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 = ?
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
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