/*! 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 102 Calculate the solubility of \(\m... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 102

Calculate the solubility of \(\mathrm{Mg}_{\mathrm{g}}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH} \mathrm{NCl}_{4}\).

Short Answer

Expert verified
The solubility of Mg(OH)鈧� in 0.50 M NH鈧凜l can be calculated by first finding the Ksp and Kf values, setting up an ICE table, and solving for the concentration of Mg虏鈦� ions (s). The Ksp value for Mg(OH)鈧� is \(5.61 \times 10^{-12}\) and the Kf value for NH鈧凮H is \(1.70 \times 10^{7}\). Using these values and simplifying the equilibrium expressions and solubility relations, you can solve for the value of s numerically. Finally, the solubility (M) of Mg(OH)鈧� is equal to the concentration of Mg虏鈦� ions, which is s.

Step by step solution

01

Write the Equilibrium Equations

First, we need to write the equilibrium equations for the dissolution of Mg(OH)鈧� and the formation of the complex ion between NH鈧勨伜 and OH鈦� ions: Mg(OH)鈧�(s) 鈬� Mg虏鈦�(aq) + 2OH鈦�(aq) - Dissolution of Mg(OH)鈧� NH鈧勨伜(aq) + OH鈦�(aq) 鈬� NH鈧凮H(aq) - Formation of the complex ion
02

Find the Values of Ksp and Kf

We need to find the values of Ksp for Mg(OH)鈧� and Kf for the complex ion. In general, these values can be found in a textbook or online. For this problem, we will use the following values: \(K_{sp} = 5.61 \times 10^{-12}\) for Mg(OH)鈧� \(K_{f} = 1.70 \times 10^{7}\) for NH鈧凮H
03

Set Up the ICE Table

Next, we will set up the ICE (Initial, Change, Equilibrium) table for the dissolution of Mg(OH)鈧� and the formation of the complex ion. The initial concentration of NH鈧勨伜 ions is 0.50 M, and the initial concentration of Mg虏鈦� and OH鈦� ions is 0. Mg虏鈦�: 0 OH鈦�: 0 NH鈧勨伜: 0.5 The changes are: Mg虏鈦�: +s OH鈦�: +2s NH鈧勨伜: -s The equilibrium values are: Mg虏鈦�: s OH鈦�: \(2s+x\) NH鈧勨伜: \(0.5-s\) We need to add x to the 2s term for the OH鈦� ion because some of the OH鈦� ions will react with the NH鈧勨伜 ions to form NH鈧凮H.
04

Write the Equilibrium Expression and Solve for s

Now, write the equilibrium expressions for Ksp and Kf: \(K_{sp} = [Mg^{2+}][OH^-]^2\) \(K_{f} = \frac{[NH_4OH]}{[NH_4^+][OH^-]}\) We can now use the ICE table values to substitute the concentrations: \(K_{sp} = s(2s+x)^2\) \(K_{f} = \frac{1.70 \times 10^7 * x}{(0.5-s)(2s+x)}\) We can solve for x and s by first assuming that s is very small compared to 0.50. This allows us to simplify our equations: \(K_{sp} = s(2s+x)^2 \approx s(4sx^2)\) - assuming s << 0.5 \(K_{f} = \frac{1.70 \times 10^7 * x}{(0.5)(2s+x)}\) Now solve for x in terms of s using the second equation: \(x = \frac{K_{f}(0.5)(2s+x)}{1.70 \times 10^7}\) Next, substitute this expression into the first equation: \(K_{sp} = s(4s(\frac{K_{f}(0.5)(2s+x)}{1.70 \times 10^7})^2)\) Finally, solve for s numerically using a calculator or a numerical analysis technique.
05

Calculate the Solubility of Mg(OH)鈧�

Once you have obtained the value of s, you can calculate the solubility of Mg(OH)鈧�: Solubility (M) = [Mg虏鈦篯 = s By following these steps and solving for s, you will find the solubility of Mg(OH)鈧� in 0.50 M NH鈧凜l.

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Key Concepts

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

Chemical Equilibrium
In the context of solubility calculations, chemical equilibrium refers to the state in which the rate of the forward reaction (dissolving of the solid into its constituent ions) is equal to the rate of the reverse reaction (recombination of the ions to form the solid). This does not mean the concentrations of the reactants and products are equal, but that their concentrations have stabilized and are not changing over time.

For the dissolution of magnesium hydroxide, Mg(OH)2, in water, the equilibrium can be disturbed by the presence of other chemical species, such as NH4+ from NH4Cl. The NH4+ can react with OH- to form NH4OH, a new compound, which shifts the equilibrium position. Understanding this concept is vital when predicting the solubility of compounds in different environments.
Solubility Product Constant (Ksp)
The solubility product constant (Ksp) is a crucial concept in solubility calculations. It is specific for every sparingly soluble compound and represents the maximum product of the molar concentrations of the ions that can exist in solution at equilibrium. The Ksp value is constant for a compound at a given temperature.

The formula for Ksp for magnesium hydroxide is given as:\[K_{sp} = [Mg^{2+}][OH^-]^2\]
As the concentration of magnesium and hydroxide ions reaches a specific point, any further dissolution will lead to a supersaturated solution, and excess solid will precipitate out to maintain the Ksp value.
Complex Ion Formation
The formation of complex ions can have a significant impact on the solubility of compounds. A complex ion consists of a central metal ion surrounded by molecules or anions called ligands. These ligands can 'coordinate' to the metal ion, effectively increasing its solubility.

In our exercise, the NH4+ acting as a ligand, reacts with OH- ions to form the complex ion NH4OH. This reaction pulls OH- ions out of equilibrium, influencing the solubility of Mg(OH)2. The equilibrium constant for complex ion formation, known as the formation constant (Kf), quantifies the extent to which complex ion formation occurs.
ICE Table
An ICE table (Initial, Change, Equilibrium) is an organizational tool used to keep track of the concentrations of species in an aqueous equilibrium system. The 'I' stands for the initial concentrations, 'C' for the change that occurs as the system moves towards equilibrium, and 'E' for the equilibrium concentrations of each species involved.

In the context of our solubility problem, the ICE table helps us determine the changes in concentration of Mg2+, OH-, and NH4+ from their initial values through to equilibrium. It simplifies the complex interactions into a format that is easier to understand and provides a systematic method for calculating the solubility of Mg(OH)2 in the presence of NH4Cl.

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Most popular questions from this chapter

\mathrm{~A}\( solution of \)\mathrm{Na}_{2} \mathrm{SO}_{4}\( is added dropwise to a solution that is \)0.010 \mathrm{M}_{\text {in } \mathrm{Ba}^{2+}}\( and \)0.010 \mathrm{M}\( in \)\mathrm{Sr}^{2+}\(. (a) What concentration of \)\mathrm{SO}_{4}^{2-}\( is necessary to begin precipitation? (Neglect volume changes, \)\mathrm{BaSO}_{4} ; K_{p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\( \)K_{s p}=3.2 \times 10^{-7}\( ) (b) Which cation precipitates first? (c) What is the concentration of \)\mathrm{SO}_{4}^{2-}$ when the second cation begins to precipitate?

Suggest how the cations in each of the following solution mixtures can be separated: (a) Na+ and \(\mathrm{Ca}^{2+}\), (b) \(\mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+}\), (c) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+}\), (d) \(\mathrm{Ag}^{+}\)and \(\mathrm{Hg}^{2+}\).

Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\) ) reacts with acids in foods to form carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\), which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(\mathrm{~g})\) forms bubbles and causes the cake to rise. (a) A rule of thumb in baking is that \(1 / 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\). Write the chemical equation for this neutralization reaction. (b) The density of baking soda is \(2.16 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of \(\mathrm{mol} / \mathrm{L}\). (One cup \(=236.6 \mathrm{~mL}=48\) teaspoons). (c) If 1/2 teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at 1 atm pressure, in an oven set to \(350^{\circ} \mathrm{F}\).

A buffer contains \(0.10\) mol of acetic acid and \(0.13\) mol of sodium acetate in \(1.00 \mathrm{~L}\). (a) What is the \(\mathrm{pH}\) of this buffer? (b) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.02 \mathrm{~mol}\) of \(\mathrm{KOH}\) ? (c) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.02 \mathrm{~mol}\) of \(\mathrm{HNO}_{3}\) ?

As shown in Figure 16.8, the indicator thymol blue has two color changes. Which color change will generally be more suitable for the titration of a weak acid with a strong base?
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