/*! 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 105 The copper(l) ion forms a comple... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 105

The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of \(\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\) in \(1.0 \mathrm{L}\) of \(1.0 \mathrm{M} \mathrm{NaCN}\) . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

Short Answer

Expert verified
a. The solubility of CuBr in 1.0 L of 1.0 M NaCN is approximately \(5.63 \times 10^{-26}\ \mathrm{mol}\). b. The concentration of Br- ions at equilibrium is approximately \(5.63 \times 10^{-26}\ \mathrm{M}\). c. The concentration of CN- ions at equilibrium is approximately \(0.9997\ \mathrm{M}\).

Step by step solution

01

Set up the equations

We are given the following equilibrium expressions: $$a. \ \mathrm{CuBr}(s) \rightleftharpoons \mathrm{Cu}^{+} (aq) + \mathrm{Br}^{-}(aq) \quad K_{sp} = 1.0 \times 10^{-5}$$ $$b. \ \mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ We need to find the solubility of CuBr in 1.0 L of 1.0 M NaCN. Let the solubility of CuBr in 1.0 L of NaCN be 'x' moles. Thus, at equilibrium, we have: - [Cu+] = x moles - [Br-] = x moles - [CN-] = (1.0 M - 3x) moles since 3 moles of CN- react with 1 mole of Cu+.
02

Use the given equilibrium constants and expressions

Using the given equilibrium constants, we have the following equilibrium expressions: $$a. \ K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Br}^{-}]$$ $$b. \ K = \frac{[\mathrm{Cu}(\mathrm{CN})_{3}^{2-}]}{[\mathrm{Cu}^{+}][\mathrm{CN}^{-}]^{3}}$$ For the \(K_{sp}\) expression, we know that: $$1.0 \times 10^{-5} = (x)(x)$$ $$x = \sqrt{1.0 \times 10^{-5}} = 1.0 \times 10^{-2.5}$$ Now, for the K expression, we have: $$1.0 \times 10^{11} = \frac{[\mathrm{Cu}(\mathrm{CN})_{3}^{2-}]}{[(1.0 \times 10^{-2.5})(1.0 - 3(1.0 \times 10^{-2.5}))]^3}$$ We can substitute [\(\mathrm{Cu}(\mathrm{CN})_{3}^{2-}\)] as 'x' in the equation above: $$1.0 \times 10^{11} = \frac{x}{[(1.0 \times 10^{-2.5})(1.0 - 3(1.0 \times 10^{-2.5}))]^3}$$
03

Solve for the solubility

Re-arrange the equation and solve for x, which represents the solubility of CuBr: $$x = 1.0 \times 10^{11} [(1.0 \times 10^{-2.5})(1.0 - 3(1.0 \times 10^{-2.5}))]^3$$ $$x \approx 5.63 \times 10^{-26}$$ a. The solubility of CuBr in 1.0 L of 1.0 M NaCN is approximately \(5.63 \times 10^{-26}\ \mathrm{mol}\).
04

Calculate the concentration of Br- ions at equilibrium

As we calculated the solubility 'x' of CuBr in 1.0 L of 1.0 M NaCN, the concentration of Br- ions at equilibrium will be the same: b. The concentration of Br- ions at equilibrium is approximately \(5.63 \times 10^{-26}\ \mathrm{M}\).
05

Calculate the concentration of CN- ions at equilibrium

Since 3 moles of CN- react with 1 mole of Cu+ as per the equation, the change in the concentration of CN- ions will be: Decrease in CN- ions = Moles of Cu+ reacted x 3 = (1.0 x 10^{-2.5}) x 3 The initial concentration of CN- ions is given as 1.0 M, so the final concentration of CN- ions at equilibrium will be: Final concentration of CN- ions = Initial concentration - Decrease in concentration = 1.0 - (1.0 x 10^{-2.5} x 3) c. The concentration of CN- ions at equilibrium is approximately \(0.9997\ \mathrm{M}\).

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Key Concepts

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

Complex Ion Formation
Complex ion formation involves the interaction between a central ion, often a metal, and surrounding molecules or ions known as ligands. In the given problem, the copper ion (\(\text{Cu}^+\)) acts as the central ion, while cyanide ions (\(\text{CN}^-\)) serve as ligands. These interactions result in the creation of a stable complex, denoted as \(\text{Cu(CN)}_3^{2-}\).
The formation of complex ions can significantly increase the solubility of sparingly soluble compounds. This happens because the ligands effectively "mask" the central ion, reducing its concentration in the solution. This promotes further dissolution of the solid compound to reach equilibrium. The complex formation is represented by an equilibrium equation with its own constant (\(K\)), highlighting the balance between the complex and its constituents.
Equilibrium Constant (K)
The equilibrium constant (\(K\)) is a vital concept in understanding chemical equilibria. It quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction. In our exercise, we have two key equilibria:
  • The solubility product (\(K_{sp}\)) for the dissociation of \(\text{CuBr}\) into \(\text{Cu}^+\) and \(\text{Br}^-\), which is \(1.0 \times 10^{-5}\).
  • The formation constant (\(K\)) for the complex ion \(\text{Cu(CN)}_3^{2-}\), which is \(1.0 \times 10^{11}\).
The huge value of \(K\) for the complex suggests that the formation of the complex is highly favored, meaning the ions will predominantly exist in this form at equilibrium. By using these constants, we can derive equilibrium concentrations and predict how much of a compound dissolves in the solution.
Common Ion Effect
The common ion effect describes the decrease in solubility of a salt when a solution contains a common ion. In this problem, the presence of \(\text{CN}^-\) ions in the solution affects the dissolution of \(\text{CuBr}\).
When added to a solution already containing \(\text{CN}^-\), the equilibrium shifts, resulting in less \(\text{Cu}^+\) and \(\text{Br}^-\) ions in solution. This is because the removal of \(\text{Cu}^+\) by complex formation (i.e., turning into \(\text{Cu(CN)}_3^{2-}\) via the large \(K\)) facilitates further dissolution until equilibrium is re-established. This interplay of equilibria is a classic illustration of the common ion effect in action, demonstrating its capability to influence solubility in solutions with existing similar ions.

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