Problem 60 - Chemistry: The Central Science

Question

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\) \(\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}{\underline{\phantom{xx}}}^{2-}\) because the solution will be quite basic.)

Step-by-Step Solution

Answer

To just dissolve 0.020 mol of NiC2O4 in 1.0 L of solution, the final concentration of NH3 must be adjusted to 2.293 M.

Step 1: Write the Chemical Equations

The dissolution of NiC2O4 in the presence of NH3 can be represented as follows: \[ NiC_{2}O_{4(s)} + 4NH_{3(aq)} \rightleftharpoons Ni(NH_{3})_{4}^{2+} + C_{2}O_{4}^{2-} \] The dissociation of Ni(NH3)4^2+ insolution can be given as: \[Ni(NH_3)_4^{2+} \rightleftharpoons Ni^{2+} + 4 NH_3\]

Step 2: Relate the Concentrations

Let x be the final concentration of NH3 needed to just dissolve 0.020 mol of NiC2O4. According to the stoichiometry of the reactions above, the increase in the concentration of Ni(NH3)4^2+ and C2O4^2- in the solution can be represented as: \[ [Ni(NH_3)_4^{2+}] = [C_{2}O_{4}^{2-}] = \frac{0.020 mol}{1 L} = 0.020 M \] This implies that: \[ [NH_3] = x - 4 * [Ni(NH_3)_4^{2+}] = x - 4 * 0.020 M = x - 0.080 M \]

Step 3: Set Up an Expression for the Ksp and Solve for the Final Concentration of NH3

Using the solubility-product constant (Ksp) of NiC2O4: \[ K_{sp} = [Ni^{2+}][C_{2}O_{4}^{2-}] = (4 \times 10^{-10}) \] And since Ni(NH3)4^2+ is in equilibrium with Ni^2+: \[ [Ni^{2+}] = \frac{[Ni(NH_3)_4^{2+}]}{[NH_3]^4} \] Combining the two equations above, we get: \[ 4 \times 10^{-10} = \frac{0.020}{(x - 0.080)^4} \] Now we can solve for the final concentration of NH3: \[x - 0.080 = \sqrt[4]{\frac{0.020}{4\times10^{-10}}}\] \[x = 0.080 + \sqrt[4]{\frac{0.020}{4\times10^{-10}}}\] Compute the value for x: \[ x = 0.080 + \sqrt[4]{\frac{0.020}{4\times10^{-10}}} = 0.080 + 2.213 \] Thus, the final concentration of NH3: \[ x = 2.293 \, M \] The solution needs to be adjusted to a final concentration of 2.293 M NH3 to just dissolve 0.020 mol of NiC2O4 in 1.0 L of solution.

Key Concepts

Complex Ion Formation

Complex ion formation is an essential concept when discussing solubility and dissolution reactions. In the context of a solution, a complex ion forms when metal ions in solution bind with molecules or ions, creating a more stable structure. Nickel(II) ion ( i^{2+}), for example, can bind with multiple ammonia (NH₃) molecules, resulting in the formation of the complex ion Ni(NH₃)₄²⁺. This reaction can be represented as follows:

\[ Ni^{2+} + 4 NH_3 \rightleftharpoons Ni(NH_3)_4^{2+} \]

Complex ions significantly influence the solubility of various compounds by stabilizing metal ions, thus reducing their free concentration in solution. This stabilization process can enhance the solubility of otherwise insoluble compounds, like NiC₂O₄, in the presence of complex-forming agents like NH₃. Understanding this concept is crucial for predicting how different compounds will behave when mixed in solution.

Equilibrium Calculations

Equilibrium calculations involve determining the concentrations of each species in a reaction at a state of balance. This balance point occurs when the rate of the forward reaction equals the rate of the reverse reaction. For the dissolution of NiC₂O₄ in the presence of NH₃, two equilibria are relevant:

• Formation of the complex ion: \[ NiC_{2}O_{4(s)} + 4NH_{3(aq)} \rightleftharpoons Ni(NH_3)_4^{2+} + C_{2}O_{4}^{2-} \]
• Dissociation in solution: \[ Ni(NH_3)_4^{2+} \rightleftharpoons Ni^{2+} + 4 NH_3 \]

To find the concentration of NH₃ needed to dissolve a certain amount of NiC₂O₄, we use the solubility product constant, Ksp. During equilibrium calculations, the Ksp provides a relationship between the concentrations of products at equilibrium. When all components are correctly inserted into the equilibrium expression, solving for the unknown can predict concentrations necessary for desired solubility.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. It helps us utilize balanced chemical equations to calculate the amounts of reactants required or products formed. In the problem of dissolving NiC₂O₄ using NH₃, stoichiometry aids in understanding how many molecules of NH₃ are necessary per Ni²⁺ to form the complex ion.

The balanced reaction shows that one mole of Ni²⁺ reacts with four moles of NH₃ to form Ni(NH₃)₄²⁺. Consequently, dissolving 0.020 mol of NiC₂O₄ requires 0.080 mol of NH₃ (i.e., 4 times 0.020 mol).

By knowing the stoichiometric relationships, you can adjust concentrations and ensure that the reaction proceeds as planned, enabling predictable dissolution outcomes. Mastering stoichiometry is vital for executing equilibrium calculations and predicting complex ion formation in chemistry.