Question

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would precipitate out of solution first?
A. Na3PO4.
B. Ag3PO4.
C. Ca3(PO4)2
When the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Answer 1

Answer:

C. $Ca_3(PO_4)_2$  will precipitate out first

the percentage of $Ca^{2+}$remaining =  12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation  of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$  is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$   is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than  in  $Ag_3PO_4$  $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$  will precipitate out first

To determine the concentration of $[Ca^+]$ when  the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second  cation starts to precipitate ; the  concentration of $[Ca^{2+}]$ in the solution is  0.00566

Therefore;

the percentage of $Ca^{2+}$  remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440  × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$remaining =  12.86%