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