Question

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 \(\mathrm{BaC}_{2} \mathrm{O}_{4}\) \(\left.K_{\mathrm{sp}}=2.3 \times 10^{-\mathrm{s}} .\right)\)

Short answer

The final concentrations of the ions in the solution are: \(\mathrm{K}^{+}(a q): 0.080\ \mathrm{M}\), \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(a q): 0.040\ \mathrm{M}\), \(\mathrm{Ba}^{2+}(a q): 5.75 \times 10^{-8}\ \mathrm{M}\), and \(\mathrm{Br}^{-}(a q): 0.190\ \mathrm{M}\).

Step-by-step solution

Identify the chemical equation for the reaction

The reaction taking place when we mix the solutions can be represented as a balanced equation: \( \mathrm{K}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(a q) + \mathrm{BaBr}_{2}(a q) \rightarrow \mathrm{BaC}_{2}\mathrm{O}_{4\downarrow} + 2\mathrm{KBr}(a q) \) This reaction occurs because barium oxalate \(\mathrm{BaC}_{2}\mathrm{O}_{4}\) forms a solid precipitate, which separates from the solution. Meanwhile, potassium bromide \(\mathrm{KBr}\) remains in the solution.

Calculate moles of reactants

Now, let's determine the initial moles of the reactants: Moles of \(\mathrm{K}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\): \( volume\times concentration= 0.100 \mathrm{~L} \times 0.200 \mathrm{M} = 0.020 \mathrm{~mol} \) Moles of \(\mathrm{BaBr}_{2}\): \( volume\times concentration= 0.150 \mathrm{~L} \times 0.250 \mathrm{M} = 0.0375 \mathrm{~mol} \)

Determine limiting reactant and moles of products

Let's find out which reactant is the limiting reactant by evaluating the stoichiometric ratio: \(\frac{moles\ of \ (\mathrm{K}_{2}\mathrm{C}_{2}\mathrm{O}_{4})}{2} = \frac{0.020\ mol}{2} = 0.010\ mol \) Since \(0.010\ mol < 0.0375\ mol\), \(\mathrm{K}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) is the limiting reactant. We consume all the moles of the limiting reactant, and from this, we can determine the moles of the products formed: Moles of \(\mathrm{BaC}_{2}\mathrm{O}_{4}\) formed: \( 0.010\ \mathrm{mol} \) Moles of \(\mathrm{KBr}\) formed: \( 2 \times 0.010\ \mathrm{mol} = 0.020\ \mathrm{mol} \)

Calculate moles of reactants left and the total volume of the solution

We can now calculate the moles of \(\mathrm{BaBr}_{2}\) left in the solution after the reaction: Moles of \(\mathrm{BaBr}_{2}\) left: \( 0.0375\ \mathrm{mol} - 0.010\ \mathrm{mol} = 0.0275\ \mathrm{mol} \) Next, let's determine the total volume of the mixed solution: Total volume: \(0.100\ \mathrm{L} + 0.150\ \mathrm{L} = 0.250\ \mathrm{L} \)

Calculate final concentrations of ions in the solution

We can now determine the final concentrations of the ions: Final concentration of \(\mathrm{K}^{+}\) ions: \( \frac{0.020\ \mathrm{mol}(\mathrm{KBr})}{0.250\ \mathrm{L}} = 0.080\ \mathrm{M} \) Final concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions: \( \frac{0.010\ \mathrm{mol}(\mathrm{BaC}_{2}\mathrm{O}_{4})}{0.250\ \mathrm{L}} = 0.040\ \mathrm{M} \) Final concentration of \(\mathrm{Ba}^{2+}\) ions: Since BaC2O4 is a precipitate, the concentration of \(\mathrm{Ba}^{2+}\) will depend on the minimum value for which the \(K_{sp}\) equation is not surpassed. Final concentration of \(\mathrm{Br}^{-}\) ions: \( \frac{0.0275\ \mathrm{mol}(\mathrm{BaBr}_{2}) + 0.020\ \mathrm{mol}(\mathrm{KBr})}{0.250\ \mathrm{L}} = 0.190 \mathrm{M} \) Now recall that \(K_{sp} = [Ba^{2+}][C2O4^{2-}] = 2.3 \times 10^{-9}\) Given that the concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-} = 0.040\ \mathrm{M}\), we will substitute it into the equation to find the final concentration of \(\mathrm{Ba}^{2+}\). Final concentration of \(\mathrm{Ba}^{2+}\) ions: \( \frac{2.3 \times 10^{-9}}{0.040} = 5.75 \times 10^{-8}\ \mathrm{M} \) So, the final concentrations of the ions in the solution are: - \(\mathrm{K}^{+}(a q):\ 0.080\ \mathrm{M}\) - \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(a q):\ 0.040\ \mathrm{M}\) - \(\mathrm{Ba}^{2+}(a q):\ 5.75 \times 10^{-8}\ \mathrm{M}\) - \(\mathrm{Br}^{-}(a q):\ 0.190\ \mathrm{M}\)

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. By using the coefficients from a balanced chemical equation, you can determine the proportions of reactants needed and products formed. It's essentially like a recipe for chemistry, ensuring that you have the right amounts of ingredients (reactants) to get your desired outcome (products).

In the exercise, we used stoichiometry to calculate moles of reactants and products. The balanced equation \( \text{K}_2\text{C}_2\text{O}_4(aq) + \text{BaBr}_2(aq) \rightarrow \text{BaC}_2\text{O}_4\text{\textdownarrow} + 2\text{KBr}(aq) \) gave us the ratios needed to compute the amounts involved in the reaction.

Limiting Reactant

In a chemical reaction, the limiting reactant (or limiting reagent) is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant, since the reaction cannot proceed further without it. Identifying the limiting reactant is crucial because it determines the maximum amount of product that can be formed.

Through stoichiometry, we identified that \( \text{K}_2\text{C}_2\text{O}_4 \) was the limiting reactant because it would be entirely used up before \( \text{BaBr}_2 \), affecting the final concentrations of ions in the solution.

Solubility Product Constant

The solubility product constant (\( K_{sp} \)) is a special type of equilibrium constant that measures the solubility of a sparingly soluble ionic compound. It represents the level at which a compound's ions are dissolved in solution and is a fixed value at a given temperature for any solid that dissolves to an extent that its ions are well-separated in solution. The calculation of \( K_{sp} \) involves multiplying the molar concentrations of the ions each raised to the power of its coefficient in the dissolution equation.

In our exercise, \( K_{sp} \) is used to calculate the final concentration of \( \text{Ba}^{2+} \) ions. We set up the equation \( K_{sp} = [\text{Ba}^{2+}][\text{C}_2\text{O}_4^{2-}] \) and rearranged it to solve for \( [\text{Ba}^{2+}] \).

Ion Concentration

Ion concentration refers to the amount of ions of a particular element or compound in solution and is generally expressed as molarity (M), the number of moles of solute per liter of solution. Calculating ion concentrations helps you predict whether a reaction will proceed or a precipitate will form, and it is essential for understanding various aspects of chemical reactions like reactivity, color, and pH.

In the given exercise, after determining the stoichiometry and the limiting reactant, we calculated the ion concentrations of \( \text{K}^{+} \), \( \text{C}_2\text{O}_4^{2-} \), \( \text{Ba}^{2+} \), and \( \text{Br}^{-} \) in the final solution. These calculations included considering the solubility product constant to ensure the concentrations would not exceed the equilibrium limit for the insoluble barium oxalate.