Calculate the final concentrations of $\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q),\( \)\mathrm{Ba}^{2+}(a q),$ and \(\operatorname{Br}^{-}(a q)\) in a solution prepared by adding 0.100 \(\mathrm{L}\) of \(0.200M\) \(\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to 0.150 \(\mathrm{L}\) of \(0.250 M\) \(\mathrm{BaBr}_{2}\) . (For $\mathrm{BaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.3 \times 10^{-8} . )$

Before mixing the solutions, we need to determine the moles of each ion in the individual solutions, using the given concentrations and volumes. We can do this by using the formula: moles = concentration × volume For \(\mathrm{K}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\) solution: - Initial moles of \(\mathrm{K}^{+}\) : \( 2 \times 0.200~\text{M} \times 0.100~\text{L} = 0.0400~\text{moles} \) (multiplied by 2 as there are 2 potassium ions in \(\mathrm{K}_2\mathrm{C}_2\mathrm{O}_4\)). - Initial moles of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) : \( 0.200~\text{M} \times 0.100~\text{L} = 0.0200~\text{moles} \) For \(\mathrm{BaBr}_{2}\) solution: - Initial moles of \(\mathrm{Ba}^{2+}\) : \( 0.250~\text{M} \times 0.150~\text{L} = 0.0375~\text{moles} \) - Initial moles of \(\mathrm{Br}^{-}\) : \( 2 \times 0.250~\text{M} \times 0.150~\text{L} = 0.0750~\text{moles} \) (multiplied by 2 as there are 2 bromide ions in \(\mathrm{BaBr}_2\)).

We will now check if \(\mathrm{BaC}_{2}\mathrm{O}_{4}\) will form a precipitate when the two solutions are mixed. To do this, we calculate the reaction quotient \(Q\) (the initial ion product without a precipitate) and compare it with the solubility product constant \(K_{sp}\). For \(\mathrm{BaC}_{2}\mathrm{O}_{4}\), we have: \(Q = [\mathrm{Ba}^{2+}] [\mathrm{C}_2\mathrm{O}_4^{2-}]\). Since the total volume after mixing is \(0.100~\text{L} + 0.150~\text{L} = 0.250~\text{L},\) we can find the initial concentrations of the ions: - Initial concentration of \(\mathrm{Ba}^{2+}\) : \(\frac{0.0375~\text{moles}}{0.250~\text{L}} = 0.150~\text{M}\) - Initial concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) : \(\frac{0.0200~\text{moles}}{0.250~\text{L}} = 0.0800~\text{M}\) Now, we can calculate the reaction quotient: \(Q = (0.150)(0.0800) = 0.0120\). Since \(Q > K_{sp} (2.3 \times 10^{-8})\), \(\mathrm{BaC}_{2}\mathrm{O}_{4}\) will precipitate.

Knowing that \(\mathrm{BaC}_{2}\mathrm{O}_{4}\) will precipitate, we can calculate the new equilibrium concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) using the \(K_{sp}\) expression: \(K_{sp} = [\mathrm{Ba}^{2+}] [\mathrm{C}_2\mathrm{O}_4^{2-}]\). Since all of \(\mathrm{Ba}^{2+}\) will react, we can simply use \(K_{sp}\) and the initial concentration of \(\mathrm{Ba}^{2+}\) to find the new equilibrium concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\): \[ [\mathrm{C}_2\mathrm{O}_4^{2-}] = \frac{K_{sp}}{[\mathrm{Ba}^{2+}]} = \frac{2.3 \times 10^{-8}}{0.150} = 1.53 \times 10^{-7}~\text{M} \]

Now that the precipitate has formed and the new equilibrium concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) is known, we can determine the final concentrations of all ions present in the solution. Since all of the \(\mathrm{Ba}^{2+}\) ions reacted, the final concentration of \(\mathrm{Ba}^{2+}\) is zero. - Final concentration of \(\mathrm{K}^{+}\) : \(\frac{0.0400~\text{moles}}{0.250~\text{L}} = 0.160~\text{M}\) - Final concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) : \(1.53 \times 10^{-7}~\text{M}\) - Final concentration of \(\mathrm{Ba}^{2+}\) : \(0~\text{M}\) - Final concentration of \(\mathrm{Br}^{-}\) : \(\frac{0.0750~\text{moles}}{0.250~\text{L}} = 0.300~\text{M}\) The final concentrations of the ions present in the solution are: \(\mathrm{K}^{+}(a q) = 0.160~\text{M}\), \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(a q) = 1.53 \times 10^{-7}~\text{M}\), \(\mathrm{Ba}^{2+}(a q) = 0~\text{M}\), and \(\mathrm{Br}^{-}(a q) = 0.300~\text{M}\).