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Chapter 22: Problem 25

Copper(I) ions in aqueous solution react with \(\mathrm{NH}_{3}(a q)\) according to $$ \mathrm{Cu}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \leftrightharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) $$ with \(K_{\mathrm{f}}=6.3 \times 10^{10} \mathrm{M}^{-2}\). Calculate the solubility in grams per liter of \(\operatorname{CuBr}(s)\) in \(0.50 \mathrm{M} \mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility of CuBr in 0.50 M NH鈧� at 25掳C is extremely small due to complex formation, effectively making it negligible.

Step by step solution

01

Write the Equation for CuBr Dissolution

First, write the dissociation equation for copper bromide in an aqueous solution:\[\text{CuBr}(s) \rightleftharpoons \text{Cu}^+(aq) + \text{Br}^-(aq)\]
02

Establish the Solubility Product

The solubility product \(K_{sp}\) for CuBr is:\[K_{sp} = [\text{Cu}^+][\text{Br}^-]\]Assuming solubility, \(s\), calculate as the concentration of \(\text{Cu}^+\) and \(\text{Br}^-\) that dissolve from CuBr.
03

Consider the Complex Formation

Given that ammonia coordinates with Cu+ ions, the complexation reaction is:\[\text{Cu}^+(aq) + 2\text{NH}_3(aq) \leftrightharpoons [\text{Cu(NH}_3\text{)}_2]^+(aq)\]The formation constant is \(K_{f} = 6.3 \times 10^{10} \text{ M}^{-2}\).
04

Solve the Equilibrium Condition

The copper ions in solution create complexes instead of contributing to the solubility product:\[[\text{Cu}^+] = \frac{s}{1 + K_f \times [\text{NH}_3]^2}\]Given \([\text{NH}_3] = 0.50 \text{ M}\), calculate:\[[\text{Cu}^+] = \frac{s}{1 + (6.3 \times 10^{10}) \times (0.50)^2}\]
05

Calculate the Solubility of CuBr

Assuming \([\text{Cu}^+] \approx s\), solve:\[s^2 \approx K_{sp}\]In this scenario since the effect of ammonia is dominant:\[K_{f} \gg \frac{1}{[\text{NH}_3]^2}\quad \Rightarrow \quad s = \frac{1}{\sqrt{K_{f}}} \cdot \text{Molar Solubility of Copper}\]Convert it into grams per liter by adjusting with the molar mass of CuBr.
06

Express Solubility in Grams per Liter

Finally, use the molar mass of CuBr (143.4 g/mol) to convert from molarity to g/L:\[\text{Solubility } = s \times 143.4 \text{ g/mol}\]Calculate this using the determined solubility magnitude.

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Key Concepts

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

Copper(I) Complexation
Complexation of Copper(I) ions occurs when they interact with other molecules or ions in solution, forming more stable compounds, known as complexes. In this case, Copper(I) ions ( Cu鈦� ) react with ammonia ( NH鈧� ) to form a complex, [Cu(NH鈧�)鈧俔鈦� . Complexation reactions can significantly impact the chemical behavior of ions in solutions, often affecting their concentrations.

The reaction under consideration is:
  • Cu鈦�(aq) + 2 NH鈧�(aq) 鈬� [Cu(NH鈧�)鈧俔鈦�(aq)
This reaction equilibrium is characterized by a formation constant ( K_f ), which in this case, is 6.3 x 10鹿鈦� M鈦宦�. A high K_f value indicates a strong tendency for the Cu鈦� ions to complex with ammonia, drastically reducing the free Cu鈦� concentration in the solution.

Understanding these complexation reactions is crucial in analyzing solubility as complexes are often more soluble than their constituent ions alone. This property is significant when considering the overall solubility of compounds like copper bromide (CuBr), as well as the potential release or capture of ions within different chemical contexts.
Solubility Product Constant
The Solubility Product Constant, K_{sp} , is a crucial aspect of determining how much of a solute can dissolve in a solution before it reaches saturation. For the dissolution of copper bromide ( CuBr ), the equilibrium expression can be written as:
  • CuBr(s) 鈬� Cu鈦�(aq) + Br鈦�(aq)
Here, K_{sp} represents the product of the molar concentrations of the ions in the solution raised to the power of their stoichiometric coefficients:
  • K_{sp} = [Cu鈦篯[Br鈦籡
The value of K_{sp} defines the maximum product of the concentrations of the dissolved ions in a saturated solution, indicating at what point no more of the solute can dissolve without precipitating back out.

In scenarios where complexing agents (like ammonia) are present, the K_{sp} value alone might not fully describe solubility. Complex formation effectively "removes" free ions from this equilibrium, often increasing the apparent solubility of a compound. This happens because the formation of stable complexes reduces the concentration of free ions that contribute to the K_{sp} value.
Chemical Equilibrium Calculations
Chemical equilibrium calculations involve assessing the balance between reactants and products in a reversible reaction. When calculating parameters like solubility, it is essential to consider both K_{sp} and formation constants like K_f , as they provide insight into the ions' behavior in solution.

To determine the solubility of CuBr in a solution of 0.50 M ammonia at 25掳C, we must account for both the dissociation of CuBr into Cu鈦� and Br鈦� and the complexation of Cu鈦� with ammonia.
  • Determine the concentration of free Cu鈦� by considering the complexation:
    • Cu鈦� = s / (1 + K_f[NH鈧僝虏)
  • Solve for solubility, s , ensuring that complexation is the dominant factor influencing Cu鈦� concentration.
    • Use the initial expression for K_{sp} = s 脳 s, noting that without further calculation details, K_{sp} needs to be provided.
  • Finally, express the solubility in grams per liter using the molar mass of CuBr (143.4 g/mol).
    • Solubility (g/L) = s 脳 143.4 g/mol
In complex systems like these, detailed analysis and systematic calculations are vital for accurately evaluating how equilibrium and complexation influence solubility.

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Most popular questions from this chapter

Excess \(\mathrm{HgI}_{2}(s)\) was equilibrated with a solution that is \(0.10 \mathrm{M}\) in \(\mathrm{KI}(a q) .\) Calculate the solubility of \(\mathrm{HgI}_{2}(s)\) in this solution at \(25^{\circ} \mathrm{C}\) given \(\mathrm{HgI}_{2}(s) \leftrightharpoons \mathrm{Hg}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \quad K_{\mathrm{sp}}=2.9 \times 10^{-29} \mathrm{M}^{3}\) \(\operatorname{HgI}_{2}(s)+2 \mathrm{I}^{-}(a q) \leftrightharpoons\left[\mathrm{HgI}_{4}\right]^{2-}(a q) \quad K_{i}=0.79 \mathrm{M}^{-1}\).

Suppose we mix \(50.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{AgNO}_{3}(a q)\) with \(50.0 \mathrm{~mL}\) of \(1.00 \times 10^{-4} \mathrm{M} \mathrm{NaBr}(a q) .\) Does \(\mathrm{AgBr}(s)\) precipitate from the solution? If yes, then calculate how many moles of \(\operatorname{AgBr}(s)\) precipitate and the values of \(\left[\mathrm{Ag}^{+}\right],\left[\mathrm{Br}^{-}\right],\left[\mathrm{Na}^{+}\right]\), and \(\left[\mathrm{NO}_{3}^{-}\right]\) at \(25^{\circ} \mathrm{C}\) at equilibrium.

Why is it possible to separate a mixture of \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Hg}_{2}^{2+}(a q)\) by selectively precipitating out the mercury as \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s) ;\) but not possible to separate the same mixture by selectively precipitating out the lead as \(\mathrm{PbI}_{2}(s) ?\)

Calculate the solubility in grams per liter of AgI \((s)\) in \(0.60 \mathrm{M} \mathrm{NH}_{3}(a q) .\) Take \(K_{\mathrm{f}}=2.0 \times 10^{7} \mathrm{M}^{-2}\) for the complexation reaction of \(\mathrm{Ag}^{+}(a q)\) to produce \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q)\). Assume a temperature of \(25^{\circ} \mathrm{C} .\)

Calculate the solubility of \(\mathrm{AgC}_{6} \mathrm{H}_{5} \mathrm{COO}(s)\) in grams per liter in an aqueous solution buffered at \(\mathrm{pH}=4.00\) at \(25^{\circ} \mathrm{C} .\) Given \(K_{\mathrm{a}}=6.3 \times 10^{-5} \mathrm{M}\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q)\) and that \(K_{\mathrm{sp}}=2.5 \times 10^{-5} \mathrm{M}^{2}\) for \(\mathrm{AgC}_{6} \mathrm{H}_{5} \mathrm{COO}(s)\).
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