Question

The Solubility product of $CaS{O_4}$ is $2.4 \times {10^{ - 5}}$. When $100ml$ of $0.01M$ $CaC{l_2}$ and $100ml$ of $0.02M$ $N{a_2}S{O_4}$ are mixed then:
A. $N{a_2}S{O_4}$ will precipitate
B.Both will precipitate
C.$CaS{O_4}$ will precipitate
D.None will precipitate

Answer 1

According to the option given in the questions we are asked to find which one will precipitate sodium sulfate $N{a_2}S{O_4}$, calcium sulfate or both. To find which one will precipitate we can calculate ionic product first and then compare it with solubility product.
Formula Used:
Solubility product $\left( {{K_{SP}}} \right)$
$AB\overset {} \leftrightarrows {A^ + } + {B^ - }$
${K_{SP}} = [{A^ + }][{B^ - }]$
Individual Molarity in a mixture,
$M' = \dfrac{{{M_1}{V_1}}}{{{V_1} + {V_2}}}$
$M$ Represents Molarity,
${V_1},{V_2}$ Represents respected volume of given solutions.

Complete step by step answer:
The best way to approach a numerical is to write the given quantities first. So, the given quantities are :
${K_{SP}}(CaS{O_4}) = 2.4 \times {10^{ - 5}},{M_1} = 0.01M,{M_2} = 0.02M,{V_1} = {V_2} = 100ml$
Condition for Precipitation:
If the Solubility product ${K_{SP}}$ is greater than the ionic product ${Q_{IP}}$ then no precipitate will form on adding excess solute.
If the Solubility product ${K_{SP}}$ is smaller than the ionic product ${Q_{IP}}$ then the excess solute will precipitate out.
If the Solubility product ${K_{SP}}$ is equal to the ionic product ${Q_{IP}}$ then the mixture will be in saturation form.
Above we mentioned the condition for precipitation. So now we will calculate ${Q_{IP}}$ for the mixture and then we will compare it with ${K_{SP}}$ given to us on the basis of above conditions.
Now, ${Q_{IP}}$ for $CaS{O_4}$ can be calculated the same as the formula used for ${K_{SP}}$.
$CaS{O_4}\underset {} \leftrightarrows C{a^{2 + }} + SO_4^{2 - }$,
So, ${Q_{IP}}$ for $CaS{O_4}$ can be written as,
${Q_{IP}}(CaS{O_4}) = [C{a^{2 + }}][SO_4^{2 - }]$
Now we need to calculate the concentration of $[C{a^{2 + }}]$ and $[SO_4^{2 - }]$.
So, in this step we will calculate concentration of $[C{a^{2 + }}]$ and $[SO_4^{2 - }]$.
Molarity of $[C{a^{2 + }}] = \dfrac{{{M_1}{V_1}}}{{{V_1} + {V_2}}}$ , after substituting values ${M_1} = 0.01M,{V_1} = {V_2} = V$ we get,
$\Rightarrow [C{a^{2 + }}] = \dfrac{{0.01MV}}{{2V}} = 0.005M$
$\Rightarrow [C{a^{2 + }}] = 0.005M$
Similarly for $[SO_4^{2 - }]$ we get, $[SO_4^{2 - }] = \dfrac{{{M_2}{V_2}}}{{{V_1} + {V_2}}}$ and then substitute values ${M_2} = 0.02M,{V_1} = {V_2} = V$
$\Rightarrow [SO_4^{2 - }] = \dfrac{{0.02MV}}{{2V}} = 0.01M$
$\Rightarrow [SO_4^{2 - }] = 0.01M$
After calculating concentrations of $[C{a^{2 + }}]$ and $[SO_4^{2 - }]$ we will substitute the concentration in the formula,
${Q_{IP}}(CaS{O_4}) = [C{a^{2 + }}][SO_4^{2 - }]$
$\Rightarrow {Q_{IP}}(CaS{O_4}) = 0.005 \times 0.01 = 5 \times {10^{ - 5}}$
$\Rightarrow {Q_{IP}}(CaS{O_4}) = 5 \times {10^{ - 5}}$
Now we can conclude that Ionic product ${Q_{IP}}(CaS{O_4}) = 5 \times {10^{ - 5}}$ is greater than solubility product ${K_{SP}}(CaS{O_4}) = 2.4 \times {10^{ - 5}}$ .
Simply, ${K_{SP}}(CaS{O_4}) < {Q_{SP}}(CaS{O_4})$
Final result: From above conditions and the result obtained ${K_{SP}}(CaS{O_4}) < {Q_{SP}}(CaS{O_4})$. We can conclude that precipitate of $CaS{O_4}$ will be formed.

The correct option is (C).

Note:
The important thing to note is that “Solubility of any sparingly soluble bivalent electrolyte is equal to the square root of the solubility product”. Let $S$ be the solubility and ${K_{SP}}$ be the solubility product then the above statement can be expressed as $S = \sqrt {{K_{SP}}}$.