A solution is 0.1 M with respect to Ag +, Ca 2+, Mg 2+ and Al 3+, which will precipitate at lowest concentration of [ PO 43-] when solution of Na 3 PO 4 is added?

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 [ PO 43-] when solution of Na 3 PO 4 is added? (A) Ag 3 PO 3(Ks p=1 × 10-6) (B) Ca 3( PO 4)2(Ks p=1 × 10-33) (C) M g3(P O4)2(Ks p=1 × 10-24) (D) AlPO4(Ks p=1 × 10-20)

Tardigrade Admin · Accepted Answer

(a) Ag 3 PO 4 leftharpoons 3 Ag ++ PO 43- [ PO 43-]=(Ks p/(0.1)3)=(1 × 10-6/10-3)=10-3 (b) Ca 3( PO 4)2 leftharpoons 3 Ca 2++2 PO 43- 4 x2(0.1)2=Ks p x=√(Ks p/4 × 10-3)=√(10-33/4 × 10-3) =√2.5 × 1-31 P O43-=5 × 10-16 (c) M g3(P O4)2 leftharpoons underset0.13 M g2++ underset2x2 P O43- (2 x)2(0.1)2=10-24 4 x2=(10-24/0.001)=10-21 x=√2.5 × 10-22 =5 × 10-11.5 (d) A l P O4 leftharpoons underset0.1A l3++ undersetxP O43- x=(Ks p/0.1)=(10-20/0.1)=10-19

Q. 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 $[P O_{4}^{3 -}]$ when solution of $N a_{3} P O_{4}$ is added?

$A g_{3} P O_{3} (K_{s p} = 1 \times 1 0^{- 6})$

$C a_{3} (P O_{4})_{2} (K_{s p} = 1 \times 1 0^{- 33})$

$M g_{3} (P O_{4})_{2} (K_{s p} = 1 \times 1 0^{- 24})$

$A lP O_{4} (K_{s p} = 1 \times 1 0^{- 20})$

Q. A solution is 0.1M with respect to Ag+,Ca2+,Mg2+ and Al3+, which will precipitate at lowest concentration of [PO43−​] when solution of Na3​PO4​ is added?

Ag3​PO3​(Ksp​=1×10−6)

Ca3​(PO4​)2​(Ksp​=1×10−33)

Mg3​(PO4​)2​(Ksp​=1×10−24)

AlPO4​(Ksp​=1×10−20)

Q. 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 $[P O_{4}^{3 -}]$ when solution of $N a_{3} P O_{4}$ is added?

$A g_{3} P O_{3} (K_{s p} = 1 \times 1 0^{- 6})$

$C a_{3} (P O_{4})_{2} (K_{s p} = 1 \times 1 0^{- 33})$

$M g_{3} (P O_{4})_{2} (K_{s p} = 1 \times 1 0^{- 24})$

$A lP O_{4} (K_{s p} = 1 \times 1 0^{- 20})$