The solubility of MX2–type electrolytes is 0.5 × 10–4 Mole/lit. then find out the Ksp of electrolytes :
The solubility of MX2–type electrolytes is 0.5 × 10–4 Mole/lit. then find out the Ksp of electrolytes :
5 × 10–12
25 × 10–10
1 × 10–13
5 × 10–13
The Correct Option is D
Solution and Explanation
The correct option is (D) : 5 × 10–13
An electrolyte MX2 undergoes dissociation as follows :
MX2\(\rightleftharpoons\)M+2+2X− Concentration MX2 M+2 X− Initial concentration 1 0 0 Concentration at Equilibrium 1-s s 2s Thus from the above condition we can say that, Ksp=s×(2s)2=4×(s)3 Here, s (the solubility ) is 0.5×10−4mole/lit. ∴ Ksp =4×(0.5×10−4)3 ∴ Ksp=5×10−13
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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.