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(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubility-product constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: AgI, \(\mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2},\) and \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\)

Short Answer

Expert verified
(a) The concentration of undissolved solid is not included in the solubility-product constant (Ksp) expression because its concentration remains nearly constant during the course of dissolution and doesn't affect the equilibrium. Ksp only considers the concentrations of the dissolved ions in the saturated solution. (b) The Ksp expressions for the given electrolytes are as follows: AgI: Ksp(AgI) = [Ag\(^+\)][I\(^-\)] SrSO4: Ksp(SrSO4) = [Sr\(^{2+}\)][SO4\(^{2-}\)] Fe(OH)2: Ksp(Fe(OH)2) = [Fe\(^{2+}\)][OH\(^-\)]^2 Hg2Br2: Ksp(Hg2Br2) = [Hg2\(^{2+}\)][Br\(^-\)]^2

Step by step solution

01

Understand the equilibrium

The solubility-product constant is concerned with the equilibrium between an ionic solid and its dissolved ions in a saturated solution. At equilibrium, the rate of dissolution equals the rate of precipitation, and the concentrations of the dissolved ions remain constant.
02

Constant concentration of solid

In a saturated solution, the solid remaining undissolved is in excess and doesn't change much during the course of dissolution. Therefore, its concentration remains nearly constant and doesn't affect the equilibrium. (b) Ksp expressions for given electrolytes
03

AgI

The dissociation of AgI in water can be represented as: AgI(s) \<=> Ag\(^+\)(aq) + I\(^-\)(aq) The Ksp expression for AgI is: Ksp(AgI) = [Ag\(^+\)][I\(^-\)]
04

SrSO4

The dissociation of SrSO4 in water can be represented as: SrSO4(s) \<=> Sr\(^{2+}\)(aq) + SO4\(^{2-}\)(aq) The Ksp expression for SrSO4 is: Ksp(SrSO4) = [Sr\(^{2+}\)][SO4\(^{2-}\)]
05

Fe(OH)2

The dissociation of Fe(OH)2 in water can be represented as: Fe(OH)2(s) \<=> Fe\(^{2+}\)(aq) + 2OH\(^-\)(aq) The Ksp expression for Fe(OH)2 is: Ksp(Fe(OH)2) = [Fe\(^{2+}\)][OH\(^-\)]^2
06

Hg2Br2

The dissociation of Hg2Br2 in water can be represented as: Hg2Br2(s) \<=> Hg2\(^{2+}\)(aq) + 2Br\(^-\)(aq) The Ksp expression for Hg2Br2 is: Ksp(Hg2Br2) = [Hg2\(^{2+}\)][Br\(^-\)]^2

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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, chemical equilibrium refers to the dynamic balance between the opposing processes of a solid solute dissolving in a solvent and the dissolved solute coming out of the solution to form a solid. At this point, the concentrations of the reactants and products no longer change with time. This equilibrium is crucial for calculating the solubility-product constant (Ksp), which helps us measure how much solute can dissolve in a given solvent at a specific temperature.

In a saturated solution, this balance is struck when the dissolution rate (the process by which the solid becomes ions in the solution) matches the precipitation rate (the process by which ions in the solution form the solid). This point of balance does not mean the reactions stop, but that they continue at equal rates, maintaining constant concentrations of the dissolved ions.
Dissociation of Electrolytes
Electrolytes are substances that produce ions when dissolved in a solvent such as water. The dissociation of electrolytes is a process where an ionic compound separates into its ions. This is fundamental to understanding the solubility of ionic solids. For example, strong electrolytes completely dissociate into their ions, providing the ions that take part in the equilibria used to describe solubility.

An ionic compound like silver iodide (AgI) breaks down into its respective ions, silver ions (Ag+) and iodide ions (I-), each contributing to the conductivity of the solution. The degree to which these compounds dissociate influences their solubility and, consequently, their Ksp values. By understanding the dissociation behavior of the compounds, we not only predict their solubility but also calculate the ion concentrations in a solution.
Ksp Expressions
The solubility-product constant (Ksp) is an expression that relates to the solubility of sparingly soluble ionic compounds. The Ksp is a direct consequence of the dissociation behavior of electrolytes. For a general compound AB that dissociates into An+ and Bm-, the expression for its solubility-product would be:n+]n[Bm-]m where the square brackets indicate the molarity, or concentration, of the ions in solution.

For example, the Ksp for AgI is derived from its dissociation into Ag+ and I-, and the expression is Ksp(AgI) = [Ag+][I-]. This mathematical formula allows us to calculate the maximum concentration of ions in a saturated solution, which is an intrinsic property of a substance, dependent on temperature and the particular ionic compound. The Ksp value is thus essential in predicting whether a precipitate will form under certain conditions and is used in a variety of applications, including water treatment and analytical chemistry.

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Most popular questions from this chapter

You are asked to prepare a \(\mathrm{pH}=3.00\) buffer solution starting from \(1.25 \mathrm{~L}\) of a \(1.00 \mathrm{M}\) solution of hydrofluoric acid (HF) and an excess of sodium fluoride (NaF). (a) What is the \(\mathrm{pH}\) of the hydrofluoric acid solution prior to adding sodium fluoride? (b) How many grams of sodium fluoride should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium fluoride is added.

Explain why a mixture formed by mixing \(100 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(50 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) will act as a buffer.

Show that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{p} K_{a}\) for the acid.

A \(1.00-\mathrm{L}\) solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{PbI}_{2} .\) Calculate the solubility-product constant for this salt at \(25^{\circ} \mathrm{C}\).

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}, 0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\) and the pH measured. (a) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{p} K_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)

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