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Question: The solubility products of MA, MB, MC and MD are \(1.8 \times {10^{ - 10}}\), \(4 \times {10^{ - 3}}...

The solubility products of MA, MB, MC and MD are 1.8×10−101.8 \times {10^{ - 10}}, 4×10−34 \times {10^{ - 3}}, 4×10−84 \times {10^{ - 8}} and 6×10−56 \times {10^{ - 5}} respectively. If a 0.01 M0.01{\text{ M}} solution of MX is added dropwise to a mixture containing A−{{\text{A}}^ - }, B−{{\text{B}}^ - }, C−{{\text{C}}^ - } and D−{{\text{D}}^ - } ions, then the one to be precipitated first will be:
A) MA
B) MB
C) MC
D) MD

Explanation

Solution

We know that the solubility product of any salt at any temperature is the product of the molar concentration of its constituent ions. The concentration of ions is raised to the number of ions produced on dissociation of one molecule of the salt.

Complete answer:
We know that the solubility product of any salt at any temperature is the product of the molar concentration of its constituent ions. The concentration of ions is raised to the number of ions produced on dissociation of one molecule of the salt.
We are given that the solubility products of MA, MB, MC and MD are 1.8×10−101.8 \times {10^{ - 10}}, 4×10−34 \times {10^{ - 3}}, 4×10−84 \times {10^{ - 8}} and 6×10−56 \times {10^{ - 5}} respectively. And 0.01 M0.01{\text{ M}} solution of MX is added dropwise to a mixture containing A−{{\text{A}}^ - }, B−{{\text{B}}^ - }, C−{{\text{C}}^ - } and D−{{\text{D}}^ - } ions.
Now, we will calculate the concentration of A−{{\text{A}}^ - }, B−{{\text{B}}^ - }, C−{{\text{C}}^ - } and D−{{\text{D}}^ - } ions as follows:
For the given salts, the expression of solubility product is as follows:
KSP=[M+][Anion]{{\text{K}}_{{\text{SP}}}} = [{{\text{M}}^ + }][{\text{Anion}}]
Where KSP{{\text{K}}_{{\text{SP}}}} is the solubility product.
In 0.01 M0.01{\text{ M}} solution of MX, [M+]=0.01 M[{{\text{M}}^ + }] = 0.01{\text{ M}}. Thus,
For MA:
KSP=[M+][A−]{{\text{K}}_{{\text{SP}}}} = [{{\text{M}}^ + }][{{\text{A}}^ - }]
[A−]=KSP[M+]=1.8×10−100.01=1.8×10−8 M[{{\text{A}}^ - }] = \dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{[{{\text{M}}^ + }]}} = \dfrac{{1.8 \times {{10}^{ - 10}}}}{{0.01}} = 1.8 \times {10^{ - 8}}{\text{ M}}
For MB:
KSP=[M+][B−]{{\text{K}}_{{\text{SP}}}} = [{{\text{M}}^ + }][{{\text{B}}^ - }]
[B−]=KSP[M+]=4×10−30.01=4×10−1 M[{{\text{B}}^ - }] = \dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{[{{\text{M}}^ + }]}} = \dfrac{{4 \times {{10}^{ - 3}}}}{{0.01}} = 4 \times {10^{ - 1}}{\text{ M}}
For MC:
KSP=[M+][C−]{{\text{K}}_{{\text{SP}}}} = [{{\text{M}}^ + }][{{\text{C}}^ - }]
[C−]=KSP[M+]=4×10−80.01=4×10−5 M[{{\text{C}}^ - }] = \dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{[{{\text{M}}^ + }]}} = \dfrac{{4 \times {{10}^{ - 8}}}}{{0.01}} = 4 \times {10^{ - 5}}{\text{ M}}
For MD:
KSP=[M+][D−]{{\text{K}}_{{\text{SP}}}} = [{{\text{M}}^ + }][{{\text{D}}^ - }]
[D−]=KSP[M+]=6×10−50.01=6×10−3 M[{{\text{D}}^ - }] = \dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{[{{\text{M}}^ + }]}} = \dfrac{{6 \times {{10}^{ - 5}}}}{{0.01}} = 6 \times {10^{ - 3}}{\text{ M}}
The anion having lowest concentration will combine with the M+{{\text{M}}^ + } ion in the 0.01 M0.01{\text{ M}} solution of MX and precipitate out first.
The anion having lowest concentration is [A−][{{\text{A}}^ - }].
Thus, the salt that will precipitate out first will be MA.

So the correct answer is option A.

Note: The solubility product is calculated using the concentration of the ions in which the salt has dissociated. Solubility factor depends on various factors such as temperature, pressure and nature of the electrolyte. The concentration of ions is affected by these factors and thus, the solubility product gets affected.