Question

Question #62539

Answer 1

The answer is A) only `CaCO_3` and `FeCO_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* 10^(-4)"M"` in carbonate ion.

If the pool originally contained `4* 10^(-3)"M"` of `Mg^(2+)` ions, `6*10^(-4)"M"` of `Ca^(2+)` ions, and `8* 10^(-7)"M"` of `Fe^(2+)` ions, then the precipitate that was formed consisted of ....?

A) only `CaCO_3` and `FeCO_3`
B) `MgCO_3`, `CaCO_3`, and `FeCO_3`
C) only `CaCO_3`
D) only `FeCO_3`
E) only `MgCO_3`

`K_(sp)` values are:

`CaCO_3: 4.7 xx 10^(-9)`
`MgCO_3: 4.0 xx 10^(-5)`
`FeCO_3: 2.0 xx 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 `CaCO_3` and `FeCO_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_(sp)`.

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

`CaCO_(3(s)) rightleftharpoons Ca_((aq))^(2+) + CO_(3(aq))^(2-)`

By definition, the solubility product constant will be

`K_(sp) = [Ca^(2+)] * [CO_3^(2-)]`

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

`[CO_3^(2-)] = 2 * 10^(-4)"M" > 7.83 * 10^(-6)"M" ->` `CaCO_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.

`[CO_3^(2-)] = 2 * 10^(-4) - 7.83 * 10^(-6) = 1.92 * 10^(-4)"M"`

Now take a look at iron (II) carbonate, `FeCO_3`.

`FeCO_(3(s)) rightleftharpoons Fe_((aq))^(2+) + CO_(3(aq))^(2-)`

`K_(sp) = [Fe^(2+)] * [CO_3^(2-)]`

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

`[CO_3^(2-)] = 1.92 * 10^(-4)"M" > 2.5 * 10^(-5)"M"` `->` `FeCO_3` precipitates.

This will leave you with

`[CO_3^(2-)] = 1.92 * 10^(-4) - 2.5 * 10^(-5) = 1.67 * 10^(-4)"M"`

Finally, take a look at magnesium carbonate, `MgCO_3`.

`MgCO_(3(s)) rightleftharpoons Mg_((aq))^(2+) + CO_(3(aq))^(2-)`

`K_(sp) = [Mg^(2+)] * [CO_3^(2-)]`

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.

`[CO_3^(2-)] = 1.67 * 10^(-4)"M" color(red)(<) 1.0 * 10^(-2)"M" -> MgCO_3` does not precipitate

So there you have it. Your solution will contain two precipitates, `CaCO_3` and `FeCO_3`, carbonate anions, `CO_3^(2-)`, and magnesium cations, `Mg^(2+)`.