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 ${K}_{s p}$ for said compound, but is it possible to work out a similarly structured process with that given "constant" ?

Jun 30, 2018

$s = 3.92 \times {10}^{- 13} \text{M}$.

Using the ${K}_{s p}$ of calcium phosphate, which is $2.07 \times {10}^{- 33}$, we first note that calcium nitrate introduces the ${\text{Ca}}^{2 +}$ common ion.

However, since calcium nitrate is a strong electrolyte, AND it contains only one ${\text{Ca}}^{2 +}$ per formula unit, it introduces 100% of its concentration as ${\text{Ca}}^{2 +}$ and 200% of its concentration as ${\text{NO}}_{3}^{-}$.

Hence, that becomes the initial concentration of ${\text{Ca}}^{2 +}$ (whereas the nitrate just sits around), instead of zero:

${\text{Ca"_3("PO"_4)_2(s) rightleftharpoons color(red)(3)"Ca"^(2+)(aq) + color(red)(2)"PO}}_{4}^{3 -} \left(a q\right)$

$\text{I"" "-" "" "" "" "" "" "0.0015" "" "" } 0$
$\text{C"" "-" "" "" "" "" "+color(red)(3)s" "" "" } + \textcolor{red}{2} s$
$\text{E"" "-" "" "" "" "" "0.0015+color(red)(3)s" "" } \textcolor{red}{2} s$

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.

${K}_{s p} = 2.07 \times {10}^{- 33} = {\left[{\text{Ca"^(2+)]^color(red)(3)["PO}}_{4}^{3 -}\right]}^{\textcolor{red}{2}}$

$= {\left(0.0015 + 3 \textcolor{red}{s}\right)}^{\textcolor{red}{3}} {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}}$

Now, since ${K}_{s p}$ is so small (even the one given in the problem), $3 s$ $\text{<<}$ $0.0015$, so:

$2.07 \times {10}^{- 33} = {\left(0.0015\right)}^{3} {\left(2 s\right)}^{2}$

$= 3.38 \times {10}^{- 9} \cdot 4 {s}^{2}$

$= 1.35 \times {10}^{- 8} {s}^{2}$

Now you can solve for the molar solubility of calcium phosphate without working with a fifth order polynomial.

$\textcolor{b l u e}{s} = \sqrt{\frac{2.07 \times {10}^{- 33}}{1.35 \times {10}^{- 8}}}$

$= \textcolor{b l u e}{3.92 \times {10}^{- 13} \text{M}}$

What would then be the molar solubility of ${\text{Ca}}^{2 +}$ in terms of $s$? What about ${\text{PO}}_{4}^{3 -}$ in terms of $s$?

On the other hand, without "Ca"("NO"_3)_2 in solution,

${K}_{s p} = {\left(3 s\right)}^{3} {\left(2 s\right)}^{2} = 108 {s}^{5} = 2.07 \times {10}^{- 33}$

And this molar solubility in pure water is then:

$s = {\left({K}_{s p} / 108\right)}^{1 / 5} = 1.14 \times {10}^{- 7} \text{M}$

This is about $1000000$ times less soluble in $\text{0.0015 M}$ 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 $C {a}_{3} \left(P {O}_{4}\right) 2$ is $8.6 \times {10}^{-} 6$ M

Explanation:

ICE table:
$\text{ " " | " } C {a}_{3} P {O}_{4} \left(s\right) \setminus r i g h t \le f t h a r p \infty n s 3 C {a}^{2 +} \left(a q\right) + 2 P {O}_{4}^{3 -} \left(a q\right)$
$\setminus \textrm{I} \text{ " | " [solid] " " " " " " " 0.0015 " " " " } 0$
$\setminus \textrm{C} \text{ | " -s " " " " " " " " " " +3s " " " } + 2 s$
$\setminus \textrm{E} \text{ | [solid]-s " " " 0.0015+3s " " " } 2 s$

K_(sp)=1.0xx10^(-18)\rArr\color(red)(108x^5)?? (I don't know what this red part means, really)
$\text{ " " } \setminus \Rightarrow {\left[0.0015 + 3 x\right]}^{3} {\left[2 x\right]}^{2}$
$\text{ " " } \setminus \cong {\left(0.0015\right)}^{3} \cdot 4 {x}^{2}$
$\setminus \therefore x = \setminus \sqrt{\frac{1.0 \times {10}^{- 18}}{{0.0015}^{3} \cdot 4}} \setminus \approx 8.6 \times {10}^{-} 6$ M

And as $s \setminus \Leftrightarrow x = \left[C {a}_{3} {\left(P {O}_{4}\right)}_{2}\right]$, so the molar solubility of $C {a}_{3} \left(P {O}_{4}\right) 2$ is $8.6 \times {10}^{-} 6$ M