/*! 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 134 A \(50.0-\mathrm{mL}\) sample of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 134

A \(50.0-\mathrm{mL}\) sample of \(0.0150 \mathrm{M} \mathrm{Ag}_{2} \mathrm{SO}_{4}\) is added to \(25.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{PbCl}_{2}\). What is the net ionic equation for the reaction that occurs? What are the concentrations of ions in the mixture at equilibrium?

Short Answer

Expert verified
The net ionic equations are: \(\text{Pb}^{2+} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4(s)\) and \(2 \text{Ag}^+ + 2 \text{Cl}^- \rightarrow 2 \text{AgCl}(s)\). All ionic concentrations become zero at equilibrium due to complete reactions.

Step by step solution

01

Identify the Possible Reaction

When mixing solutions of Ag鈧係O鈧� and PbCl鈧�, we need to identify if any double displacement reactions occur. The possible exchange reactions are: 1. Ag鈧係O鈧� + PbCl鈧� 鈫� PbSO鈧� + 2 AgCl We need to determine which, if any, of these products are insoluble.
02

Determine the Solubility of Potential Products

Using a solubility chart: - PbSO鈧� is known to be poorly soluble in water. - AgCl is also known to be poorly soluble in water. Therefore, both products could precipitate out of solution.
03

Write the Ionic Equation

The total ionic equation for the potential formation of precipitates is:\[2 \text{Ag}^+ + \text{SO}_4^{2-} + \text{Pb}^{2+} + 2\text{Cl}^- \rightarrow \text{PbSO}_4(s) + 2\text{AgCl}(s)\]
04

Write the Net Ionic Equation

The net ionic equation removes spectator ions (those that do not participate in the reaction). Here, all ions are involved in forming a solid, so:\[\text{Pb}^{2+} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4(s) \text{ and } 2 \text{Ag}^+ + 2 \text{Cl}^- \rightarrow 2 \text{AgCl}(s)\]This expresses the precipitation reactions.
05

Calculate Initial Concentrations

Calculate initial concentrations before mixing:- For Ag鈧係O鈧� in 75.0 mL :: \([\text{Ag}^+] = 0.0150 \times \frac{50}{75} = 0.0100 \text{M}\)- For SO鈧劼测伝 in 75.0 mL :: \([\text{SO}_4^{2-}] = 0.0075 \text{M}\)- For Pb虏鈦� in 75.0 mL :: \([\text{Pb}^{2+}] = 0.0100 \times \frac{25}{75} = 0.00333 \text{M}\)- For Cl鈦� in 75.0 mL :: \([\text{Cl}^-] = 0.00667 \text{M}\)
06

Determine Limiting Reactant

Use stoichiometry to find the limiting reactant. - Balancing act for Pb虏鈦� with SO鈧劼测伝 gives 0.00333 M Pb虏鈦� utilized. - For Cl鈦� with Ag鈦�, 0.00667 M must balance against twice that of Ag鈦�. It requires every 0.0 M Cl鈦� consumed per mole. Given that all ion concentrations precipitate, all 0.0100 M Ag鈦� will precipitate making it excess. This indicates Ag鈦� and SO鈧劼测伝 are completely consumed.
07

Calculate Ions at Equilibrium

Since the limitations have indicated precisely how much of each reactant is consumed, the remaining concentrations in the mixture are:\[[\text{Ag}^+] = 0.000 M , [\text{SO}_4^{2-}] = 0.000 M ,[\text{Pb}^{2+}] = 0.000 M [\text{Cl}^-] = 0.000 M \] The reactions' theoretical complete precipitation makes any simulation not feasibility resulting in an equilibrium mixture of all zeros.

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 91影视!

Key Concepts

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

Solubility Rules
Understanding solubility rules is crucial for predicting if a reaction will occur when mixing different solutions. Solubility rules help us determine whether compounds are soluble or insoluble in water. This is essential in identifying precipitates formed in a reaction.

Here are some basic rules:
  • Compounds containing alkali metal ions (like K鈦�, Na鈦�) and ammonium ion (NH鈧勨伜) are generally soluble.
  • Nitrates (NO鈧冣伝) and acetates (CH鈧僀OO鈦�) are usually soluble.
  • Sulfates (SO鈧劼测伝) are soluble, except when combined with lead (Pb虏鈦�), barium (Ba虏鈦�), and strontium (Sr虏鈦�).
  • Chlorides (Cl鈦�), bromides (Br鈦�), and iodides (I鈦�) are soluble except when bound to silver (Ag鈦�), lead (Pb虏鈦�), and mercury (Hg鈧偮测伜).
In our example, such rules tell us that both AgCl and PbSO鈧� will precipitate since they are not soluble in water. Knowing these rules enables chemists to make predictions about whether a precipitate will form when two ionic compounds are mixed.
Precipitation Reactions
Precipitation reactions occur when two soluble salts combine in solution to form one or more insoluble products, known as precipitates. This is a typical double displacement reaction where the cations and anions exchange partners.

For our case study involving \( ext{Ag}_2 ext{SO}_4\) and \( ext{PbCl}_2\), a double replacement reaction suggests:
  • Ag鈦� from silver sulfate combines with Cl鈦� from lead chloride.
  • Pb虏鈦� from lead chloride combines with SO鈧劼测伝 from silver sulfate.
The chemical reactions are:
  • \[2 ext{Ag}^+ + 2 ext{Cl}^- ightarrow 2 ext{AgCl}(s)\]
  • \[ ext{Pb}^{2+} + ext{SO}_4^{2-} ightarrow ext{PbSO}_4(s)\]
Precipitation occurs when the product of the reaction forms an insoluble solid that settles out of the solution. Observing whether a solid forms helps confirm the occurrence of a precipitation reaction.
Chemical Stoichiometry
Chemical stoichiometry involves the calculation of reactants and products in chemical reactions. It provides a quantitative means to predict the amounts of reactants consumed and products formed.

Using initial concentrations and balanced equations in stoichiometry helps identify limiting reactants鈥攖hose entirely consumed in the reaction, thus restricting the amount of product formed.
  • In our scenario, we determine initial concentrations for each ion.
  • Perform stoichiometric calculations to find the limiting reactant, which affects the completion of the reaction.
  • By calculating the stoichiometry, we identify that all ions in the mixture are completely precipitated, resulting in zero concentration at equilibrium.
In precise stoichiometric terms, knowing the balanced equations allows for the prediction of the quantities of all substances involved, ensuring that calculations match real-life quantities achieved when reactions run to completion.

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

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 $$ \begin{array}{l} \quad \mathrm{A}_{3} \mathrm{~B}_{2}(s) \rightleftharpoons 3 \mathrm{~A}^{2+}(a q)+2 \mathrm{~B}^{3-}(a q) \\ \text { a) } 3.7 \times 10^{-17} \\ \text {b) } 9.1 \times 10^{-25} \\ \text {c) } 9.1 \times 10^{-40} \\ \text {d) } 3.7 \times 10^{-32} \\ \text {e) } 1.4 \times 10^{-33} \end{array} $$

A solution is \(1.8 \times 10^{-4} M \mathrm{Co}^{2+}\) and \(0.20 \mathrm{M} \mathrm{HSO}_{4}\) The solution also contains \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). What should be the minimum molarity of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) to prevent the precipitation of cobalt(II) sulfide when the solution is saturated with hydrogen sulfide \(\left(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\right)\) ?

You are given a solution of the ions \(\mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}\), and \(\mathrm{Ba}^{2+}\). Devise a scheme to separate these ions using sodium sulfate. Note that magnesium sulfate is soluble.

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

You are given a saturated solution of lead(II) chloride. Which one of the following solutions would be most effective in yielding a precipitate when added to the lead(II) chloride solution? a) \(0.1 M \mathrm{NaCl}(a q)\) b) saturated \(\mathrm{PbS}(a q)\) c) \(0.1 M \mathrm{NaSO}_{4}(a q)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation
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.