What is the molar solubility of $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ in 0.0015 M $C a {(N O_{3})}_{2}$ $Ca(NO_3)_2$ if $K_{s p}$ $K_(sp)$ for $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ is $1.0 \times 10^{- 18}$ $1.0xx10^(-18)$ ?

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What is the molar solubility of $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ in 0.0015 M $C a {(N O_{3})}_{2}$ $Ca(NO_3)_2$ if $K_{s p}$ $K_(sp)$ for $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ is $1.0 \times 10^{- 18}$ $1.0xx10^(-18)$ ?

I'm aware that's probably not the accurate $K_{s p}$ $K_(sp)$ for said compound, but is it possible to work out a similarly structured process with that given "constant" ?

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Answer 1 · 2018-06-30T23:55:04

$s = 3.92 \times 10^{- 13} M$ $s = 3.92 xx 10^(-13) "M"$ .

Using the $K_{s p}$ $K_(sp)$ of calcium phosphate, which is $2.07 \times 10^{- 33}$ $2.07 xx 10^(-33)$ , we first note that calcium nitrate introduces the ${Ca}^{2 +}$ $"Ca"^(2+)$ common ion.

However, since calcium nitrate is a strong electrolyte, AND it contains only one ${Ca}^{2 +}$ $"Ca"^(2+)$ per formula unit, it introduces $100 %$ $100%$ of its concentration as ${Ca}^{2 +}$ $"Ca"^(2+)$ and $200 %$ $200%$ of its concentration as ${NO}_{3}^{-}$ $"NO"_3^(-)$ .

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

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

$I - 0.0015 0$ $"I"" "-" "" "" "" "" "" "0.0015" "" "" "0$
$C - + 3 s + 2 s$ $"C"" "-" "" "" "" "" "+color(red)(3)s" "" "" "+color(red)(2)s$
$E - 0.0015 + 3 s 2 s$ $"E"" "-" "" "" "" "" "0.0015+color(red)(3)s" "" "color(red)(2)s$

where $s$ $s$ 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.

$K_{s p} = 2.07 \times 10^{- 33} = {[{Ca}^{2 +}]}^{3} {[{PO}_{4}^{3 -}]}^{2}$ $K_(sp) = 2.07 xx 10^(-33) = ["Ca"^(2+)]^color(red)(3)["PO"_4^(3-)]^color(red)(2)$

$= {(0.0015 + 3 s)}^{3} {(2 s)}^{2}$ $= (0.0015 + 3color(red)(s))^color(red)(3)(color(red)(2)s)^color(red)(2)$

Now, since $K_{s p}$ $K_(sp)$ is so small (even the one given in the problem), $3 s$ $3s$ $<<$ $"<<"$ $0.0015$ $0.0015$ , so:

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

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

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

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

$s = \sqrt{\frac{2.07 \times 10^{- 33}}{1.35 \times 10^{- 8}}}$ $color(blue)(s) = sqrt((2.07 xx 10^(-33))/(1.35 xx 10^(-8)))$

$= 3.92 \times 10^{- 13} M$ $= color(blue)(3.92 xx 10^(-13) "M")$

What would then be the molar solubility of ${Ca}^{2 +}$ $"Ca"^(2+)$ in terms of $s$ $s$ ? What about ${PO}_{4}^{3 -}$ $"PO"_4^(3-)$ in terms of $s$ $s$ ?

On the other hand, without $Ca {({NO}_{3})}_{2}$ $"Ca"("NO"_3)_2$ in solution,

$K_{s p} = {(3 s)}^{3} {(2 s)}^{2} = 108 s^{5} = 2.07 \times 10^{- 33}$ $K_(sp) = (3s)^3(2s)^2 = 108s^5 = 2.07 xx 10^(-33)$

And this molar solubility in pure water is then:

$s = {(\frac{K_{s p}}{108})}^{1 / 5} = 1.14 \times 10^{- 7} M$ $s = (K_(sp)/108)^(1//5) = 1.14 xx 10^(-7) "M"$

This is about $1000000$ $1000000$ times less soluble in $0.0015 M$ $"0.0015 M"$ calcium nitrate than in pure water, so the calcium common ion suppresses the solubility of calcium phosphate.

That is the common ion effect.

Answer 2 · 2018-07-01T11:35:35

The molar solubility of $C a_{3} (P O_{4}) 2$ $Ca_3(PO_4)2$ is $8.6 \times 10^{- 6}$ $8.6xx10^-6$ M

Explanation:

ICE table:
$| C a_{3} P O_{4} (s) ⇌ 3 C a^{2 +} (a q) + 2 P O_{4}^{3 -} (a q)$ $" " " | " " Ca_3PO_4(s)\rightleftharpoons3Ca^(2+)(aq)+2PO_4^(3-)(aq)$
$I ∣ [solid] 0.0015 0$ $\text(I) " " | " [solid] " " " " " " " 0.0015 " " " " " 0$
$C | - s + 3 s + 2 s$ $\text(C) " | " -s " " " " " " " " " " +3s " " " " +2s$
$E | [solid]-s 0.0015 + 3 s 2 s$ $\text(E) " | [solid]-s " " " 0.0015+3s " " " " 2s$

$K_{s p} = 1.0 \times 10^{- 18} \Rightarrow 108 x^{5} ? ?$ $K_(sp)=1.0xx10^(-18)\rArr\color(red)(108x^5)??$ (I don't know what this red part means, really)
$\Rightarrow {[0.0015 + 3 x]}^{3} {[2 x]}^{2}$ $" " " "\rArr[0.0015+3x]^3[2x]^2$
$≅ {(0.0015)}^{3} \cdot 4 x^{2}$ $" " " "\cong(0.0015)^3*4x^2$
$\thereforex=\sqrt((1.0xx10^(-18))/(0.0015^3*4))\approx8.6xx10^-6$ 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

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What is the molar solubility of $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ in 0.0015 M $C a {(N O_{3})}_{2}$ $Ca(NO_3)_2$ if $K_{s p}$ $K_(sp)$ for $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ is $1.0 \times 10^{- 18}$ $1.0xx10^(-18)$ ?

I'm aware that's probably not the accurate $K_{s p}$ $K_(sp)$ for said compound, but is it possible to work out a similarly structured process with that given "constant" ?

2 Answers

Explanation:

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What is the molar solubility of Ca_3(PO_4)_2Ca3(PO4)2Ca_3(PO_4)_2 in 0.0015 M Ca(NO_3)_2Ca(NO3)2Ca(NO_3)_2 if K_(sp)KspK_(sp) for Ca_3(PO_4)_2Ca3(PO4)2Ca_3(PO_4)_2 is 1.0xx10^(-18)1.0×10−181.0xx10^(-18)?

I'm aware that's probably not the accurate K_(sp)KspK_(sp) for said compound, but is it possible to work out a similarly structured process with that given "constant" ?

2 Answers

Explanation:

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What is the molar solubility of $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ in 0.0015 M $C a {(N O_{3})}_{2}$ $Ca(NO_3)_2$ if $K_{s p}$ $K_(sp)$ for $C a_{3} {(P O_{4})}_{2}$ $Ca_3(PO_4)_2$ is $1.0 \times 10^{- 18}$ $1.0xx10^(-18)$ ?

I'm aware that's probably not the accurate $K_{s p}$ $K_(sp)$ for said compound, but is it possible to work out a similarly structured process with that given "constant" ?