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Question: A saturated solution of \[C{a_3}{\left( {P{O_4}} \right)_2}\] contains \(2.0 \times {10^{ - 8}}{\tex...

A saturated solution of Ca3(PO4)2C{a_3}{\left( {P{O_4}} \right)_2} contains 2.0×10−8M2.0 \times {10^{ - 8}}{\text{M}} Ca2+C{a^{2 + }} and 1.6×10−5M1.6 \times {10^{ - 5}}{\text{M}} PO43−PO_4^{3 - } at a certain temperature. The solubility product Ksp{K_{sp}} of Ca3(PO4)2C{a_3}{\left( {P{O_4}} \right)_2}at that temperature is-
A. 2.048×10−342.048 \times {10^{ - 34}}
B. 2.04×10−332.04 \times {10^{ - 33}}
C. 3.20×10−343.20 \times {10^{ - 34}}
D. 8×10−348 \times {10^{ - 34}}

Explanation

Solution

The solubility product is the product of concentration of ions in water. To find the solubility product, dissociate the compoundCa3(PO4)2C{a_3}{\left( {P{O_4}} \right)_2} into ions and then multiply the concentration of the ions, with each concentration raised to the power equal to the coefficient of that ion in the balanced equation.

Formula used: Ksp=product of concentration of ions{{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}

Step-by-step Solution:
We are given that A saturated solution of Ca3(PO4)2C{a_3}{\left( {P{O_4}} \right)_2} contains 2.0×10−8M2.0 \times {10^{ - 8}}{\text{M}} Ca2+C{a^{2 + }} and 1.6×10−5M1.6 \times {10^{ - 5}}{\text{M}} PO43−PO_4^{3 - } at a certain temperature.
We have to find its solubility product at that temperature. Solubility Product is the product of concentration of ions, with each concentration raised to the power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium. It is denoted by Ksp{{\text{K}}_{{\text{sp}}}} . Now the given salt will dissociate into following ions-
⇒Ca3(PO4)2→3Ca2++2PO43−\Rightarrow C{a_3}{\left( {P{O_4}} \right)_2} \to 3C{a^{2 + }} + 2PO_4^{3 - }
Then using formula Ksp=product of concentration of ions{{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}
⇒Ksp=[Ca2+]3×[PO43−]2\Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {C{a^{2 + }}} \right]^3} \times {\left[ {PO_4^{3 - }} \right]^2}
We already know the concentration of the given ions so putting the values of their concentration in the formula we get,
⇒Ksp=[2.0×10−8]3×[1.6×10−5M]2\Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {2.0 \times {{10}^{ - 8}}} \right]^3} \times {\left[ {1.6 \times {{10}^{ - 5}}{\text{M}}} \right]^2}
On solving we get,
⇒Ksp=8×2.56×10−10−24\Rightarrow {{\text{K}}_{{\text{sp}}}} = 8 \times 2.56 \times {10^{ - 10 - 24}} {as we know that am.an=am+n{a^m}.{a^n} = {a^{m + n}} }
On simplifying we get,
⇒Ksp=20.4×10−34\Rightarrow {{\text{K}}_{{\text{sp}}}} = 20.4 \times {10^{ - 34}}
We can also write it as-
⇒Ksp=2.04×10−33\Rightarrow {{\text{K}}_{{\text{sp}}}} = 2.04 \times {10^{ - 33}}

Answer- Hence the correct answer is B.

Note: Difference between solubility and solubility product is-
The solubility of a substance is the total amount of the solute that can be dissolved in the solvent at equilibrium.
Solubility product constant (Ksp{{\text{K}}_{{\text{sp}}}} ) is an equilibrium constant which represents the level at which the solute dissolves in a solution or we can say it provides insight into the equilibrium between the solute and its constituent ions that are dissociated across the solution.