Question #62539

1 Answer
Jun 22, 2015

Answer:

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+)#.