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Chemistry Question on Equilibrium

A solution is 0.1M0.1 M with respect to Ag+,Ca2+,Mg2+Ag ^{+}, Ca ^{2+}, Mg ^{2+} and Al3+Al ^{3+}, which will precipitate at lowest concentration of [PO43−]\left[ PO _{4}^{3-}\right] when solution of Na3PO4Na _{3} PO _{4} is added?

A

Ag3PO3(Ksp=1×10−6)Ag _{3} PO _{3}\left(K_{s p}=1 \times 10^{-6}\right)

B

Ca3(PO4)2(Ksp=1×10−33)Ca _{3}\left( PO _{4}\right)_{2}\left(K_{s p}=1 \times 10^{-33}\right)

C

Mg3(PO4)2(Ksp=1×10−24)M g_{3}\left(P O_{4}\right)_{2}\left(K_{s p}=1 \times 10^{-24}\right)

D

AlPO4(Ksp=1×10−20)AlPO_{4}\left(K_{s p}=1 \times 10^{-20}\right)

Answer

AlPO4(Ksp=1×10−20)AlPO_{4}\left(K_{s p}=1 \times 10^{-20}\right)

Explanation

Solution

(a) Ag3PO4⇌3Ag++PO43−Ag _{3} PO _{4} \rightleftharpoons 3 Ag ^{+}+ PO _{4}^{3-}
[PO43−]=Ksp(0.1)3=1×10−610−3=10−3{\left[ PO _{4}^{3-}\right]=\frac{K_{s p}}{(0.1)^{3}}=\frac{1 \times 10^{-6}}{10^{-3}}=10^{-3}}
(b) Ca3(PO4)2⇌3Ca2++2PO43−Ca _{3}\left( PO _{4}\right)_{2} \rightleftharpoons 3 Ca ^{2+}+2 PO _{4}^{3-}
4x2(0.1)2=Ksp4 x^{2}(0.1)^{2}=K_{s p}
x=Ksp4×10−3=10−334×10−3x=\sqrt{\frac{K_{s p}}{4 \times 10^{-3}}}=\sqrt{\frac{10^{-33}}{4 \times 10^{-3}}}
=2.5×1−31=\sqrt{2.5 \times 1^{-31}}
PO43−=5×10−16P O_{4}^{3-}=5 \times 10^{-16}
(c) Mg3(PO4)2⇌3Mg2+0.1+2PO43−2x M g_{3}\left(P O_{4}\right)_{2} \rightleftharpoons \underset{0.1}{3 M g^{2+}}+ \underset{2x}{2 P O_{4}^{3-}}
(2x)2(0.1)2=10−24(2 x)^{2}(0.1)^{2}=10^{-24}
4x2=10−240.001=10−214 x^{2}=\frac{10^{-24}}{0.001}=10^{-21}
x=2.5×10−22x=\sqrt{2.5 \times 10^{-22}}
=5×10−11.5=5 \times 10^{-11.5}
(d) AlPO4⇌Al3+0.1+PO43−xA l P O_{4} \rightleftharpoons \underset{0.1}{A l^{3+}}+ \underset{x}{P O_{4}^{3-}}
x=Ksp0.1=10−200.1=10−19x=\frac{K_{s p}}{0.1}=\frac{10^{-20}}{0.1}=10^{-19}