/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 Calculate the equilibrium concen... [FREE SOLUTION] | 91影视

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Calculate the equilibrium concentrations of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+},\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{mL}\) of \(3.00M\) \(\mathrm{NH}_{3}\) with \(500.0 \mathrm{mL}\) of \(2.00 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) . The step wise equilibria are $$\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q) \quad K_{1}=1.86 \times 10^{4}$$ $$\mathrm{CuNH}_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q) \quad K_{2}=3.88 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{2}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{3}^{2+}(a q) \quad K_{3}=1.00 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{4}=1.55 \times 10^{2}$$

Short Answer

Expert verified
The final equilibrium concentrations of each species are: \(\mathrm{NH}_{3}: 1.50-x-y-z-w\) \(\mathrm{Cu}^{2+}: 1.00\times10^{-3} - x\) \(\mathrm{CuNH}_{3}^{2+}: x - y\) \(\mathrm{Cu(NH鈧)鈧倉^{2+}: y - z\) \(\mathrm{Cu(NH鈧)鈧儅^{2+}: z - w\) \(\mathrm{Cu(NH鈧)鈧剗^{2+}: w\) To find the values of x, y, z, and w, you can use numerical methods with a calculator, computer software, or even an online solver for nonlinear systems of equations.

Step by step solution

01

Calculate initial concentrations

First, we need to find the initial concentrations of NH鈧 and Cu虏鈦 ions when the solutions are mixed. The volume of the mixed solution is 1000.0 mL (500.0 mL of each solution). To find the initial concentration, use the formula: Initial Concentration = (Moles of solute before mixing) / (Total volume of solution after mixing) For NH鈧: \( \mathrm{Initial Concentrations}=\dfrac{3.00\; \mathrm{M} \times 500.0 \; \mathrm{mL}}{1000.0 \;\mathrm{mL}} = 1.50\; \mathrm{M}\) For Cu虏鈦: \( \mathrm{Initial\; Concentrations}=\dfrac{2.00 \times 10^{-3}\; \mathrm{M} \times 500.0\; \mathrm{mL}}{1000.0\; \mathrm{mL}} = 1.00 \times 10^{-3}\;\mathrm{M}\) Now that we have initial concentrations, we can start working on solving the stepwise equilibria.
02

First equation for Cu(NH鈧)虏鈦

We'll let x be the change in concentration of Cu虏鈦 and NH鈧 after the first reaction, so the final concentrations are: For Cu虏鈦: \(1.00\times10^{-3} - x\) For NH鈧: \(1.50 - x\) For Cu(NH鈧)虏鈦: \(x\) Then using the first equilibrium constant: \(K_1 = \dfrac{[Cu(NH鈧)虏鈦篯}{[Cu虏鈦篯[NH鈧僝} \Rightarrow 1.86 \times 10^{4} = \dfrac{x}{(1.00\times 10^{-3} - x)(1.50 - x)}\)
03

Second equation for Cu(NH鈧)鈧偮测伜

Let y be the change in concentration of Cu(NH鈧)虏鈦 and NH鈧 after the second reaction, so the final concentrations are: For NH鈧: \(1.50-x-y\) For Cu(NH鈧)虏鈦: \(x-y\) For Cu(NH鈧)鈧偮测伜: \(y\) Then using the second equilibrium constant: \(K_2 = \dfrac{[Cu(NH鈧)_2^{2+}]}{[Cu(NH鈧)^{2+}][NH鈧僝} \Rightarrow 3.88 \times 10^{3} = \dfrac{y}{(x-y)(1.50-x-y)}\)
04

Third equation for Cu(NH鈧)鈧兟测伜

Let z be the change in concentration of Cu(NH鈧)鈧偮测伜 and NH鈧 after the third reaction, so the final concentrations are: For NH鈧: \(1.50-x-y-z\) For Cu(NH鈧)鈧偮测伜: \(y-z\) For Cu(NH鈧)鈧兟测伜: \(z\) Then using the third equilibrium constant: \(K_3 = \dfrac{[Cu(NH鈧)_3^{2+}]}{[Cu(NH鈧)_2^{2+}][NH鈧僝} \Rightarrow 1.00 \times 10^{3} = \dfrac{z}{(y-z)(1.50-x-y-z)}\)
05

Fourth equation for Cu(NH鈧)鈧劼测伜

Let w be the change in concentration of Cu(NH鈧)鈧兟测伜 and NH鈧 after the fourth reaction, so the final concentrations are: For NH鈧: \(1.50-x-y-z-w\) For Cu(NH鈧)鈧兟测伜: \(z-w\) For Cu(NH鈧)鈧劼测伜: \(w\) Then using the fourth equilibrium constant: \(K_4 = \dfrac{[Cu(NH鈧)_4^{2+}]}{[Cu(NH鈧)_3^{2+}][NH鈧僝} \Rightarrow 1.55 \times 10^{2} = \dfrac{w}{(z-w)(1.50-x-y-z-w)}\)
06

Solve the system of equations

We've got a system of four nonlinear equations in four variables (x, y, z, and w). To solve this, we can use numerical methods, such as the Newton-Raphson method or other iterative methods. This process is beyond the scope of high school solutions, but you can use a calculator, computer software like MATLAB or WolframAlpha, or even an online solver for nonlinear systems of equations. Once you have the values for x, y, z, and w, you can find the equilibrium concentrations of all species by substituting back into the concentration expressions in each step. The final equilibrium concentrations of each species are: \(\mathrm{NH}_{3}: 1.50-x-y-z-w\) \(\mathrm{Cu}^{2+}: 1.00\times10^{-3} - x\) \(\mathrm{CuNH}_{3}^{2+}: x - y\) \(\mathrm{Cu(NH鈧)鈧倉^{2+}: y - z\) \(\mathrm{Cu(NH鈧)鈧儅^{2+}: z - w\) \(\mathrm{Cu(NH鈧)鈧剗^{2+}: w\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Concentrations
In chemical reactions, especially those involving complex ions, it's important to determine the equilibrium concentrations of different species. At equilibrium, the concentration of reactants and products will remain constant over time. When you mix solutions, as in the exercise with ammonia (\( \mathrm{NH}_{3} \)) and copper(II) ions (\( \mathrm{Cu}^{2+} \)), the concentrations are not uniform. You calculate them based on the reaction container's total volume after mixing.For instance, if you mix two 500 mL solutions of varying concentrations, the combined solution volume is 1000 mL. The initial concentration of each reactant can then be calculated using the formula:\[ \text{Initial Concentration} = \frac{\text{Moles of solute before mixing}}{\text{Total volume of solution after mixing}} \]The equilibrium concentrations must take into account how much reactant and product is present at the end of the reaction. This involves solving system equations based on equilibrium constants from the reaction steps. In the exercise, after solving these equations, the equilibrium concentrations are derived for multiple copper-ammonia complexes, indicating how the reaction progresses with the given conditions.
Complex Ion Formation
Complex ion formation plays a crucial role in reactions involving metal ions and ligands, like copper(II) ions (\( \mathrm{Cu}^{2+} \)) reacting with ammonia (\( \mathrm{NH}_{3} \)). A complex ion is formed when a metal ion binds with one or more ligands. Ligands are molecules that can donate electron pairs to the metal ion.
  • The binding of copper with ammonia forms several complexes: \( \mathrm{CuNH}_{3}^{2+} \), \( \mathrm{Cu(NH_3)_2^{2+}} \), \( \mathrm{Cu(NH_3)_3^{2+}} \), and \( \mathrm{Cu(NH_3)_4^{2+}} \).
  • This sequential process involves the addition of ammonia molecules to the copper ion, progressively creating more complex ions. The ligand donates a pair of electrons to the empty orbitals of the metal, stabilizing the inner sphere.
  • Complex ion formation can significantly modify the solution's properties, often increasing solubility and affecting color due to changes in the electronic environment of the metal ion.
Understanding how these complexes form step by step is important for predicting chemical behavior in solution.
Stepwise Equilibrium Reactions
Stepwise equilibrium reactions are a hallmark of complex formation, particularly when dealing with metal ions like \( \mathrm{Cu}^{2+} \) and ligands such as ammonia. In such cases, each step has its own equilibrium constant, denoting the extent to which that particular stage of the reaction occurs.
  • Each stage of ammonia binding to copper ions has a specific equilibrium constant (\( K_1 \) through \( K_4 \)). These constants indicate how strongly each successive ammonia molecule binds to the complex, with typically larger constants indicating stronger binding.
  • For example, the formation of \( \mathrm{CuNH_3}^{2+} \) from \( \mathrm{Cu}^{2+} \) and \( \mathrm{NH_3} \) has a high \( K_1 \), suggesting a favorable initial reaction. As more ligands are added, the equilibrium constants decrease, showing how each subsequent binding is slightly less favored than the last.
  • Solving the related equations involves finding the concentration changes through these steps, defining the balance of each species in reaction. This understanding is vital for manipulating reactions, for instance, to enhance a desired complex ion's production.
By understanding these stepwise reactions and their equilibria, scientists can predict and control complex ion behaviors in various chemical processes.

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Most popular questions from this chapter

Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to 0.50 \(\mathrm{L}\) of \(3.0 M\) \(\mathrm{NH}_{3}\) . Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text { overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

On a hot day, a 200.0 -mL sample of a saturated solution of \(\mathrm{PbI}_{2}\) was allowed to evaporate until dry. If 240 mg of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{PbI}_{2}\) on this hot day.

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} \mathrm{M} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{PbBr}_{2}.\)

Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.

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