/*! 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 37 What is the solubility of \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the solubility of \(\mathrm{PbF}_{2}\) in water? The \(K_{s p}\) for \(\mathrm{PbF}_{2}\) is \(2.7 \times 10^{-8}\).

Short Answer

Expert verified
The solubility of \(\mathrm{PbF}_2\) in water is approximately \(1.88 \times 10^{-3}\) mol/L.

Step by step solution

01

Understanding the Dissolution Equation

The dissolution of lead(II) fluoride, \( \text{PbF}_2 \), in water can be represented by the equation: \[ \text{PbF}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{F}^- (aq) \] This equation will help us to understand the relationship between the ions in solution and the solubility product constant, \(K_{sp}\).
02

Writing the Expression for the Solubility Product Constant

The solubility product constant \(K_{sp}\) is a special kind of equilibrium constant used for the solubility of sparingly soluble salts. For lead(II) fluoride, the \(K_{sp}\) expression is: \[ K_{sp} = [\text{Pb}^{2+}][\text{F}^-]^2 \] Substituting the known value, we have \(K_{sp} = 2.7 \times 10^{-8}\).
03

Defining the Variables for Ion Concentrations

Let the solubility of \(\text{PbF}_2\) be \(s\) mol/L. When \(\text{PbF}_2\) dissolves, it releases \(1\) mole of \(\text{Pb}^{2+}\) and \(2\) moles of \(\text{F}^-\) ions into the solution, meaning \([\text{Pb}^{2+}] = s\) and \([\text{F}^-] = 2s\).
04

Substituting and Simplifying the Expression

Substitute \([\text{Pb}^{2+}] = s\) and \([\text{F}^-] = 2s\) into the \(K_{sp}\) expression: \[ 2.7 \times 10^{-8} = s(2s)^2 = s \cdot 4s^2 \] This simplifies to: \[ 2.7 \times 10^{-8} = 4s^3 \]
05

Solving for Solubility \(s\)

To find \(s\), divide both sides of the equation by \(4\): \[ s^3 = \frac{2.7 \times 10^{-8}}{4} \] Now, calculate \(s^3\): \[ s^3 = 6.75 \times 10^{-9} \] Next, take the cube root of both sides to solve for \(s\): \[ s = \sqrt[3]{6.75 \times 10^{-9}} \approx 1.88 \times 10^{-3} \text{ mol/L} \]
06

Conclusion

The solubility of \(\text{PbF}_2\) in water at equilibrium is approximately \(1.88 \times 10^{-3}\) mol/L.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Solubility Product Constant (Ksp)
The solubility product constant, often denoted as \(K_{sp}\), is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts in water. When these salts dissolve, they dissociate into their constituent ions. Unlike fully soluble compounds, sparingly soluble salts only partially dissolve, reaching a state of equilibrium. At this point, the rate of dissolution equals the rate at which the ions precipitate back into solid form.
The \(K_{sp}\) expression is derived from the concentrations of the ions in this saturated solution. For a general salt \(AB_2\), which dissociates into \(A^{+}\) and \(2B^{-}\), the expression is written as:
  • \(K_{sp} = [A^+][B^-]^2\)
By knowing \(K_{sp}\), we can determine the extent to which a salt can dissolve in water at a given temperature.
Sparingly Soluble Salts
Sparingly soluble salts are compounds with limited solubility in water. Unlike highly soluble salts, which dissolve completely, sparingly soluble salts reach a saturation point quite quickly. Once this point is reached, any additional solid added to the solution will not dissolve, remaining as a precipitate.
These salts are an important topic in chemistry because they help us understand how precipitation occurs and how such processes are influenced by factors like temperature and the presence of other ions in a solution.
  • An example of such a salt is lead(II) fluoride (PbF\(_2\)), which dissolves minimally in water.
This behavior is dictated largely by the solubility product constant, \(K_{sp}\), of the salt.
Equilibrium Expressions
Equilibrium expressions are vital for understanding reactions that do not go to completion. For sparingly soluble salts, equilibrium is reached when the rate at which ions dissolve into solution matches the rate at which they precipitate out. This delicate balance is captured mathematically in the solubility product expression.
For a salt like PbF\(_2\), the equilibrium expression derived from its dissociation is:
  • \(PbF_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2F^- (aq)\)
  • Leads to: \(K_{sp} = [Pb^{2+}][F^-]^2\)
This expression is vital for calculating the solubility or concentration of ions in a saturated solution, and it demonstrates the relationship between solubility and \(K_{sp}\).
Ion Concentrations
Ion concentrations denote the amount of individual ions present in a solution at equilibrium. For sparingly soluble salts, the dissociated ions generate characteristic concentrations that are connected to their solubility.
In the case of PbF\(_2\), let \(s\) represent its solubility in \(\text{mol/L}\), meaning:
  • \([Pb^{2+}] = s\)
  • \([F^-] = 2s\)
These concentrations allow us to calculate the solubility, \(s\), by substituting them back into the \(K_{sp}\) expression and solving for \(s\). This ensures we understand how much of the salt has been dissolved, and it confirms our calculations align with the known \(K_{sp}\) for the compound.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Lead chloride at first precipitates when sodium chloride is added to a solution of lead nitrate. Later, when the solution is made more concentrated in chloride ion, the precipitate dissolves. Explain what is happening. What equilibria are involved? Note that lead ion forms the complex ion \(\mathrm{PbCl}_{4}^{2-}\).

The solubility of the hypothetical salt \(\mathrm{A}_{3} \mathrm{~B}_{2}\) is \(6.1 \times\) \(10^{-9} \mathrm{~mol} / \mathrm{L}\) at a certain temperature. What is \(K_{s p}\) for the salt? \(\mathrm{A}_{3} \mathrm{~B}_{2}\) dissolves according to the equation $$ \begin{array}{l} \quad \mathrm{A}_{3} \mathrm{~B}_{2}(s) \rightleftharpoons 3 \mathrm{~A}^{2+}(a q)+2 \mathrm{~B}^{3-}(a q) \\ \text { a) } 3.7 \times 10^{-17} \\ \text {b) } 9.1 \times 10^{-25} \\ \text {c) } 9.1 \times 10^{-40} \\ \text {d) } 3.7 \times 10^{-32} \\ \text {e) } 1.4 \times 10^{-33} \end{array} $$

Consider three hypothetical ionic solids: \(\mathrm{AX}, \mathrm{AX}_{2}\), and \(\mathrm{AX}_{3}\) (each \(\mathrm{X}\) forms \(\mathrm{X}^{-}\) ). Each of these solids has the same \(K_{s p}\) value, \(5.5 \times 10^{-7}\). You place 0.25 mol of each compound in a separate container and add enough water to bring the volume to \(1.0 \mathrm{~L}\) in each case. a. Write the chemical equation for each of the solids dissolving in water. b. Would you expect the concentration of each solution to be \(0.25 M\) in the compound? Explain, in some detail, why or why not. c) Would you expect the concentrations of the A cations \(\left(\mathrm{A}^{+}, \mathrm{A}^{2+},\right.\) and \(\left.\mathrm{A}^{3+}\right)\) in the three solutions to be the same? Does just knowing the stoichiometry of each reaction help you determine the answer, or do you need something else? Explain your answer in detail, but without doing any arithmetic calculations. d. Of the three solids, which one would you expect to have the greatest molar solubility? Explain in detail, but without doing any arithmetic calculations. e. Calculate the molar solubility of each compound.

A 45-mL sample of \(0.015 M\) calcium chloride, \(\mathrm{CaCl}_{2}\), is added to \(55 \mathrm{~mL}\) of \(0.010 \mathrm{M}\) sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). Is a precipitate expected? Explain.

A precipitate forms when a small amount of sodium hydroxide is added to a solution of aluminum sulfate. This precipitate dissolves when more sodium hydroxide is added. Explain what is happening.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.