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

Answer:

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