Question

How many grams of iron (II) sulphide will dissolve per litre in $0.20{\text{ M}}$ sodium sulphide solution? ${K_{sp}} = 4.9 \times {10^{ - 18}}$

Answer 1

Hint :In a solution, several species are associated with each other via chemical equilibrium. When an electrolyte which consists of a common ion as present in the solution is added, an increase in the concentration of particular species leads to increase in degree of dissociation of ions and due to which solubility of ionic solid decreases. This effect is known as the common ion effect.

Complete Step By Step Answer:
Iron (II) sulphide also known as ferrous sulphide are black water insoluble solids i.e., when iron sulphide is dissolved in water then a dissociation equilibrium is established between undissolved solid and dissociated ions.
Dissociation of ferrous sulphide in water takes place as follows:
$FeS \rightleftharpoons F{e^{2 + }} + {S^{2 - }}$
It is clearly observed that on dissolving one mole of solid ferrous sulphide, one mole of cation and one mole of anion is formed. So, the solubility product for the given dissociation reaction can be represented as follows:
${K_{sp}} = \left[ {F{e^{2 + }}} \right]\left[ {{S^{2 - }}} \right]$
Assuming the solubility of ions formed after dissociation $ = s$
Therefore, the solubility product can be expressed in terms of solubility of ions as follows:
$ \Rightarrow {K_{sp}} = s \times s$
$ \Rightarrow {K_{sp}} = {s^2}$
Now, it is given in the question that iron (II) disulphide is added to sodium sulphide solution. So, an increase in the concentration of sulphide ions will be observed. The dissociation of sodium sulphide solution takes place as follows:
$N{a_2}S \rightleftharpoons 2N{a^ + } + {S^{2 - }}$
During the reaction, one mole of sodium sulphide dissociates into two moles of sodium ion and one mole of sulphide ion. So, the concentration of sulphide ion formed after reaction is $0.20{\text{ M}}$. So, we can conclude that the concentration of sulphide ions increased by the value of $0.20{\text{ M}}$.
Therefore, the new concentration of sulphide ions in the solution will be as follows:
$ \Rightarrow s' = s + 0.20$
So, in this case the solubility product can be expressed as follows:
${K_{sp}} = s \times (s + 0.20)$
Substituting the given value of ${K_{sp}}$ in the expression:
$ \Rightarrow 4.9 \times {10^{ - 18}} = s \times (s + 0.20)$
$ \Rightarrow {s^2} + 0.20s - 4.9 \times {10^{ - 18}} = 0$
On solving above quadratic equation, the value of s is as follows:
$ \Rightarrow s = 2.45 \times {10^{ - 17}}\;mol{L^{ - 1}}$
But according to the question, we need to find the solubility of iron (II) sulphide in grams per litre, so we need to convert moles into grams by multiplying the value with its molar mass.
$ \Rightarrow s = 88 \times 2.45 \times {10^{ - 17}}$
$ \Rightarrow s = 2.15 \times {10^{ - 15}}\;g{L^{ - 1}}$
Hence, solubility of iron (II) sulphide in $0.20{\text{ M}}$ sodium sulphide solution is $2.15 \times {10^{ - 15}}\;g{L^{ - 1}}$.

Note :
It is important to note that if a solid dissociates into ions which consist of more than one mole, then the solubility of resulting ions will be greater than the solubility of the solid. For example $Ca{F_2}$ dissociates into one mole of calcium ion and two moles of fluoride ions, so the solubility of $C{a^{2 + }}$ and ${F^ - }$ ions will be $s$ and $2s$ respectively. Also, if the solubility product of two compounds is similar then the molar solubilities may not necessarily be the same.