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

We will find the moles of solute in each solution by multiplying their molarity with the volume of the solution: Moles of \(\mathrm{K}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\): \(0.200 \mathrm{M} \times 0.100 \mathrm{~L} = 0.020 \mathrm{mol}\) Moles of \(\mathrm{BaBr}_{2}\): \(0.250 \mathrm{M} \times 0.150 \mathrm{~L} = 0.0375 \mathrm{mol}\)

Next, we will find the initial moles of each ion in the combined solution: Moles of \(\mathrm{K}^{+} = 2 \times\) moles of \(\mathrm{K}_{2}\mathrm{C}_{2}\mathrm{O}_{4} = 2 \times 0.020 \mathrm{mol} = 0.040 \mathrm{mol}\) Moles of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-} = 0.020 \mathrm{mol}\) Moles of \(\mathrm{Ba}^{2+} = 0.0375 \mathrm{mol}\) Moles of \(\mathrm{Br}^{-} = 2 \times\) moles of \(\mathrm{BaBr}_{2} = 2 \times 0.0375 \mathrm{mol} = 0.075 \mathrm{mol}\)

The final volume of the combined solution will be the sum of individual solution volumes: Total volume \(= 0.100 \mathrm{~L} + 0.150 \mathrm{~L} = 0.250 \mathrm{~L}\)

Using the total volume and moles calculated above, we can find the initial concentration of each ion: \([\mathrm{K}^{+}] = \frac{0.040 \mathrm{mol}}{0.250 \mathrm{~L}} = 0.160 \mathrm{M}\) \([\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}] = \frac{0.020 \mathrm{mol}}{0.250 \mathrm{~L}} = 0.080 \mathrm{M}\) \([\mathrm{Ba}^{2+}] = \frac{0.0375 \mathrm{mol}}{0.250 \mathrm{~L}} = 0.150 \mathrm{M}\) \([\mathrm{Br}^{-}] = \frac{0.075 \mathrm{mol}}{0.250 \mathrm{~L}} = 0.300 \mathrm{M}\)

Since we are given the value of \(K_{sp}\) for \(\mathrm{BaC}_{2}\mathrm{O}_{4}\) (\(K_{sp} = 2.3 \times 10^{-s}\)), we can use this equation: \(K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}]^2\) Now, we substitute the given value of \(K_{sp}\) and the initial concentrations we found in the previous steps: \(2.3 \times 10^{-s} = (0.150 - x)(0.080 - x)^2\) The above equation can be solved to find the value of \(x\). Although this equation may appear to be a quadratic, the value of Ksp is so small compared to the initial concentrations that the change in concentration (\(x\)) will be negligible. Therefore, the final concentrations, taking into account the solubility equilibrium, will be approximately equal to the initial concentrations: Final concentrations : \(\mathrm{K}^{+}(a q) \approx 0.160 \mathrm{M}\) \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}(a q) \approx 0.080 \mathrm{M}\) \(\mathrm{Ba}^{2+}(a q) \approx 0.150 \mathrm{M}\) \(\mathrm{Br}^{-}(a q) \approx 0.300 \mathrm{M}\)