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Question: Solubility product of \({ BaCl }_{ 2 }\) is \({ 4\times 10 }^{ -9 }\) . Its solubility would be: A...

Solubility product of BaCl2{ BaCl }_{ 2 } is 4×10−9{ 4\times 10 }^{ -9 } . Its solubility would be:
A. 1×10−27{ 1\times 10 }^{ -27 }
B. 1×10−3{ 1\times 10 }^{ -3 }
C. 1×10−7{ 1\times 10 }^{ -7 }
D. 1×10−2{ 1\times 10 }^{ -2 }

Explanation

Solution

Solubility product,Ksp{ K }_{ sp }, is the product of a substance’s dissolved ion concentrations raised to the power of their stoichiometric coefficients. Ksp{ K }_{ sp } is related to the solubility as one can be used to derive the other, as the solubility product is the product of solubility of each ion in moles per liter.

Complete step by step solution :
It is given that,
Ksp=4×10−9{ K }_{ sp }= { 4\times 10 }^{ -9 }
S =?
where, Ksp{ K }_{ sp } = solubility product constant
S = solubility
The salt BaCl2{ BaCl }_{ 2 } will dissociate as shown below;
BaCl2⇌Ba+2+2Cl−{ BaCl }_{ 2 }{ \rightleftharpoons Ba }^{ +2 }{ +2Cl }^{ - }

Barium chloride dissociates into a barium ion and two chloride ions. Let, the solubility of BaCl2{ BaCl }_{ 2 } is S then the solubility of ions will be S and 2S respectively.

Let, the solubility of BaCl2{ BaCl }_{ 2 } is S, then the solubility of the products shown above.
As we know, solubility products are the products of substances raised to their power of coefficients.
So, Ksp=Ba+2×(Cl−1)2{ K }_{ sp }{ =Ba }^{ +2 }{ \times (Cl }^{ -1 }{ ) }^{ 2 }
= S×2S2{ S\times 2S }^{ 2 }
Ksp=4S3{ K }_{ sp }= { 4S }^{ 3 }

Now, put the value of Ksp{ K }_{ sp }, we get
4×10−9=4S3{ 4\times 10 }^{ -9 }= { 4S }^{ 3 }
S3=10−9{ S }^{ 3 }{ =10 }^{ -9 }
S=10−3{ S }{ =10 }^{ -3 }
Hence, its solubility would be 1×10−3{ 1\times 10 }^{ -3 }.

The correct option is B.

Additional Information:
Ionic product is the product of the concentrations of ions, each raised to a power equal to the stoichiometric number for a solution of salt at a specified concentration.
Ionic product equals the solubility product for a saturated solution.

Note :The possibility to make a mistake is that when barium chloride dissociates into barium ion and two chloride ions. As the solubility product is the product of substances raised to their power of coefficients, so the solubility of chloride ion will be 2S not S.