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.