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%.
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