When 100 mL of 0.60 M Na2CO3

(aq) is combined with 500 mL of 0.060 M Ba(NO3)2(aq), a precipitate is observed. What is the equilibrium concentration of Ba2+(aq) at 25°C?Ksp(BaCO3) = 8.1 x 10-9

millimols CO3^= = 100 x 0.6 = 60

millimols Ba^++ = 500 x 0.06 = 30

......Ba^++(aq) + CO3^=(aq) --> BaCO3(s)
I......30.........................
added..............60............
C.....-30.........-30............+30
E.......0..........30.............30

So this results in a common ion problem along with Ksp. You have a solid ppt of BaCO3 and an excess of 30 mmols CO3^=.
Ksp = (Ba^++)(CO3^=)
8.7E-9 = (Ba^++)(30/600).
Solve for Ba^2+.
Note: 30/600 comes from this.
You have 30 millimols CO3^= in 600 mL solution so the (CO3^=) is millimols/mL = 30/600 = ?

To find the equilibrium concentration of Ba2+(aq), we need to consider the solubility product constant, Ksp, for BaCO3.

The equation for the reaction between Ba(NO3)2 and Na2CO3 is:

Ba(NO3)2(aq) + Na2CO3(aq) -> BaCO3(s) + 2NaNO3(aq)

From the balanced equation, we can see that the mole ratio between Ba(NO3)2 and BaCO3 is 1:1. This means that the concentration of Ba2+(aq) is equal to the concentration of BaCO3(s) that precipitates.

First, let's calculate the number of moles of Ba(NO3)2 and Na2CO3:

Moles of Ba(NO3)2 = Volume of Ba(NO3)2(aq) x Molarity of Ba(NO3)2
= 500 mL x 0.060 M
= 30 mmol

Moles of Na2CO3 = Volume of Na2CO3(aq) x Molarity of Na2CO3
= 100 mL x 0.60 M
= 60 mmol

From the balanced equation, we can see that 1 mole of Ba(NO3)2 reacts with 1 mole of Na2CO3 to form 1 mole of BaCO3. So, the maximum number of moles of BaCO3(s) that is formed will be the same as the number of moles of Ba(NO3)2.

Moles of BaCO3 formed = Moles of Ba(NO3)2 = 30 mmol

Now, let's calculate the concentration of Ba2+(aq):

Volume of solution after mixing = Volume of Ba(NO3)2(aq) + Volume of Na2CO3(aq)
= 500 mL + 100 mL
= 600 mL

Concentration of Ba2+(aq) = Moles of Ba2+(aq) / Volume of solution
= Moles of BaCO3 formed / Volume of solution
= 30 mmol / 600 mL
= 0.05 M

Therefore, the equilibrium concentration of Ba2+(aq) at 25°C is 0.05 M.