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Question: The solubility product,K\(_{sp}\) , of sparingly soluble salt MX at 25\(^\circ\)C is 2.5\(\times\)10...

The solubility product,Ksp_{sp} , of sparingly soluble salt MX at 25∘^\circC is 2.5×\times10−9^{-9}. The solubility of the salt in molL−1^{-1} at this temperature is:
(A)- 1.5×\times10−14^{-14}
(B)- 5×\times10−8^{-8}
(C)- 1.25×\times10−9^{-9}
(D)- 5×\times10−5^{-5}

Explanation

Solution

The solubility product represents a type of equilibrium constant, and it is dependent on the temperature. On dissolving, it dissociates into the ions, and the solubility can be calculated of sparingly soluble salt with the help of given data.

Complete step by step answer:
Now, let us first know about the solubility product. If the solubility increases with the increase in temperature, then the value of solubility product constant increases.
So, we can say that solubility product, and solubility are interrelated to each other.
According to the question, we are given with the sparingly soluble salt MX, it will dissociate into M, and X ions. It can be represented as
MX ⇌\rightleftharpoons M+^{+} + X−^{-}
Now, if we say the solubility of MX when it is not dissolved is S, but when it is dissolved solubility of ions becomes S.
As we know the solubility product constant is the product of concentration of ions on the right hand side. Thus, Ksp_{sp}= S ×\times S
Now, we can say the value of s is equal to the square root of solubility product constant.
Therefore, S = 2.5×10−9\sqrt{2.5\times10^{-9}} (given)
So, we get S = 5×\times10−5^{-5} mol/L, or 5×\times10−5^{-5} molL−1^{-1}
In the last, we can conclude that the solubility of salt is 5×\times10−5^{-5} molL−1^{-1}.

The correct option is (D).

Note: Don’t get confused between the solubility, and solubility product constant. As mentioned these both terms are interrelated to each other. The solubility is kind of property which defines the solute when it dissolves in a solvent to form a solution.