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" ?

2 Answers
Jun 30, 2018

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


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?


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.

Jul 1, 2018

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