# 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)#?

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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" ?

I'm aware that's probably not the accurate

##### 2 Answers

Using the

However, since calcium nitrate is a **strong electrolyte**, AND it contains **only one** **per** formula unit, it introduces

Hence, that becomes the **initial concentration** of

#"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 themolar solubilityof 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

#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

On the other hand, without

#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 **suppresses** the solubility of calcium phosphate.

That is the **common ion effect**.

#### Answer:

The molar solubility of

#### Explanation:

ICE table:

And as