/*! 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 53 Separate samples of a solution o... [FREE SOLUTION] | 91影视

91影视

Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\). A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

Short Answer

Expert verified
The cations that could be present in the unknown soluble ionic compound are barium (Ba虏鈦) and strontium (Sr虏鈦), as they form a precipitate with SO鈧劼测伝 but not with Cl鈦 or OH鈦.

Step by step solution

01

Identify possible precipitation reactions

When KCl, Na2SO4, and NaOH are added to the solution, the ions may react with the anions/cations to form precipitates. The precipitation reactions can be defined as: 1. KCl added: K鈦 + unknown anion 鈫 precipitate 2. Na2SO4 added: 2Na鈦 + unknown cation 鈫 precipitate 3. NaOH added: Na鈦 + unknown anion 鈫 precipitate
02

Analyze possible cations in the given scenarios

In scenario 1 and 3, no precipitate forms, meaning the unknown anion doesn't react with K鈦 or OH鈦 ions. In scenario 2, the formation of a precipitate indicates that the unknown cation reacts with SO鈧劼测伝 ions. The common cations that react with SO鈧劼测伝 ions to form precipitates are Ba虏鈦, Sr虏鈦, Pb虏鈦, and Ca虏鈦. Thus, the unknown soluble ionic compound must have a cation that forms a precipitate when SO鈧劼测伝 is added but not when Cl鈦 or OH鈦 is added. Let's analyze the possible precipitation reactions for each cation.
03

Check the solubility rules for the possible cations

Ba虏鈦 - BaCl鈧: soluble (Ba虏鈦 has no precipitate with Cl鈦) - BaSO鈧: insoluble (formation of precipitate) - Ba(OH)鈧: soluble (Ba虏鈦 has no precipitate with OH鈦) Sr虏鈦 - SrCl鈧: soluble (Sr虏鈦 has no precipitate with Cl鈦) - SrSO鈧: insoluble (formation of precipitate) - Sr(OH)鈧: soluble (Sr虏鈦 has no precipitate with OH鈦) Pb虏鈦 - PbCl鈧: insoluble (Pb虏鈦 forms a precipitate with Cl鈦) - PbSO鈧: insoluble (formation of precipitate) - Pb(OH)鈧: insoluble (Pb虏鈦 forms a precipitate with OH鈦) Ca虏鈦 - CaCl鈧: soluble (Ca虏鈦 has no precipitate with Cl鈦) - CaSO鈧: slightly insoluble (formation of precipitate) - Ca(OH)鈧: insoluble (Ca虏鈦 forms a precipitate with OH鈦)
04

Match the cations with the observed reactions

Observations: 1. No precipitate with KCl (no precipitate with Cl鈦) 2. Precipitate with Na2SO4 (precipitate with SO鈧劼测伝) 3. No precipitate with NaOH (no precipitate with OH鈦) From the solubility rules analyzed in step 3, the cations that match the observations are Ba虏鈦 and Sr虏鈦. Ba虏鈦 1. No precipitate with Cl鈦 2. Precipitate with SO鈧劼测伝 3. No precipitate with OH鈦 Sr虏鈦 1. No precipitate with Cl鈦 2. Precipitate with SO鈧劼测伝 3. No precipitate with OH鈦
05

Conclusion

The cations that could be present in the unknown soluble ionic compound are barium (Ba虏鈦) and strontium (Sr虏鈦).

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.

Ionic Compounds
Ionic compounds are formed from the electrostatic attraction between positively charged ions, known as cations, and negatively charged ions, called anions. These bonds occur between metals and non-metals. Upon dissolving in water, ionic compounds dissociate into their constituent ions. For instance, a compound like sodium chloride (NaCl) will separate into sodium ions (Na鈦) and chloride ions (Cl鈦) when dissolved.
Recognizing ionic compounds is crucial because their behavior in solution is fundamental to understanding reactions like precipitation. Ionic compounds have distinct properties, such as high melting and boiling points, and they often conduct electricity when dissolved in water. This conductive property is key to many chemical processes and analyses, such as cation identification in solutions.
When working with ionic compounds, knowledge of their dissociation can help predict outcomes of reactions, like whether a precipitate will form or what ions are present in solution. This is particularly beneficial when dealing with unknown solutions, as it aids in identifying potential ions through reactions.
Precipitation Reactions
Precipitation reactions involve the formation of an insoluble solid, or precipitate, when solutions containing soluble salts are mixed. These reactions are a subset of double replacement reactions, where ions switch partners to form new compounds. A classic example is the reaction between barium chloride (BaCl鈧) and sodium sulfate (Na鈧係O鈧), which results in the formation of barium sulfate (BaSO鈧) as a precipitate.
In the context of identifying unknown ionic compounds, precipitation reactions are a useful tool. By adding specific reagents to a solution, we can determine which ions are present based on whether a precipitate forms. This requires knowing which combinations of ions result in insoluble compounds.
  • A precipitate with Na鈧係O鈧 suggests the possible presence of cations like Ba虏鈦 or Sr虏鈦 that form insoluble sulfates.
  • No precipitate with Cl鈦 or OH鈦 further narrows down the possibilities.
By monitoring which mixtures form precipitates, chemists can use these reactions strategically to analyze and identify unknown solutions.
Solubility Rules
Solubility rules are essential guidelines that help predict the solubility of ionic compounds in water. These rules aren't hard and fast, but they offer a good approximation for most circumstances. For example:
  • Sulfates (SO鈧劼测伝) are generally soluble, with exceptions like BaSO鈧, SrSO鈧, and PbSO鈧, which are insoluble.
  • Chlorides (Cl鈦) are typically soluble, but not when paired with Ag鈦, Pb虏鈦, and Hg鈧偮测伜.
  • Hydroxides (OH鈦) tend to be insoluble, except for those of alkali metals and some alkaline earth metals like Ba虏鈦 and Sr虏鈦.
Understanding these rules allows us to predict whether a given ionic compound will dissolve or form a precipitate. For instance, the formation of a precipitate when Na鈧係O鈧 is added indicates the presence of cations like Ba虏鈦 or Sr虏鈦.
These rules are fundamental to performing accurate cation analysis and solving chemical problems involving solubility and precipitation. They aid in narrowing down potential ions in a solution by showing how they interact with known anions.
Cation Analysis
Cation analysis involves identifying which cations are present in a sample through the use of various chemical reactions, including precipitation reactions. This technique is crucial in areas such as quality control, pollution monitoring, and laboratory analysis.
In the given exercise, the absence or presence of a precipitate provides clues about the cations involved:
  • No precipitate with KCl or NaOH rules out cations that form insoluble compounds with Cl鈦 or OH鈦.
  • A precipitate with Na鈧係O鈧 suggests a reaction with SO鈧劼测伝, indicating cations like Ba虏鈦 or Sr虏鈦 could be present.
By understanding the solubility behavior and reaction tendencies of cations, such as their proclivity to react with sulfate ions to form a precipitate, chemists can systematically identify the ions in a solution. Cation analysis often utilizes a sequence of specific tests, each contributing to a more conclusive identification of unknown cations. This methodical approach enables accurate and reliable determination of ionic compositions in various chemical samples.

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

What mass of silver chloride can be prepared by the reaction of \(100.0 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) silver nitrate with \(100.0 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The \(\mathrm{VO}^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q)\) To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

What mass of solid \(\mathrm{AgBr}\) is produced when \(100.0 \mathrm{~mL}\) of \(0.150 \mathrm{MAgNO}_{3}\) is added to \(20.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaBr} ?\)

A \(230 .-\mathrm{mL}\) sample of a \(0.275-M \mathrm{CaCl}_{2}\) solution is left on a hot plate overnight; the following morning, the solution is \(1.10 M\). What volume of water evaporated from the \(0.275 \mathrm{M} \mathrm{CaCl}_{2}\) solution?

Using the general solubility rules given in Table 4.1, name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions. a. chloride ion b. calcium ion c. iron(III) ion d. sulfate ion e. mercury(I) ion, \(\mathrm{Hg}_{2}{ }^{2+}\) f. silver ion

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.