Question

A solution is $0.1$M with respect to $A{g}^{+},C{a}^{2+},M{g}^{2+}$ and $A{l}^{3+}$ which will precipitate at lowest concentration of $\left[P{O}_4^{3-}\right]$ 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. $C{a}_3\left(P{O}_4{)}_2\right(K_{sp}=1\times {10}^{-33})$
• C. $M{g}_3\left(P{O}_4{)}_2\right(K_{sp}=1\times {10}^{-24})$
• D. $AlP{O}_4(K_{sp}=1\times {10}^{-20})$

Answer 1

Answer: (D)

$AlP{O}_4(K_{sp}=1\times {10}^{-20})$

A) ${Ag}_3{PO}_4\rightleftharpoons {3Ag}^{+}+{PO}_4^{3-}$

$K_{sp}={\left[{Ag}^{+}\right]}^3\left[{PO}_4^{3-}\right]=1\times {10}^{-6}$

$\Rightarrow \left[{PO}_4^{3-}\right]=\frac{1\times {10}^{-6}}{{\left(0.1\right)}^3}={10}^{-3}$

B) ${Ca}_3{\left({PO}_4\right)}_2\rightleftharpoons {3Ca}^{2+}+2{PO}_4^{3-}$

$\therefore$ $K_{sp}={\left[{Ca}^{2+}\right]}^3\times {\left[{PO}_4^{3-}\right]}^2={10}^{-33}$

$\therefore$ $\left[{PO}_4^{3-}\right]={\left(\frac{{10}^{-33}}{{\left(0.1\right)}^3}\right)}^{1/2}={\left({10}^{-30}\right)}^{1/2}={10}^{-15}$

C) ${Mg}_3{\left({PO}_4\right)}_2\rightleftharpoons {3Mg}^{2+}+{2PO}_4^{3-}$

$K_{sp}={\left[{Mg}^{2+}\right]}^3{\left[{PO}_4^{3-}\right]}^2={10}^{-24}$

$\therefore$ $\left[{PO}_4^{3-}\right]={\left\{\frac{{10}^{-24}}{{\left(0.1\right)}^3}\right\}}^{1/2}={10}^{-10.5}$

D) ${AlPO}_4\rightleftharpoons {Al}^{3+}+{PO}_4^{3-}$

$K_{sp}=\left[{Al}^{3+}\right]\left[{PO}_4^{3-}\right]={10}^{-20}$

$\Rightarrow \left[{PO}_4^{3-}\right]=\frac{{10}^{-20}}{0.1}={10}^{-19}$

Since lowest concentration of ${PO}_4^{3-}$ needed for ${AlPO}_4$ to precipitate, ${AlPO}_4$ will precipitate.

$\therefore$ Correct answer will be D.