{F}^{-}}$ ions are made halved.
We should have a basic idea about the reversible reaction and about the equilibrium, what changes affect the equilibrium condition etc.
- Now let’s see what a reversible reaction is, it is a reaction in which the reaction could occur in both sides i.e. forward reaction and backward reaction takes place.
If the system is said to be in equilibrium then the rate of forward reaction and backward reaction is happening at the same rate i.e. the rate product formation and decomposition of the products is in having the same rate.
- There are many conditions which affect the equilibrium of reaction. Ike the concentration of reactants, temperature etc.
And all the effects can be explained using the Le- Chatelier’s principle.
Le-Chatelier’s principle could explain in which direction the chemical reaction will be carried out if any changes in the equilibrium conditions happens and the reaction will occur in the direction at which the system can reintroduce the equilibrium state.
- So here let’s see the conditions by which the equilibrium may be redefined if the change in concentration occurs.
So basic the concept is first write the balanced chemical equation of the reaction, then write its equilibrium constant equation.
From the equilibrium constant equation we could predict the changes happening.
- The balanced equation is, $Ca{{F}_{2\left( s \right)}}\rightleftharpoons C{{a}^{2+}}_{(aq)}+2{{F}^{-}}_{(aq)}$
The equilibrium constant is written as,${{K}_{sp}}=\frac{\left[ C{{a}^{2+}} \right]{{\left[ {{F}^{-}} \right]}^{2}}}{\left[ Ca{{F}_{2}} \right]}$
Since the reaction is a heterogeneous equilibrium existing between the solid phase and liquid phase.
- We won’t write the concentration of solid phase in equilibrium constant equation, hence,
${{K}_{sp}}=\left[ C{{a}^{2+}} \right]{{\left[ {{F}^{-}} \right]}^{2}}$
Now it’s given that the concentration of fluoride ion is made half.
- If we see the equilibrium constant if the concentration of the fluoride ion is decreased then the concentration of Ca ions must be increased to maintain the equilibrium constant. Now the question is how much it should be increased.
- Let’s apply very basic idea,
Concentration of ${{F}^{-}}$ ions made half i.e., ${{\left[ {{F}^{-}} \right]}^{2}}=\dfrac{1}{2}$
But since in the equation the fluoride ions have coefficients with them, it is written as the power of the concentration of fluoride ions.
- So in actual sense the concentration of ${{F}^{-}}$ ions are decreased by four times,
${{\left[ {{F}^{-}} \right]}^{2}}={{\left[ \dfrac{1}{2} \right]}^{2}}=\left[ \dfrac{1}{4} \right]$
So the $C{{a}^{2+}}$ ions should increase by four times so that the equilibrium is restated.
The correct option is option “C” .

Note: Students often make mistakes in writing the equilibrium constant equation for the species that have coefficients associated with them. The coefficient should be written as the power for the concentration of the respective species.