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Question: The \({{\text{K}}_{{\text{SP}}}}\) of salts \({\text{AB}}\), \({\text{A}}{{\text{B}}_2}\) and \({{\t...

The KSP{{\text{K}}_{{\text{SP}}}} of salts AB{\text{AB}}, AB2{\text{A}}{{\text{B}}_2} and A3B{{\text{A}}_3}{\text{B}} are 4.0×10−84.0 \times {10^{ - 8}}, 3.2×10−143.2 \times {10^{ - 14}} and 2.7×10−152.7 \times {10^{ - 15}} respectively at temperature T{\text{T}}. The solubility order of these salts in water at temperature T{\text{T}} (in mole litere−1{\text{mole liter}}{{\text{e}}^{ - 1}}) is:
A) AB{\text{AB}} > AB2{\text{A}}{{\text{B}}_2} > A3B{{\text{A}}_3}{\text{B}}
B) A3B{{\text{A}}_3}{\text{B}} > AB2{\text{A}}{{\text{B}}_2} > AB{\text{AB}}
C) AB2{\text{A}}{{\text{B}}_2} > A3B{{\text{A}}_3}{\text{B}} > AB{\text{AB}}
D) AB{\text{AB}} > A3B{{\text{A}}_3}{\text{B}} > AB2{\text{A}}{{\text{B}}_2}

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 of a salt at any temperature is calculated from its solubility product.
Now, consider a salt AxBy{{\text{A}}_x}{{\text{B}}_y}. The salt dissociates as follows:
AxBy⇌xAy++yBx−{{\text{A}}_x}{{\text{B}}_y} \rightleftharpoons x{{\text{A}}^{y + }} + y{{\text{B}}^{x - }}
The solubility product of the salt AxBy{{\text{A}}_x}{{\text{B}}_y} is given as follows:
KSP=[Ay+]x[Bx−]y{{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^{y + }}]^x}{[{{\text{B}}^{x - }}]^y}
Where KSP{{\text{K}}_{{\text{SP}}}} is the solubility product.
We are given three salts AB{\text{AB}}, AB2{\text{A}}{{\text{B}}_2} and A3B{{\text{A}}_3}{\text{B}}. We will calculate the solubility and solubility products of the given salts.
Consider salt AB{\text{AB}} which dissociates as follows:
AB⇌A++B−{\text{AB}} \rightleftharpoons {{\text{A}}^ + } + {{\text{B}}^ - }
The solubility product of the salt AB{\text{AB}} is as follows:
KSP=[A+][B−]{{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^ + }][{{\text{B}}^ - }]
For the salt AB{\text{AB}}, [A+]=[B−][{{\text{A}}^ + }] = [{{\text{B}}^ - }]. And [A+]=[B−]=s[{{\text{A}}^ + }] = [{{\text{B}}^ - }] = s, where ss is the solubility of the ions. Thus,
KSP=[A+][B−]=s×s=s2{{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^ + }][{{\text{B}}^ - }] = s \times s = {s^2}
We are given that the solubility product of salt AB{\text{AB}} is 4.0×10−84.0 \times {10^{ - 8}}. Thus,
s2=4.0×10−8{s^2} = 4.0 \times {10^{ - 8}}
s=2.0×10−4s = 2.0 \times {10^{ - 4}}
Thus, the solubility of salt AB{\text{AB}} is 2.0×10−4 mole litre−12.0 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}.
Consider salt AB2{\text{A}}{{\text{B}}_2} which dissociates as follows:
AB2⇌A2++2B−{\text{A}}{{\text{B}}_2} \rightleftharpoons {{\text{A}}^{2 + }} + 2{{\text{B}}^ - }
The solubility product of the salt AB2{\text{A}}{{\text{B}}_2} is as follows:
KSP=[A2+][B−]2{{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^{2 + }}]{[{{\text{B}}^ - }]^2}
For the salt AB2{\text{A}}{{\text{B}}_2}.
KSP=[A2+][B−]2=s×(2s)2=4s3{{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^{2 + }}]{[{{\text{B}}^ - }]^2} = s \times {\left( {2s} \right)^2} = 4{s^3}
Where ss is the solubility of the ions.
We are given that the solubility product of salt AB2{\text{A}}{{\text{B}}_2} is 3.2×10−143.2 \times {10^{ - 14}}. Thus,
4s3=3.2×10−144{s^3} = 3.2 \times {10^{ - 14}}
s3=8×10−15{s^3} = 8 \times {10^{ - 15}}
s=2×10−5=0.2×10−4s = 2 \times {10^{ - 5}} = 0.2 \times {10^{ - 4}}
Thus, the solubility of salt AB2{\text{A}}{{\text{B}}_2} is 0.2×10−4 mole litre−10.2 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}.
Consider salt A3B{{\text{A}}_3}{\text{B}} which dissociates as follows:
A3B⇌3A++B3−{{\text{A}}_3}{\text{B}} \rightleftharpoons 3{{\text{A}}^ + } + {{\text{B}}^{3 - }}
The solubility product of the salt A3B{{\text{A}}_3}{\text{B}} is as follows:
KSP=[A+]3[B−]{{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^ + }]^3}[{{\text{B}}^ - }]
For the salt A3B{{\text{A}}_3}{\text{B}}.
KSP=[A+]3[B−]=(3s)3×s=27s4{{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^ + }]^3}[{{\text{B}}^ - }] = {\left( {3s} \right)^3} \times s = 27{s^4}
Where ss is the solubility of the ions.
We are given that the solubility product of salt A3B{{\text{A}}_3}{\text{B}} is 2.7×10−152.7 \times {10^{ - 15}}. Thus,
27s4=2.7×10−1527{s^4} = 2.7 \times {10^{ - 15}}
s4=1×10−16{s^4} = 1 \times {10^{ - 16}}
s=1×10−4s = 1 \times {10^{ - 4}}
Thus, the solubility of salt A3B{{\text{A}}_3}{\text{B}} is 1×10−4 mole litre−11 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}.
Thus, we have calculated that,
The solubility of salt AB{\text{AB}} is 2.0×10−4 mole litre−12.0 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}.
The solubility of salt AB2{\text{A}}{{\text{B}}_2} is 0.2×10−4 mole litre−10.2 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}.
The solubility of salt A3B{{\text{A}}_3}{\text{B}} is 1×10−4 mole litre−11 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}.
From this we can write the solubility order as follows:
AB{\text{AB}} > A3B{{\text{A}}_3}{\text{B}} > AB2{\text{A}}{{\text{B}}_2}

Thus, the correct answer is option (D) AB{\text{AB}} > A3B{{\text{A}}_3}{\text{B}} > AB2{\text{A}}{{\text{B}}_2}.

Note: We know that solubility is the amount of solute that dissolves in a solvent. The solubility of many solutes increases with the temperature of the solvent. We can see that the units of solubility and molar concentration are the same. Solve for the solubility of each of the given salt carefully to avoid errors.