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Question: Which of the following is most soluble in water? (A) \(B{{i}_{2}}{{S}_{3}}\left( {{K}_{sp}}=1\time...

Which of the following is most soluble in water?
(A) Bi2S3(Ksp=1×10−70M5)B{{i}_{2}}{{S}_{3}}\left( {{K}_{sp}}=1\times {{10}^{-70}}{{M}^{5}} \right)
(B) MnS(Ksp=7×10−16M2)MnS\left( {{K}_{sp}}=7\times {{10}^{-16}}{{M}^{2}} \right)
(C) CuS(Ksp=8×10−37M2)CuS\left( {{K}_{sp}}=8\times {{10}^{-37}}{{M}^{2}} \right)
(D) Ag2S(Ksp=6×10−51M3)A{{g}_{2}}S\left( {{K}_{sp}}=6\times {{10}^{-51}}{{M}^{3}} \right)

Explanation

Solution

The solubility product constant describes the equilibrium between the constituent ions in the solution and as solid. The value of constant is the degree to which the compound can dissociate in water and is directly proportional. Thus, higher the solubility product constant, the compound will be more soluble.

Complete answer:
Here, we could have directly answered the question as who has a higher solubility product constant will be more soluble in water. But the solubility products have different units, so we won’t compare them directly.
We will calculate solubility using solubility-solubility product constant formula and then whichever solubility will be maximum; that would be our required answer.
The formula is given as;
Ksp=xsn{{K}_{sp}}=x{{s}^{n}}
where,
s = solubility
Ksp{{K}_{sp}} = solubility product constant
x = numerical constant
n = degree of ionisation
Now, let us solve one by one option to check the solubility;
(A) For, Bi2S3(Ksp=1×10−70M5)B{{i}_{2}}{{S}_{3}}\left( {{K}_{sp}}=1\times {{10}^{-70}}{{M}^{5}} \right)
The dissociation will take place as-
Bi2S3→2Bi+3+3S−2B{{i}_{2}}{{S}_{3}}\to 2B{{i}^{+3}}+3{{S}^{-2}}
Therefore, the solubility product will be,
Ksp=(2s)2(3s)3=108s5{{K}_{sp}}={{\left( 2s \right)}^{2}}{{\left( 3s \right)}^{3}}=108{{s}^{5}}
Thus, we can calculate s as;
s=1×10−701085 s=3.92×10−15M \begin{aligned} & s=\sqrt[5]{\frac{1\times {{10}^{-70}}}{108}} \\\ & s=3.92\times {{10}^{-15}}M \\\ \end{aligned}
(B) For, MnS(Ksp=7×10−16M2)MnS\left( {{K}_{sp}}=7\times {{10}^{-16}}{{M}^{2}} \right)
The dissociation will take place as-
MnS→Mn+2+S−2MnS\to M{{n}^{+2}}+{{S}^{-2}}
Therefore, the solubility product will be,
Ksp=s×s=s2{{K}_{sp}}=s\times s={{s}^{2}}
Thus, we can calculate s as;
s=7×10−16 s=2.64×10−8M \begin{aligned} & s=\sqrt{7\times {{10}^{-16}}} \\\ & s=2.64\times {{10}^{-8}}M \\\ \end{aligned}
(C) For, CuS(Ksp=8×10−37M2)CuS\left( {{K}_{sp}}=8\times {{10}^{-37}}{{M}^{2}} \right)
The dissociation will take place as-
CuS→Cu+2+S−2CuS\to C{{u}^{+2}}+{{S}^{-2}}
Therefore, the solubility product will be,
Ksp=s×s=s2{{K}_{sp}}=s\times s={{s}^{2}}
Thus, we can calculate s as;
s=8×10−37 s=8.95×10−19M \begin{aligned} & s=\sqrt{8\times {{10}^{-37}}} \\\ & s=8.95\times {{10}^{-19}}M \\\ \end{aligned}
(D) For, Ag2S(Ksp=6×10−51M3)A{{g}_{2}}S\left( {{K}_{sp}}=6\times {{10}^{-51}}{{M}^{3}} \right)
The dissociation will take place as-
Ag2S→2Ag+1+S−2A{{g}_{2}}S\to 2A{{g}^{+1}}+{{S}^{-2}}
Therefore, the solubility product will be,
Ksp=(2s)2(s)=4s3{{K}_{sp}}={{\left( 2s \right)}^{2}}\left( s \right)=4{{s}^{3}}
Thus, we can calculate s as;
s=6×10−5143 s=1.15×10−17M \begin{aligned} & s=\sqrt[3]{\frac{6\times {{10}^{-51}}}{4}} \\\ & s=1.15\times {{10}^{-17}}M \\\ \end{aligned}

According to above calculations, we can say that the option (B) is correct.

Note:
Most ionic compounds are soluble in water thus, the relationship between the solubility and solubility product constant is important to be known. Use of this equation is important as the units of solubility products constant are inconsistent.