Question

If ${{\text{K}}_{{\text{sp}}}}\left( {{\text{BaS}}{{\text{O}}_{\text{4}}}} \right)$ is $1.1 \times {10^{ - 10}}$ then in which of the following cases ${\text{BaS}}{{\text{O}}_{\text{4}}}$ is precipitated out?
A.100 ml of $4 \times {10^{ - 3}}{\text{M}}$ ${\text{BaC}}{{\text{l}}_2}$ $+$ 300 ml of $6 \times {10^{ - 4}}{\text{M}}$ ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$
B.100 ml of $4 \times {10^{ - 4}}{\text{M}}$ ${\text{BaC}}{{\text{l}}_2}$ $+$ 300 ml of $6 \times {10^{ - 8}}{\text{M}}$ ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$
C.300 ml of $4 \times {10^{ - 3}}{\text{M}}$ ${\text{BaC}}{{\text{l}}_2}$ $+$ 100 ml of $6 \times {10^{ - 8}}{\text{M}}$ ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$
D.All of the above

Answer 1

If ionic product is greater than the solubility product, the solution is supersaturated and precipitation will take place.
Thus, the essential condition for precipitation of an electrolyte is that the ionic product, i.e., the product of the concentration of its ions present in a solution should exceed the solubility product of the substance and this can be used in predicting ionic reactions.

Complete step by step answer:
If two solutions are mixed in which the ions combine to form a precipitate, then the concentration of ions after mixing are calculated. From this, the ionic product can be calculated. If the ionic product exceeds the solubility product, only then the precipitation will take place. In this way, we can find out whether precipitation will take place or not.
After mixing 100 ml of $4 \times {10^{ - 3}}{\text{M}}$ ${\text{BaC}}{{\text{l}}_2}$ and 300 ml of $6 \times {10^{ - 4}}{\text{M}}$ ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ , the total volume of the solution is 400 ml.
Since ${\text{BaC}}{{\text{l}}_2}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ are strong electrolytes, they are completely dissociated. So, we have
$\left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right] = 4 \times {10^{ - 3}}{\text{M}} \times \dfrac{{100}}{{400}} \\\ \Rightarrow \left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right] = {10^{ - 3}}{\text{M}} \\\$
And
$\left[ {{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right] = 6.0 \times {10^{ - 4}}{\text{M}} \times \dfrac{{300}}{{400}} \\\ \Rightarrow \left[ {{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right] = 4.5 \times {10^{ - 4}}{\text{M}} \\\$
Hence, the ionic product is:
$= \left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right]\left[ {{\text{S}}{{\text{O}}_4}^{{\text{2 - }}}} \right] \\\ = {10^{ - 3}}{\text{M}} \times 4.5 \times {10^{ - 4}}{\text{M}} \\\ = 4.5 \times {10^{ - 7}}{{\text{M}}^2} \\\$
This exceeds the solubility product of ${\text{BaS}}{{\text{O}}_{\text{4}}}$ which is $1.1 \times {10^{ - 10}}$ and so it will be precipitated.
After mixing 100 ml of $4 \times {10^{ - 4}}{\text{M}}$ ${\text{BaC}}{{\text{l}}_2}$ and 300 ml of $6 \times {10^{ - 8}}{\text{M}}$ ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ , the total volume of the solution is 400 ml.
Since ${\text{BaC}}{{\text{l}}_2}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ are strong electrolytes, they are completely dissociated. So, we have
$\left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right] = 4 \times {10^{ - 4}}{\text{M}} \times \dfrac{{100}}{{400}} \\\ \Rightarrow \left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right] = {10^{ - 4}}{\text{M}} \\\$
And
$\left[ {{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right] = 6.0 \times {10^{ - 8}}{\text{M}} \times \dfrac{{300}}{{400}} \\\ \Rightarrow \left[ {{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right] = 4.5 \times {10^{ - 8}}{\text{M}} \\\$
Hence, the ionic product is:
$= \left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right]\left[ {{\text{S}}{{\text{O}}_4}^{{\text{2 - }}}} \right] \\\ = {10^{ - 4}}{\text{M}} \times 4.5 \times {10^{ - 8}}{\text{M}} \\\ = 4.5 \times {10^{ - 12}}{{\text{M}}^2} \\\$
This is less than the solubility product of ${\text{BaS}}{{\text{O}}_{\text{4}}}$ and so it will not be precipitated.
After mixing 300 ml of $4 \times {10^{ - 3}}{\text{M}}$ ${\text{BaC}}{{\text{l}}_2}$ and 100 ml of $6 \times {10^{ - 8}}{\text{M}}$ ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ , the total volume of the solution is 400 ml.
Since ${\text{BaC}}{{\text{l}}_2}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ are strong electrolytes, they are completely dissociated. So, we have
$\left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right] = 4 \times {10^{ - 3}}{\text{M}} \times \dfrac{{300}}{{400}} \\\ \Rightarrow \left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right] = 3 \times {10^{ - 3}}{\text{M}} \\\$
And
$\left[ {{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right] = 6.0 \times {10^{ - 8}}{\text{M}} \times \dfrac{{100}}{{400}} \\\ \Rightarrow \left[ {{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right] = 1.5 \times {10^{ - 8}}{\text{M}} \\\$
Hence, the ionic product is:
$= \left[ {{\text{B}}{{\text{a}}^{{\text{2 + }}}}} \right]\left[ {{\text{S}}{{\text{O}}_4}^{{\text{2 - }}}} \right] \\\ = 3 \times {10^{ - 3}}{\text{M}} \times 1.5 \times {10^{ - 8}}{\text{M}} \\\ = 4.5 \times {10^{ - 11}}{{\text{M}}^2} \\\$
This is less than the solubility product of ${\text{BaS}}{{\text{O}}_{\text{4}}}$ and so it will not be precipitated.

Therefore, option A is correct.

Note: Solubility and solubility product are two different terms. The term solubility is applicable for all kinds of solutes but the solubility product is only applicable for electrolytes. The solubility product has a constant value at a given temperature while solubility of an electrolyte can be reduced by the addition of common ions.