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Question: What is the molarity of \({{F}^{-}}\) in a saturated solution of \(In{{F}_{3}}\)? \(\left( {{K}_{sp}...

What is the molarity of F−{{F}^{-}} in a saturated solution of InF3In{{F}_{3}}? (Ksp=7.9×10−10)\left( {{K}_{sp}}=7.9\times {{10}^{-10}} \right)
A) 2.3×10−32.3\times {{10}^{-3}}
B) 8.3×10−38.3\times {{10}^{-3}}
C) 1.0×10−31.0\times {{10}^{-3}}
D) 7.0×10−37.0\times {{10}^{-3}}

Explanation

Solution

Ksp{{K}_{sp}} is solubility product constant, which can be considered as the equilibrium constant in the case of solids i.e. it is the amount of solid dissolving in the aqueous solution.
For,cC(s)⇌aA(aq)+bB(aq)c{{C}_{\left( s \right)}}\rightleftharpoons a{{A}_{\left( aq \right)}}+b{{B}_{\left( aq \right)}}
Ksp=[A]a[B]b{{K}_{sp}}={{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}

Complete answer:
If we analyze the question, the requirement of the question is to find the molarity of F−{{F}^{-}}in the saturated solution of InF3In{{F}_{3}} if the Ksp{{K}_{sp}} value is given. Before going into the solution part of the question, we should have the basic idea about the calculation of Ksp{{K}_{sp}} and molarity from the givenKsp{{K}_{sp}}.
-Ksp{{K}_{sp}} can be calculated by multiplying the molarities or concentration of the products, if there is coefficient in the given equilibrium equation then it should be taken as the power of the concentration of the element.
-So for an equilibrium reaction say,
cC(s)⇌aA(aq)+bB(aq)c{{C}_{\left( s \right)}}\rightleftharpoons a{{A}_{\left( aq \right)}}+b{{B}_{\left( aq \right)}}
-For the above reaction the solubility product constant equation is written as,Ksp=[A]a[B]b{{K}_{sp}}={{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}
-So now let’s check the data given in the question.
-It’s given that,Ksp=7.9×10−10{{K}_{sp}}=7.9\times {{10}^{-10}}so solubility product is given and we have to find the molarity ofF−{{F}^{-}}
-Now let’s write the equilibrium equation for the given compound, how it dissociates in the aqueous solution.

The equation is: InF3(S)⇄In3+(aq)+3F(aq))In{{F}_{{{3}_{(S)}}}}\rightleftarrows I{{n}^{3+}}_{(aq)}+3{{F}_{\left( aq) \right)}}

When time, t=0 | c 0 0
When time,t=t | c-s s 3s

Where s is the solubility or can say as concentration of the solute dissolved (molarity).
Here in the question, it is given thatKsp=7.9×10−10{{K}_{sp}}=7.9\times {{10}^{-10}}, therefore, Ksp{{K}_{sp}} can be written as,Ksp=s×(3s)3{{K}_{sp}}=s\times {{(3s)}^{3}}
Since,7.9×10−10=27s47.9\times {{10}^{-10}}=27{{s}^{4}}
Therefore, s4=2.92×10−11{{s}^{4}}=2.92\times {{10}^{-11}}
When we take fourth power we will get solubility s as,
s=2.32×10−3mol/Ls=2.32\times {{10}^{-3}}mol/L, where mol/Lmol/L is equal to molarity.
As we got the value of s, now we have to find 3s to find the molarityF−{{F}^{-}}.
F−=3s=3×2.32×10−3{{F}^{-}}=3s=3\times 2.32\times {{10}^{-3}}
F−=7.0×10−3M{{F}^{-}}=7.0\times {{10}^{-3}}M

Note:
Caution should be taken when taking the root of fourth power and doing the multiplications with the power raised to 10.