!! LONG ANSWER !!
So, you have a solution that contains barium carbonate, #BaCO_3#, to which you add calcium carbonate, #CaCO_3#. The concentration of the carbonate ions will be affected by the fact that you have 2 insoluble compounds that release carbonate ions in solution.
More specifically, the saturated barium carbonate solution will influence the molar solubility of the calcium carbonate solution. This happens because of the common ion effect - the carbonate ions released into solution by the barium carbonate will change the solubility of the added calcium carbonate.
So, you know that you're dealing with a 1-L, saturated solution of barium carbonate, an insoluble compound. This means that your solution contains all the dissolved barium carbonate it can possibly support at that temperature.
This means that you can use the solubility product constant, #K_(sp)#, to determine the molar solubility of your compound, i.e. how many moles of barium carbonate actually dissolve in a 1-L solution.
Use an ICE table to help you with that
#" "BaCO_(3(s)) -> Ba_((aq))^(2+) + CO_(3(aq))^(2-)#
I........#-#.....................0...............0
C......#-#.....................(+x)...........(+x)
E.......#-#.......................x...............x
By definition, #K_(sp)# will be
#K_(sp) = [Ba^(2+)] * [CO_3^(2-)] = x * x = x^2#
Therefore,
#6.61 * 10^(-9) = x^2 => x = sqrt(6.61 * 10^(-9)) = 8.1 * 10^(-5)#
This means that concentration of the carbonate ion, #CO_3^(2-)#, is equal to #8.1 * 10^(-5)"M"#.
Now you add enough calcium carbonate to the solution. Once again, use an ICE table to help you with the calculations. Keep in mind that the initial concentration of carbonate ions is #8.1*10^(-5)"M"#.
#" "CaCO_(3(s)) -> Ca_((aq))^(2+) + CO_(3(aq))^(2-)#
I........#-#......................0..............#8.1* 10^(-5)#
C......#-#.....................(+x)................(+x)
E.......#-#.......................x...............#8.1*10^(-5)+x#
The solubility equilibrium constant of the calcium carbonate will be
#K_(sp) = [Ca^(2+)] * [CO_3^(2-)]#
#K_(sp) = x * (8.1 * 10^(-5)+x) = 5.49 * 10^(-10)#
This is equivalent to
#x^2 + 8.1 * 10^(-5)x - 5.49 * 10^(-10) = 0#
Solving this quadratic equation for #x# will produce only one suitable value, which is
#x = 6.29 * 10^(-6)"M"#
The molar solubility of the calcium carbonate decreases because of the presence of the carbonate ions.
The final concentration ofcarbonate ions in solution will be
#[CO_3^(2-)] = 8.1 * 10^(-5) + 6.29 * 10^(-6) = 8.73 * 10^(-5)"M"#
Now, in aquesous solution, the carbonate ion will react with water to form the bicarbonate ion, #HCO_3^(-)#, and hydroxide ions, #OH^(-)#. Once again, use an ICE table to help you with the calculations
#" "CO_(3(aq))^(2-) + H_2O_((l)) rightleftharpoons HCO_(3(aq))^(-) + OH_((aq))^(-)#
I...#8.73 * 10^(-5)#.............................0.....................0
C.......(-x).....................................(+x).................(+x)
E...#8.73*10^(-5)-x#.....................x......................x
Since the carbonate ion results from the second dissociation of carbonic acid, #H_2CO_3#, you're going to have to use #K_(a2)# to determine #K_b#.
#K_b = 10^(-14)/K_(a2) = 10^(-14)/(5.6 * 10^(-11)) = 1.79 * 10^(-4)#
Therefore, you have
#K_b = ([HCO_3^(-)] * [OH^(-)])/([CO_3^(2-)]) = (x * x)/(8.73 * 10^(-5) - x)#
This will eventually get you
#x^2 + 1.79 * 10^(-4)x - 1.56 * 10^(-8) = 0#
The only acceptable solution for this quadratic equation is
#x = 6.41 * 10^(-5)#
The concentration of hydroxide ions will thus be
#[OH^(-)] = x = 6.41 * 10^(-5)"M"#
The pOH of the solution will be
#pOH = - log([OH^(-)]) = -log(6.41 * 10^(-5)) = 4.20#
The pH of the solution will thus be
#pH_"sol" = 14 - pOH = 14 - 4.20 = color(green)(9.80)#
The concentrations of the other compounds will be
#[HCO_3^(-)] = 6.41 * 10^(-5)"M"#
#[CO_3^(2-)] = 2.32 * 10^(-5)"M"#
#[Ba^(2+)] = 8.1 * 10^(-5)"M"#
#[Ca^(2+)] = 8.73 * 10^(-5)"M"#
#[OH^(-)] = 6.41 * 10^(-5)"M"#
