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Chapter 4: Problem 27

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could be present in the unknown salt solution: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+} ?\)

Short Answer

Expert verified
The cation present in the unknown salt solution is Pb\(^{2+}\) because it forms precipitates when reacting with all three given solutions (HBr, H\(_2\)SO\(_4\), and NaOH) according to their solubility rules.

Step by step solution

01

Analyze the solubility rules

To identify which cation is present in the unknown salt, we will examine the solubility rules for the possible combinations of cations and anions. This will help us to eliminate any cations that do not form precipitates with the given anions.
02

Examine possible reactions with HBr

Analyze the possible reactions between the cations and the anion from HBr (Br-) and determine which would form a precipitate: - K+ and Br- : When combined, they form KBr, which is soluble and does not form a precipitate (according to solubility rules). - Pb2+ and Br- : When combined, they form PbBr2, which is insoluble and forms a precipitate. - Ba2+ and Br- : When combined, they form BaBr2, which is soluble and does not form a precipitate. Thus, only the combination of Pb2+ and Br- forms a precipitate in this case.
03

Examine possible reactions with H2SO4

Now, analyze the possible reactions between the cations and the anion from H2SO4 (SO42-) and determine which would form a precipitate: - K+ and SO42- : When combined, they form K2SO4, which is soluble and does not form a precipitate. - Pb2+ and SO42- : When combined, they form PbSO4, which is insoluble and forms a precipitate. - Ba2+ and SO42- : When combined, they form BaSO4, which is insoluble and forms a precipitate. Both Pb2+ and Ba2+ form precipitates in this case.
04

Examine possible reactions with NaOH

Lastly, analyze the possible reactions between the cations and the anion from NaOH (OH-) and determine which would form a precipitate: - K+ and OH- : When combined, they form KOH, which is soluble and does not form a precipitate. - Pb2+ and OH- : When combined, they form Pb(OH)2, which is insoluble and forms a precipitate. - Ba2+ and OH- : When combined, they form Ba(OH)2, which is soluble (at high concentrations) and does not form a precipitate. In this case, only the combination of Pb2+ and OH- forms a precipitate.
05

Identify the cation present in the unknown salt

Since the problem states that a precipitate forms in all three cases (with HBr, H2SO4, and NaOH), we can now conclude that the cation present in the unknown salt solution is Pb2+ because it is the only cation that forms precipitates in all three reactions.

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Most popular questions from this chapter

Calculate the concentration of each ion in the following solutions obtained by mixing: (a) \(32.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{KMnO}_{4}\) (b) \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{ZnCl}_{2}^{+}\) with \(15.0 \mathrm{~mL}\) of \(0.60 \mathrm{MKMnO}_{4}\) with \(5.0 \mathrm{~mL}\) of $0.200 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) 4.2 \mathrm{~g}$ of \(\mathrm{CaCl}_{2}\) in \(150.0 \mathrm{~mL}\) of \(0.02 M \mathrm{KCl}\) solution. Assume that the volumes are additive.

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or neutralization reactions. (a) $\mathrm{P}_{4}(s)+10 \mathrm{HClO}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ (b) \(\mathrm{Br}_{2}(l)+2 \mathrm{~K}(s) \longrightarrow 2 \mathrm{KBr}(s)\) (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)$ (d) $\mathrm{ZnCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+$

(a) A caesium hydroxide solution is prepared by dissolving \(3.20 \mathrm{~g}\) of \(\mathrm{CsOH}\) in water to make \(25.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Then, the caesium hydroxide solution prepared in part (a) is used to titrate a hydroiodic acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between the caesium hydroxide and hydroiodic acid solutions. (c) If \(18.65 \mathrm{~mL}\) of the caesium hydroxide solution was needed to neutralize a $42.3 \mathrm{~mL}$ aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?

Determine the oxidation number for the indicated element in each of the following substances: (a) \(\mathrm{N}\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\), (b) \(\mathrm{Nin} \mathrm{NO}_{2}\), (c) \(\mathrm{Mn}\) in \(\mathrm{MnCl}_{3}\) (d) Fe in \(\mathrm{FeSO}_{4^{\prime}}\) (e) \(\mathrm{Pt}\) in \(\mathrm{PtCl}_{4}\) (f) \(\mathrm{Cl}\) in \(\mathrm{NaClO}_{4}\).

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing $1.00 \times 10^{2} \mathrm{~mL}\( of \)0.0105 \mathrm{MHCl} .\( The \)\mathrm{NH}_{3}$ reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?
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