Question

A solution is $0.1 M$ with respect to $Ag ^{+}, Ca ^{2+}, Mg ^{2+}$ and $Al ^{3+}$, which will precipitate at lowest concentration of $\left[ PO _{4}^{3-}\right]$ when solution of $Na _{3} PO _{4}$ is added?

• A. $Ag _{3} PO _{3}\left(K_{s p}=1 \times 10^{-6}\right)$
• B. $Ca _{3}\left( PO _{4}\right)_{2}\left(K_{s p}=1 \times 10^{-33}\right)$
• C. $M g_{3}\left(P O_{4}\right)_{2}\left(K_{s p}=1 \times 10^{-24}\right)$
• D. $AlPO_{4}\left(K_{s p}=1 \times 10^{-20}\right)$

Answer 1

Answer: (D)

(a) $Ag _{3} PO _{4} \rightleftharpoons 3 Ag ^{+}+ PO _{4}^{3-} $
${\left[ PO _{4}^{3-}\right]=\frac{K_{s p}}{(0.1)^{3}}=\frac{1 \times 10^{-6}}{10^{-3}}=10^{-3}} $
(b) $Ca _{3}\left( PO _{4}\right)_{2} \rightleftharpoons 3 Ca ^{2+}+2 PO _{4}^{3-} $
$4 x^{2}(0.1)^{2}=K_{s p} $
$x=\sqrt{\frac{K_{s p}}{4 \times 10^{-3}}}=\sqrt{\frac{10^{-33}}{4 \times 10^{-3}}}$
$=\sqrt{2.5 \times 1^{-31}} $
$P O_{4}^{3-}=5 \times 10^{-16} $
(c) $ M g_{3}\left(P O_{4}\right)_{2} \rightleftharpoons \underset{0.1}{3 M g^{2+}}+ \underset{2x}{2 P O_{4}^{3-}}$
$(2 x)^{2}(0.1)^{2}=10^{-24} $
$4 x^{2}=\frac{10^{-24}}{0.001}=10^{-21}$
$x=\sqrt{2.5 \times 10^{-22}} $
$=5 \times 10^{-11.5}$
(d) $A l P O_{4} \rightleftharpoons \underset{0.1}{A l^{3+}}+ \underset{x}{P O_{4}^{3-}}$
$x=\frac{K_{s p}}{0.1}=\frac{10^{-20}}{0.1}=10^{-19}$