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Chapter 17: Problem 107

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14}\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what would the initial concentration of \(\mathrm{NaBr}\) need to be in order to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3}\) moles per liter?

Short Answer

Expert verified
The molar solubility of Cd(OH)₂ in water is approximately \(7.95 \times 10^{-5}\) mol/L. To increase the molar solubility of Cd(OH)₂ to \(1.0 \times 10^{-3}\) mol/L by forming the \(CdBr_4^{2-}\) complex ion, the initial concentration of NaBr needed is approximately \(4.2 \times 10^{-3}\) mol/L.

Step by step solution

01

Calculate the solubility of Cd(OH)₂ in water

To do this, we need to set up an equilibrium expression for the following dissociation reaction of Cd(OH)₂ in water: \[ \text{Cd(OH)}_2(s) \rightleftharpoons Cd^{2+}(aq) + 2 OH^-(aq) \] Let s be the molar solubility of Cd(OH)₂ in water, then in equilibrium, we have [Cd²⁺] = s mol/L, and [OH⁻] = 2s mol/L. The solubility product constant, Kₛₚ, is given by: \[ K_{sp} = [Cd^{2+}][OH^-]^2 \] Using the given value of Kₛₚ = 2.5 x 10⁻¹⁴, we can write the expression: \[ 2.5 \times 10^{-14} = s(2s)^2 \]
02

Solve for s, the solubility of Cd(OH)₂ in water

In order to solve for s, we first rewrite the equation as a cubic equation in s: \[ 2.5 \times 10^{-14} = 4s^3 \] Now, divide both sides by 4: \[ s^3 = \frac{2.5 \times 10^{-14}}{4} \] Next, take the cube root of both sides to find the value of s: \[ s = \sqrt[3]{\frac{2.5 \times 10^{-14}}{4}} \] \[ s \approx 7.95 \times 10^{-5} \] Thus, the molar solubility of Cd(OH)₂ in water is about 7.95 x 10⁻⁵ mol/L.
03

Calculate the initial concentration of NaBr to increase the solubility of Cd(OH)₂

First, we need to set up an equilibrium expression involving the formation of the CdBr₄²⁻ complex ion. The reaction is as follows: \[ Cd^{2+}(aq) + 4 Br^-(aq) \rightleftharpoons CdBr_4^{2-}(aq) \] The formation constant, K_f, is given as 5 x 10³. The equilibrium expression for the complex formation reaction is: \[ K_f = \frac{[CdBr_4^{2-}]}{[Cd^{2+}][Br^-]^4} \] Using the value s = 1.0 x 10⁻³ mol/L, the concentration of Cd²⁺ can be set to s. Let the initial concentration of NaBr be x mol/L, then the initial concentration of Br⁻ is also x mol/L. The equilibrium concentrations of Br⁻ and CdBr₄²⁻ are given by x-4s and s, respectively. The reaction can be expressed as: \[ 5 \times 10^3 = \frac{s}{(s)(x-4s)^4} \]
04

Solve for x, the initial concentration of NaBr

In order to solve for x, substitute 1.0 x 10⁻³ mol/L for s: \[ 5 \times 10^3 = \frac{1.0 \times 10^{-3}}{(1.0 \times 10^{-3})(x-4(1.0 \times 10^{-3})^4} \] Now, rearrange the equation to solve x: \[ 5 \times 10^3 = \frac{1}{x-4 \times 10^{-3}} \] \[ x - 4 \times 10^{-3} = \frac{1}{5 \times 10^3} \] Add 4x 10⁻³ to both sides: \[ x = \frac{1}{5 \times 10^3} + 4 \times 10^{-3} \] \[ x \approx 4.2 \times 10^{-3} \] So, the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)₂ in water to 1.0 x 10⁻³ mol/L is about 4.2 x 10⁻³ mol/L.

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Key Concepts

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

Solubility Product Constant (Ksp)
The solubility product constant, denoted as Ksp, is an important concept in chemistry that quantifies the solubility of a sparingly soluble compound under equilibrium conditions. It is defined as the product of the concentrations of the ions of a salt, each raised to the power of its coefficient in the balanced equation for the dissolution of the salt.

For instance, the solubility equilibrium for cadmium hydroxide, Cd(OH)2, can be represented as follows:
Cd(OH)2(s) \rightleftharpoons Cd2+(aq) + 2 OH-(aq)
The solubility product constant expression for this reaction is:
\[ K_{sp} = [Cd^{2+}][OH^-]^2 \]
To find the molar solubility, 's', we set up the equilibrium concentrations based on the stoichiometry of the balanced equation. The molar solubility is then obtained by solving the equation that results from inserting these values into the Ksp expression, as demonstrated in the provided exercise. It's important to note that the value of Ksp is unique for each salt and is temperature-dependent.
Complex Ion Formation
Complex ion formation is a reaction where a metal ion combines with one or more ligands to form a coordination complex. The formation of a complex ion in a solution can significantly increase the solubility of some salts because it lowers the concentration of free metal ions that contribute to the Ksp.

For example, when cadmium hydroxide reacts with bromide ions from sodium bromide (NaBr), a complex ion CdBr42- is formed. This reaction is represented by the equation:
\[ Cd^{2+}(aq) + 4 Br^-(aq) \rightleftharpoons CdBr_4^{2-}(aq) \]
The constant associated with the formation of the complex ion is called the formation constant (Kf). It describes how strongly the metal ion binds to the ligands. If a complex ion forms, the dissolved metal ion concentration decreases, which in turn increases the solubility of the original compound because the reaction shifts to dissolve more solid to re-establish equilibrium.
Equilibrium Concentration
Equilibrium concentration refers to the concentration of reactants and products in a reaction mixture when the forward and reverse reaction rates are equal, resulting in no net change over time. Equilibrium concentrations are crucial for calculating both the solubility product constant (Ksp) and the formation constant (Kf) of complex ions.

In the context of solubility equilibria, calculating the equilibrium concentrations often involves setting up an 'ICE' table (Initial, Change, Equilibrium) to track the changes from the initial concentrations to the equilibrium state. For the dissolution of a sparingly soluble salt like Cd(OH)2, the initial concentration of dissolved ions is zero, and the change is represented by 's', the solubility. When complex ions are formed, these concentrations change further, affecting the solubility of the original salt.

In the exercise, to increase the solubility of Cd(OH)2, the goal was to establish what initial concentration of NaBr is needed. Through the concept of equilibrium concentration, we could determine the initial amount of NaBr required to augment the formation of the CdBr42- complex ion and thus increase the solubility of Cd(OH)2.

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Most popular questions from this chapter

Assume that \(30.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of a weak base \(\mathrm{B}\) that accepts one proton is titrated with a \(0.10 \mathrm{M}\) solution of the monoprotic strong acid HX. (a) How many moles of \(\mathrm{HX}\) have been added at the equivalence point? (b) What is the predominant form of \(\mathrm{B}\) at the equivalence point? (c) What factor determines the \(\mathrm{pH}\) at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer \(\mathrm{A}\) is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer B is prepared by using \(0.010 \mathrm{~mol}\) of each. (a) Calculate the \(\mathrm{pH}\) of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) \(\mathrm{ZnCO}_{3},(\mathbf{b}) \mathrm{ZnS}\), (c) \(\mathrm{BiI}_{3}\) (d) \(\mathrm{AgCN},(\mathrm{e}) \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\)

The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16} .\) The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}^{2-},\) is \(4.6 \times 10^{17}\). What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

A 20.0 -mL sample of \(0.150 \mathrm{M} \mathrm{KOH}\) is titrated with \(0.125 \mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(20.0 \mathrm{~mL},\) (b) \(23.0 \mathrm{~mL},\) (c) \(24.0 \mathrm{~mL}\), (d) \(25.0 \mathrm{~mL},(\mathrm{e}) 30.0 \mathrm{~mL}\).
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