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#?
Here's what I got.
Your strategy here will be to use the molarity of the potassium cations,
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,
In order to get that molarity,
#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"_2"CO"_text(3(aq]) -> color(red)(2)"K"_text((aq])^(+) + "CO"_text(3(aq])^(2-)#
Notice that since the potassium cations carry a
This tells you that every mole of potassium carbonate will produce
Since you know how many potassium cations are needed for the target solution, you can use this
#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")#