Question

Which of the following is most soluble in water?
A. $MnS({K_{sp}} = 8 \times {10^{ - 37}})$
B. $ZnS({K_{sp}} = 7 \times {10^{ - 16}})$
C. $B{i_2}{S_2}({K_{sp}} = 8 \times {10^{ - 7}}$
D. $A{g_3}(P{O_4})({K_{sp}} = 1.8 \times {10^{ - 6}})$

Answer 1

Hint: Here, ${K_{sp}}$ is the solubility product and its value indicates the extent of dissociation in water. The relation between solubility and solubility product is given by the formula as given below:
${K_{sp}} = {X^x}{Y^y}.{S^{x + y}}$
Where X and Y are the number of moles and S be the solubility of salt mol per litre.

Complete step by step answer: We will start by calculating the solubility ‘s’ for each of the compounds given above. As we know ${K_{sp}}$ is defined as the product of the equilibrium concentration of the ions in a saturated solution. And we also know the concentration of the ion at equilibrium at saturated solution is equal to its solubility i.e. s.
Let us calculate for MnS first. On dissociation, MnS gives 1Mn2+ and 1S2- ions. If the solubility of each ion is “s”. Then, according to formula ${K_{sp}} = s \times s$ we get:
$
  {K_{sp}} = {s^2} \\
   \Rightarrow s = {({K_{sp}})^{\dfrac{1}{2}}} \\
   \Rightarrow s = {(8 \times {10^{ - 37}})^{\dfrac{1}{2}}} \\
$
Taking log on both side we get
$\log s = \dfrac{1}{2}\log (8 \times {10^{ - 37}})$
Since, $\log (A \times B) = \log A + \log B$
Hence,
$\log s = \dfrac{1}{2}(\log 8 + ( - 37)\log 10)$
Value of log 8=0.9030 and Log10=1
On putting the value, we get:
$
  \log s = \dfrac{1}{2}(0.9030 - 37 \times 1) \\
   \Rightarrow \log s = - 18.0485 \\
$
Taking antilog on both side, we get:
$s = 8.94 \times {10^{ - 19}}$.
Similarly for the case of ZnS, on dissociation it gives 1 Zn2+ and 1 S2- Ion. So
 $
  {K_{sp}} = {s^2} \\
   \Rightarrow s = \sqrt {{K_{sp}}} \\
   \Rightarrow s = {(7 \times {10^{ - 16}})^{\dfrac{1}{2}}} \\
   \Rightarrow s = 2.6 \times {10^{ - 8}} \\
$
In case of Bi2S3, there are 2Bi3+ and 3S2- ions. Hence, solubility of Bi3+ion is 2s and for sulphide ion solubility will be 3s.
$
  {K_{sp}} = {(2s)^2} \times {(3s)^3} \\
   \Rightarrow {K_{sp}} = 4{s^2} \times {27^3} \\
   \Rightarrow {K_{sp}} = 108{s^5} \\
   \Rightarrow s = {(\dfrac{{{K_{sp}}}}{{108}})^{1/5}} \\
    \\
$
Taking log on both side, we get:
$
  \log s = \dfrac{1}{5}(\log 8 + ( - 7)\log 10 - \log 108) \\
   \Rightarrow \log s = \dfrac{1}{5}(0.9030 - 7 - 2.033) \\
   \Rightarrow \log s = - 1.626 \\
$
Taking antilog on both side:
We get
$s = 0.0236$
And in case of Ag3 (PO4), there are 3 silver ion and 1 phosphate ions
Hence, solubility is calculated as:
 $
  {K_{sp}} = {(3s)^3} \times s \\
   \Rightarrow {K_{sp}} = 27{s^4} \\
   \Rightarrow {s^4} = \dfrac{1}{{27}}{K_{sp}} \\
   \Rightarrow s = {(\dfrac{1}{{27}}{K_{sp}})^{1/4}} \\
$
Taking log on both side, we get:
$
  \log s = \dfrac{1}{4}\log (\dfrac{1}{{27}} \times 1.8 \times {10^{ - 6}}) \\
  \log s = \dfrac{1}{4}(\log \dfrac{1}{{27}} + \log 1.8 + ( - 6)\log 10) \\
  \log s = \dfrac{1}{4}(\log 1 - \log 27 + \log 1.8 - 6) \\
$
$
  \log s = \dfrac{1}{4}(0 - 1.431 + 0.2552 - 6) \\
  \log s = - 1.7939 \\
$
Taking antilog on both side:
$s = 0.016073$
Hence, as we calculated the solubility of $A{g_3}(P{O_4})$ is maximum, so the correct option is D.

Note: : As we know, more will be the value of solubility product more will be solubility. So, as per options given $A{g_3}(P{O_4})$ has the maximum value of solubility product so it is most soluble in water. Hence the correct answer is D.