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" ?
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
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 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
#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
That is the common ion effect.
The molar solubility of
Explanation:
ICE table:
And as