Question

A solution with respect to $A{g^ + },C{a^{2 + }},M{g^{2 + }}and$$A{l^{3 + }}$which will precipitate at lowest concentration of $[PO_4^{3 - }]$ when solution of $N{a_3}P{O_4}$ is added?
A. $A{g_3}P{O_4}({K_{sp}} = 1 \times {10^{ - 6}})$
B. $C{a_3}{(P{O_4})_2}({K_{sp}} = 1 \times {10^{ - 33}})$
C. $M{g_2}{(P{O_4})_2}({K_{sp}} = 1 \times {10^{ - 24}})$
D. $AlP{O_4}({K_{sp}} = 1 \times {10^{ - 20}})$

Answer 1

Hint: The solubility product constant is the equilibrium constant for the dissolution of a solid substance into aqueous solution. It is denoted by the symbol ${K_{sp}}$.

Complete step by step answer: As we know, the solubility product is a kind of equilibrium constant and its value depends on temperature.
     \[{K_{sp}}\]
 usually increases with an increase in temperature due to increase in solubility.
We need to remove that
When, solubility is >0.1 M, then it is soluble salt
1.02 M < solubility < 0.1M ,slightly soluble
Solubility <0.1M , sparingly soluble
Now the question is asking for precipitation at lowest concentration.
So we need to solve all options one by one and then the option with the least concentration of $PO_4^{3 - }$ will be our answer.
- $A{g_3}P{O_4}(ksp = 1*{10^{ - 6}})$

Therefore, ${K_{sp}}$ = ${[A{g^ + }]^3}[PO_4^{3 - }]$
                  $[PO_4^{3 - }]$= $\dfrac{{{k_{sp}}}}{{{{\left[ {A{g^ + }} \right]}^3}}}$=$\tfrac{{1 \times {{10}^{ - 6}}}}{{{{(0.1)}^3}}}$=${10^{ - 3}}$M
- $C{a_3}{(P{O_4})_2}$ $\left( {{k_{sp}} = 1 \times {{10}^{ - 33}}} \right)$
$C{a_3}{(P{O_4})_2} \rightleftharpoons 3C{a^{ + 2}} + 2PO_4^{3 - }$
Therefore, ${k_{sp}} = {[C{a^{2 + }}]^3}{[PO_4^{3 - }]^2}$
$\left[ {PO_4^{3 - }} \right] = {\left( {\dfrac{{{k_{sp}}}}{{{{[C{a^{2 + }}]}^3}}}} \right)^{\dfrac{1}{2}}} = {\left( {\dfrac{{{{10}^{ - 33}}}}{{{{(0.1)}^3}}}} \right)^{\dfrac{1}{2}}} = {10^{ - 15}}M$
- $M{g_3}{(P{O_4})_2}$ $\left( {{K_{sp}} = 1 \times {{10}^{ - 24}}} \right)$
$M{g_3}{(P{O_4})_2} \rightleftharpoons 3M{g^{2 + }} + 2PO_4^{3 - }$
Therefore, ${K_{sp}} = {[M{g^{2 + }}]^3}{[PO_4^{3 - }]^2}$
$[PO_4^{3 - }] = {\left( {\dfrac{{{K_{sp}}}}{{{{[M{g^{2 + }}]}^3}}}} \right)^{\dfrac{1}{2}}} = {\left( {\dfrac{{1 \times {{10}^{ - 24}}}}{{{{(0.1)}^3}}}} \right)^{\dfrac{1}{2}}} = {10^{ - 10.5}}M$
- $AlP{O_4}$ $\left( {{K_{sp}} = 1 \times {{10}^{ - 20}}} \right)$
$AlP{O_4} \rightleftharpoons A{l^{3 + }} + PO_4^{3 - }$
Therefore, ${K_{sp}} = [A{l^{3 + }}][PO_4^{3 - }]$
$[PO_4^{3 - }] = \dfrac{{{K_{sp}}}}{{[A{l^{3 + }}]}} = \dfrac{{{{10}^{ - 20}}}}{{0.1}} = {10^{ - 19}}M$
Since, we see that the lowest concentration of $PO_4^{3 - }$ needed for $AlP{O_4}$to precipitate, $AlP{O_4}$will precipitate at lowest concentration. Therefore, the correct answer is $AlP{O_4}$ i.e. option D.

So, the correct answer is “Option D”.

Note: It is important to note that reaction quotient (Q) can be used to determine whether precipitate will form with given concentration of ions.
If Q<${K_{sp}}$, no precipitate will form.
If, Q>${K_{sp}}$, a precipitate will form.
If, Q=${K_{sp}}$, a precipitate just started to form.