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Chapter 4: Problem 87

A student is asked to identify the metal nitrate present in an aqueous solution. The cation in the solution can either be \(\mathrm{Na}^{+}, \mathrm{Ba}^{2+}, \mathrm{Ag}^{+}\), or \(\mathrm{Ni}^{2+}\). Results of solubility experiments are as follows: unknown + chloride ions-no precipitate unknown + carbonate ions-precipitate unknown + sulfate ions-precipitate What is the cation in the solution?

Short Answer

Expert verified
Answer: The cation present in the metal nitrate solution is \(\mathrm{Ba}^{2+}\) (barium).

Step by step solution

01

Write down the possible metal nitrate cations

There are four possible metal nitrate cations in the aqueous solution: \(\mathrm{Na}^{+}, \mathrm{Ba}^{2+}, \mathrm{Ag}^{+}\), or \(\mathrm{Ni}^{2+}\).
02

Analyze the first experiment (unknown + chloride ions)

In the first experiment, no precipitate forms when the unknown cation reacts with chloride ions. This indicates that the unknown cation does not form an insoluble compound with chloride ions. From solubility rules, we know that the following compound is insoluble: \(\mathrm{AgCl}\) (silver chloride). Therefore, \(\mathrm{Ag}^{+}\) can be ruled out as the unknown cation.
03

Analyze the second experiment (unknown + carbonate ions)

In the second experiment, a precipitate forms when the unknown cation reacts with carbonate ions. This indicates that the unknown cation forms an insoluble compound with carbonate ions. From solubility rules, we know that the following compounds are insoluble with carbonate ions: \(\mathrm{BaCO_3}\) (barium carbonate) and \(\mathrm{NiCO_3}\) (nickel carbonate). This means the unknown cation could be \(\mathrm{Ba}^{2+}\) or \(\mathrm{Ni}^{2+}\). Since we already ruled out \(\mathrm{Ag}^{+}\), \(\mathrm{Na}^{+}\) can be ruled out now as well, as it forms a soluble compound with carbonate ions.
04

Analyze the third experiment (unknown + sulfate ions)

In the third experiment, a precipitate forms when the unknown cation reacts with sulfate ions. This indicates that the unknown cation forms an insoluble compound with sulfate ions. From the solubility rules, we know that the following compound is insoluble with sulfate ions: \(\mathrm{BaSO_4}\) (barium sulfate). Since we already ruled out \(\mathrm{Ag}^{+}\) and \(\mathrm{Na}^{+}\), and we know that the unknown cation forms an insoluble compound with both carbonate and sulfate ions, it must be \(\mathrm{Ba}^{2+}\), as \(\mathrm{Ni}^{2+}\) forms a soluble compound with sulfate ions.
05

Identify the cation in the solution

Based on the analysis of the three solubility experiments, the unknown cation in the metal nitrate solution is \(\mathrm{Ba}^{2+}\).

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Most popular questions from this chapter

Using squares to represent atoms of one element (or cations) and circles to represent the atoms of the other element (or anions), represent the principal species in the following pictorially. (You may represent the hydroxide anion as a single circle.) (a) a solution of \(\mathrm{HCl}\) (b) a solution of HF (c) a solution of \(\mathrm{KOH}\) (d) a solution of \(\mathrm{HNO}_{2}\)

Three students titrate different samples of the same solution of \(\mathrm{HCl}\) to obtain its molarity. Below are their data. Student \(\mathrm{A}: \quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student B: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+40.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student C: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) used to titrate to the equivalence point. All the students calculated the molarities correctly. Which (if any) of the following statements are true? (a) The molarity calculated by \(A\) is half that calculated by \(B\). (b) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{C}\). (c) The molarity calculated by B is twice that calculated by C. (d) The molarity calculated by \(\mathrm{A}\) is twice that calculated by \(\mathrm{B}\). (e) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{B}\). Challenge Problems

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), in a powder for stomach upsets is found by titrating with \(0.275 M\) hydrochloric acid. If \(15.5 \mathrm{~mL}\) of hydrochloric acid is required to react with \(0.500 \mathrm{~g}\) of the sample, what is the percentage of sodium hydrogen carbonate in the sample? The balanced equation for the reaction that takes place is $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

What is the molarity of each ion present in aqueous solutions prepared by dissolving \(20.00 \mathrm{~g}\) of the following compounds in water to make 4.50 L of solution? (a) cobalt(III) chloride (b) nickel(III) sulfate (c) sodium permanganate (d) iron(II) bromide

Write the formulas of the following compounds and decide which are soluble in water. (a) sodium sulfate (b) iron(III) nitrate (c) silver chloride (d) chromium(III) hydroxide
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