Question

What is the molar solubility of `Ca_3(PO_4)_2` in 0.0015 M `Ca(NO_3)_2` if `K_(sp)` for `Ca_3(PO_4)_2` is `1.0xx10^(-18)`?

I'm aware that's probably not the accurate `K_(sp)` for said compound, but is it possible to work out a similarly structured process with that given "constant" ?

Answer 1

`s = 3.92 xx 10^(-13) "M"`.

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Using the `K_(sp)` of calcium phosphate, which is `2.07 xx 10^(-33)`, we first note that calcium nitrate introduces the `"Ca"^(2+)` common ion.

However, since calcium nitrate is a strong electrolyte, AND it contains only one `"Ca"^(2+)` per formula unit, it introduces `100%` of its concentration as `"Ca"^(2+)` and `200%` of its concentration as `"NO"_3^(-)`.

Hence, that becomes the initial concentration of `"Ca"^(2+)` (whereas the nitrate just sits around), instead of zero:

`"Ca"_3("PO"_4)_2(s) rightleftharpoons color(red)(3)"Ca"^(2+)(aq) + color(red)(2)"PO"_4^(3-)(aq)`

`"I"" "-" "" "" "" "" "" "0.0015" "" "" "0`
`"C"" "-" "" "" "" "" "+color(red)(3)s" "" "" "+color(red)(2)s`
`"E"" "-" "" "" "" "" "0.0015+color(red)(3)s" "" "color(red)(2)s`

where `s` is the molar solubility of calcium phosphate. Each product is forming in water, so they get `+` in the ICE table.

Remember that the coefficients go into the change in concentration, as well as the exponents.

`K_(sp) = 2.07 xx 10^(-33) = ["Ca"^(2+)]^color(red)(3)["PO"_4^(3-)]^color(red)(2)`

`= (0.0015 + 3color(red)(s))^color(red)(3)(color(red)(2)s)^color(red)(2)`

Now, since `K_(sp)` is so small (even the one given in the problem), `3s` `"<<"` `0.0015`, so:

`2.07 xx 10^(-33) = (0.0015)^3(2s)^2`

`= 3.38 xx 10^(-9) cdot 4s^2`

`= 1.35 xx 10^(-8)s^2`

Now you can solve for the molar solubility of calcium phosphate without working with a fifth order polynomial.

`color(blue)(s) = sqrt((2.07 xx 10^(-33))/(1.35 xx 10^(-8)))`

`= color(blue)(3.92 xx 10^(-13) "M")`

What would then be the molar solubility of `"Ca"^(2+)` in terms of `s`? What about `"PO"_4^(3-)` in terms of `s`?

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On the other hand, without `"Ca"("NO"_3)_2` in solution,

`K_(sp) = (3s)^3(2s)^2 = 108s^5 = 2.07 xx 10^(-33)`

And this molar solubility in pure water is then:

`s = (K_(sp)/108)^(1//5) = 1.14 xx 10^(-7) "M"`

This is about `1000000` times less soluble in `"0.0015 M"` calcium nitrate than in pure water, so the calcium common ion suppresses the solubility of calcium phosphate.

That is the common ion effect.

Answer 2

The molar solubility of `Ca_3(PO_4)2` is `8.6xx10^-6` M

Explanation:

ICE table:
`" " " | " " Ca_3PO_4(s)\rightleftharpoons3Ca^(2+)(aq)+2PO_4^(3-)(aq)`
`\text(I) " " | " [solid] " " " " " " " 0.0015 " " " " " 0`
`\text(C) " | " -s " " " " " " " " " " +3s " " " " +2s`
`\text(E) " | [solid]-s " " " 0.0015+3s " " " " 2s`

`K_(sp)=1.0xx10^(-18)\rArr\color(red)(108x^5)??` (I don't know what this red part means, really)
`" " " "\rArr[0.0015+3x]^3[2x]^2`
`" " " "\cong(0.0015)^3*4x^2`
`\thereforex=\sqrt((1.0xx10^(-18))/(0.0015^3*4))\approx8.6xx10^-6` M

And as `s\hArrx=[Ca_3(PO_4)_2]`, so the molar solubility of `Ca_3(PO_4)2` is `8.6xx10^-6` M