# Question 62539

Jun 22, 2015

The answer is A) only $C a C {O}_{3}$ and $F e C {O}_{3}$.

#### Explanation:

FULL QUESTION

A swimming pool was sufficiently alkaline so that the carbon dioxide absorbed from the air produced a solution
in the pool that was
$2 \cdot {10}^{- 4} \text{M}$ in carbonate ion.

If the pool originally contained $4 \cdot {10}^{- 3} \text{M}$ of $M {g}^{2 +}$ ions, $6 \cdot {10}^{- 4} \text{M}$ of $C {a}^{2 +}$ ions, and $8 \cdot {10}^{- 7} \text{M}$ of $F {e}^{2 +}$ ions, then the precipitate that was formed consisted of ....?

A) only $C a C {O}_{3}$ and $F e C {O}_{3}$
B) $M g C {O}_{3}$, $C a C {O}_{3}$, and $F e C {O}_{3}$
C) only $C a C {O}_{3}$
D) only $F e C {O}_{3}$
E) only $M g C {O}_{3}$

${K}_{s p}$ values are:

$C a C {O}_{3} : 4.7 \times {10}^{- 9}$
$M g C {O}_{3} : 4.0 \times {10}^{- 5}$
$F e C {O}_{3} : 2.0 \times {10}^{- 11}$

RIght from the get-go, you can predict which precipitate will form by inspecting the concentrations of the ions and the solubility product constants.

In your case, the only plausible candidates are $C a C {O}_{3}$ and $F e C {O}_{3}$. Here's why that is.

In order for a precipitate to form, you need the product of the concentrations of the ions that form that precipitate to exceed the value of the solubility product constant, ${K}_{s p}$.

So, for instance, in the case of calcium carbonate, you'd get

$C a C {O}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s C {a}_{\left(a q\right)}^{2 +} + C {O}_{3 \left(a q\right)}^{2 -}$

By definition, the solubility product constant will be

${K}_{s p} = \left[C {a}^{2 +}\right] \cdot \left[C {O}_{3}^{2 -}\right]$

You can use the initial concentration of calcium cations to determine the minimum concentration of carbonate ions that will allow for calcium carbonate to precipitate out of solution.

[CO_(3"min")^(2-)] = K_(sp)/([Ca^(2+)]) = (4.7 * 10^(-9))/(6 * 10^(4)) = 7.83 * 10^(-5)"M"

This is the minimum amount of carbonate ions that are needed in order to precipitate the existing calcium cations. Since you have

$\left[C {O}_{3}^{2 -}\right] = 2 \cdot {10}^{- 4} \text{M" > 7.83 * 10^(-6)"M} \to$ $C a C {O}_{3}$ precipitates.

After the calcium carbonate precipitates, the concentration of carbonate ions decreases. This happens because of the carbonate ions that are now part of the precipitate.

$\left[C {O}_{3}^{2 -}\right] = 2 \cdot {10}^{- 4} - 7.83 \cdot {10}^{- 6} = 1.92 \cdot {10}^{- 4} \text{M}$

Now take a look at iron (II) carbonate, $F e C {O}_{3}$.

$F e C {O}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s F {e}_{\left(a q\right)}^{2 +} + C {O}_{3 \left(a q\right)}^{2 -}$

${K}_{s p} = \left[F {e}^{2 +}\right] \cdot \left[C {O}_{3}^{2 -}\right]$

Once again, the minimum concentration of carbonate ions needed to precipitate iron (II) carbonate from solution is

[CO_(3"min")^(2-)] = K_(sp)/([Fe^(2+)]) = (2.0 * 10^(-11))/(8 * 10^(-7)) = 2.5 * 10^(-5)"M"

Looks like you have more carbonate ions present in solution than you'd need

$\left[C {O}_{3}^{2 -}\right] = 1.92 \cdot {10}^{- 4} \text{M" > 2.5 * 10^(-5)"M}$ $\to$ $F e C {O}_{3}$ precipitates.

This will leave you with

$\left[C {O}_{3}^{2 -}\right] = 1.92 \cdot {10}^{- 4} - 2.5 \cdot {10}^{- 5} = 1.67 \cdot {10}^{- 4} \text{M}$

Finally, take a look at magnesium carbonate, $M g C {O}_{3}$.

$M g C {O}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s M {g}_{\left(a q\right)}^{2 +} + C {O}_{3 \left(a q\right)}^{2 -}$

${K}_{s p} = \left[M {g}^{2 +}\right] \cdot \left[C {O}_{3}^{2 -}\right]$

To precipitate magnesium carbonate, you'd need

[CO_(3"min")^(2-)] = K_(sp)/([Mg^(2+)]) = (4.0 * 10^(-5))/(4 * 10^(-3)) = 1.0 * 10^(-2)"M"#

This time, the amount of carbonate ions present in solution is not enough to precipitate magnesium carbonate.

$\left[C {O}_{3}^{2 -}\right] = 1.67 \cdot {10}^{- 4} \text{M" color(red)(<) 1.0 * 10^(-2)"M} \to M g C {O}_{3}$ does not precipitate

So there you have it. Your solution will contain two precipitates, $C a C {O}_{3}$ and $F e C {O}_{3}$, carbonate anions, $C {O}_{3}^{2 -}$, and magnesium cations, $M {g}^{2 +}$.