The precipitate of C$aF_2, \, ( K_{sp} = 1.7 \times 10^{ - 10 } )$ is obtained, when equal volumes of which of the following are mixed?
- $ 10^{ - 4 } \, M \, Ca^{ 2 + } + 10^{ - 4 } \, M \, F^- $
- $ 10^{ - 2 } \, M \, Ca^{ 2 + } + 10^{ - 3 } \, M \, F^- $
- $ 10^{ - 5 } \, M \, Ca^{ 2 + } + 10^{ - 3 } \, M \, F^- $
- $ 10^{ - 3 } \, M \, Ca^{ 2 + } + 10^{ - 5 } \, M \, F^- $
The Correct Option is B
Solution and Explanation
$ Q_W = [ Ca^{ 2 +} ][ F^- ]^2 = \bigg( \frac{ 10^ { - 2 }}{ 2} \bigg) \times \bigg( \frac{ 10^ { - 3 }}{ 2} \bigg) $
= $ 1.25 \times 10^{ - 9 } > K_{ s} $ precipitate will be formed
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Concepts Used:
Law of Chemical Equilibrium
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.
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