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a solution contains 0.0440 m ca2+ and 0.0930 m ag+. if solid na3po4 is added to this mixture, which of the phosphate species would precipitate out of solution first? ca3(po4)2 ag3po4 na3po4 when the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Sagot :

qwchocolate

The percentage of the first cation that remained in the solution when the second cation starts to precipitate is 12.86%.

[tex]Ca ^{ + } in \: the \: solution \: = 0.0440[/tex]

[tex]Ag ^{ + } \: in \: the \: solution = 0.0940[/tex]

The balanced reaction for the equation is,

[tex]Ag_{3} PO _{4} →3Ag ^{3 + } + \: PO ^{3 - } _{4}[/tex]

[tex]Solubility \: product \: constant \: of[/tex]

[tex]Ag_{3} PO _{4} = 8.89 \times 10 ^{ - 17} [/tex]

[tex]k _{s} =[Ag ^{ + } ]^{3} \: [PO _{4} ^{3 - } ] [/tex]

[tex]8.89 \times 10 ^{ - 7} = (0.940) ^{3} \: [PO _{4} ^{3 - } ] [/tex]

[tex] \frac{8.89 \times 10 ^{ - 7}} { (0.940) ^{3}} = \: [PO _{4} ^{3 - } ] [/tex]

[tex] [PO _{4} ^{3 - } ] = 1.07 \times 10 ^{ - 13} [/tex]

The balanced equation for the reaction is,

[tex]Ca_{3} (PO _{4}) _{2}→3Ca ^{2 + } + 2PO _{4} ^{3 - } [/tex]

[tex]The \: solubility \: product \: constant \: of [/tex]

[tex]Ca_{3} (PO _{4}) _{2} = 2.07 \times 10 ^{ -32} [/tex]

[tex]k _{s} = [Ca ^{ 2 + } ] ^{ 3} \: (PO _{4}) _{2}[/tex]

[tex]2.07 \times 10 ^{ - 33} = (0.0440)^{3} \: (PO _{4} ^{3 - } )^{2} [/tex]

[tex] [PO _{4} ^{3 - } ] = \frac{2.07 \times 10 ^{ - 33} }{(0.0440)^{3} }[/tex]

[tex](PO _{4} ^{3 - }) = 4.93 \times 10 ^{ - 15} [/tex]

[tex]Ca_{3} (PO _{4}) _{2} \: will \: precipitate \: first.[/tex]

[tex]Concentration \: of \: Ca ^{ + 2} \: when \:[/tex]

the second cation starts to precipitate.

[tex]k _{s} = [Ca ^{ + 2} ] ^{ 3} \: (PO _{4} ^{3 - } ) ^{2} [/tex]

[tex]2.07 \times 10 ^{ - 33} = [Ca ^{ + 2}] ^{3} \: \times 10 ^{ - 33} [1.07 \times 10 ^{ - 13} ] ^{2} [/tex]

[tex] [Ca ^{ + 2}] = 0.00566[/tex]

The percentage of the first cation that remained in the solution when the second cation starts to precipitate is,

[tex]percentage \: \: of \: remaining \: Ca ^{ + 2} = \frac{concentration \: remaining}{initial \: concentration } [/tex]

[tex] = \frac{0.00566}{0.044} \times 100[/tex]

= 12.86 %

Therefore, the percentage of the first cation that remained in the solution when the second cation starts to precipitate is 12.86%.

To know more about precipitation, refer to the below link:

https://brainly.com/question/20925010

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