# Calcium phosphate will precipitate out of blood plasma when calcium concentration in blood is 9.2mg/dL, and Ksp for calcium diphosphate is 8.64x10^(-13), what minimum concentration of diphospate results in precipitation?

Nov 1, 2015

$1.6 \cdot {10}^{- 7} \text{M}$

#### Explanation:

I assume that you're actually referring to calcium diphosphate, ${\text{Ca"_2"P"_2"O}}_{7}$, since the question aks for the concentration of the disphosphate ions, ${\text{P"_2"O}}_{7}^{4 -}$.

Moreover, calcium phosphate, "Ca"_3("PO"_4)_2, has a solubility product constant of about $2.0 \cdot {10}^{- 33}$, much, much smaller than what you have there.

So, calcium diphosphate is insoluble in blood plasma, which means that an equilibrium reaction is established between the solid and the dissolved ions

${\text{Ca"_2"P"_2"O"_text(7(s]) rightleftharpoons color(red)(2)"Ca"_text((aq])^(2+) + "P"_2"O}}_{\textrm{7 \left(a q\right]}}^{4 -}$

Notice that $1$ mole of calcium diphosphate will produce $\textcolor{red}{2}$ moles of calcium cations and $1$ mole of diphosphate anions in solution.

To find the concentration of diphosphate anions that will precipitate the calcium diphoshate, write the expression for the solubility product constant, ${K}_{\text{sp}}$

K_"sp" = ["Ca"^(2+)]^color(red)(2) * ["P"_2"O"_7^(4-)]

Rearrange to solve for the cocnentration of diphosphate anions

["P"_2"O"_7^(4-)] = K_text(sp)/(["Ca"^(2+)]^color(red)(2))

You need to convert the concentration of the calcium ations to moles per liter.

Now, since you didn't specify which units you must use for the concentration of the diphosphate anions, I"ll leave the answer expressed in moles per liter as well.

This means that the concentration of the calcium cations will be

9.2(color(red)(cancel(color(black)("mg"))))/(color(red)(cancel(color(black)("dL")))) * (10color(red)(cancel(color(black)("dL"))))/"1 L" * (1color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("mg")))) * "1 mole Ca"^(2+)/(40.078color(red)(cancel(color(black)("g")))) = 2.2955 * 10^(-3)"M"

The concentration of the diphosphate anions will thus be

["P"_2"O"_7^(4-)] = (8.64 * 10^(-13))/((2.2955 * 10^(-3))^2) = 1.64 * 10^(-7)"M"

You need to round this value to two sig figs, the number of sig figs you have for the concentration of calcium cations

$\left[\text{P"_2"O"_7^(4-)] = color(green)(1.6 * 10^(-7)"M}\right)$

As practice, you can convert this value to $\text{mg/dL}$.