What is the mass of solid #K_2CO_3# which must be added to 500.0 mL of water in order to make a solution in which the #[K^+]# is #3.0# #M#?

1 Answer
Jan 16, 2016

Answer:

Here's what I got.

Explanation:

Your strategy here will be to use the molarity of the potassium cations, #"K"^(+)#, and the volume of the solution to figure out how many moles of potassium cations must be present in the solution.

SIDE NOTE The volume of the solution will probably increase with this addition of the potassium carbonate, but I think that the problem wants you to assume that it remains constant.

Once you know how many moles of potassium cations you need, you can backtrack to find the number of moles of potassium carbonate, #"K"_2"CO"_3#, that would deliver that many moles of potassium cation to the solution.

As you know, molarity is defined as moles of solute per liters of solution. In your case, you want the target solution to have a molarity of #"3.0 M"# with respect to potassium cations.

In order to get that molarity, #"500.0 mL"# of solution must contain

#color(blue)(c = n/V implies n = c * V)#

#n_(K^(+)) = "3.0 M" * 500.0 * 10^(-3)"L" = "1.50 moles K"^(+)#

Now, potassium bicarbonate is a soluble ionic compound, which means that it will dissociate completely in aqueous solution to form potassium cations, #"K"^(+)#, and carbonate anions, #"CO"_3^(2-)#.

#"K"_2"CO"_text(3(aq]) -> color(red)(2)"K"_text((aq])^(+) + "CO"_text(3(aq])^(2-)#

Notice that since the potassium cations carry a #1+# charge, you need #color(red)(2)# of them to balance the #2-# charge on the carbonate anions.

This tells you that every mole of potassium carbonate will produce #color(red)(2)# moles of potassium cations and #1# mole of carbonate anions in solution.

Since you know how many potassium cations are needed for the target solution, you can use this #1:color(red)(2)# mole ratio to find how many moles of potassium carbonate are needed

#1.5 color(red)(cancel(color(black)("moles K"^(+)))) * ("1 mole K"_2"CO"_3)/(color(red)(2)color(red)(cancel(color(black)("moles K"^(+))))) = "0.750 moles K"_2"CO"_3#

Finally, to get the mass of potassium carbonate that will contain this many moles, use the compound's molar mass

#0.750 color(red)(cancel(color(black)("moles K"_2"CO"_3))) * " 138.2 g"/(1color(red)(cancel(color(black)("mole K"_2"CO"_3)))) = "103.65 g"#

Rounded to three sig figs, the number of sig figs you have for the molarity of the potassium cations, the answer will be

#m_(K_2CO_3) = color(green)("104 g")#