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Chapter 4: Problem 28

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{BaCl}_{2}\). Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: \(\mathrm{Br}^{-}, \mathrm{CO}_{3}{ }^{2-}, \mathrm{NO}_{3}{ }^{-}\)?

Short Answer

Expert verified
The anion of the unknown salt must form precipitates with AgNO3, Pb(NO3)2, and BaCl2. Based on our analysis, Br- and CO3^2- can form precipitates with these compounds, while NO3- cannot. Thus, the anion of the unknown salt could be either Br- or CO3^2-.

Step by step solution

01

Analyze the reactivity of Br-

In the presence of AgNO3, Br- would react as follows: AgNO3(aq) + XBr(aq) 鈫� AgBr(s) + XNO3(aq) AgBr is almost insoluble and would precipitate. Br- would also react with Pb(NO3)2 and BaCl2 in similar ways: Pb(NO3)2(aq) + 2XBr(aq) 鈫� PbBr2(s) + 2XNO3(aq) BaCl2(aq) + 2XBr(aq) 鈫� BaBr2(s) + 2XCl-(aq) Both PbBr2 and BaBr2 are slightly soluble in water, allowing the formation of precipitates. So, based on these reactions, Br- could possibly be the anion of the unknown salt.
02

Analyze the reactivity of CO3^2-

In the presence of AgNO3, CO3^2- would react as follows: 2AgNO3(aq) + XCO3^2-(aq) 鈫� Ag2CO3(s) + 2XNO3(aq) Ag2CO3 is almost insoluble and would precipitate. CO3^2- would also react with Pb(NO3)2 and BaCl2 in similar ways: Pb(NO3)2(aq) + XCO3^2-(aq) 鈫� PbCO3(s) + 2XNO3(aq) BaCl2(aq) + XCO3^2-(aq) 鈫� BaCO3(s) + 2XCl-(aq) Both PbCO3 and BaCO3 are also almost insoluble in water, allowing the formation of precipitates. So, based on these reactions, CO3^2- could also possibly be the anion of the unknown salt.
03

Analyze the reactivity of NO3-

In the presence of AgNO3, NO3- would react as follows: AgNO3(aq) + XNO3(aq) 鈫� AgNO3(aq) + XNO3(aq) Since AgNO3 is soluble in water, there wouldn't be any precipitate formed with NO3-. The same will happen with reactions of NO3- with Pb(NO3)2 and BaCl2. So, NO3- cannot be the anion of the unknown salt as it doesn't form precipitates.
04

Conclusion

The anion of the unknown salt must form precipitates with AgNO3, Pb(NO3)2, and BaCl2. Based on our analysis, Br- and CO3^2- can form precipitates with these compounds, while NO3- cannot. Thus, the anion of the unknown salt could be either Br- or CO3^2-.

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Key Concepts

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

Precipitation Reactions
Precipitation reactions occur when two soluble salts react in aqueous solution to form an insoluble solid, called a precipitate. When you mix solutions containing ionic compounds, they dissociate into individual ions in water. If a combination of these ions forms an insoluble compound, a precipitate forms, visually signaling the reaction.

In the given exercise, unknown ions react with
  • 础驳狈翱鈧� to form AgBr or Ag鈧侰O鈧�
  • 笔产(狈翱鈧�)鈧� to form PbBr鈧� or PbCO鈧�
  • 叠补颁濒鈧� to form BaBr鈧� or BaCO鈧�
These reactions result in a precipitate because the products are not soluble in water. Identifying which precipitates form helps determine the anion in the unknown ionic compound.
Anion Identification
Identifying an anion involves testing reactions that can distinctly reveal the presence of specific ions. In this exercise, the goal is to determine which anion forms precipitates with silver, lead, and barium compounds.

By forming a precipitate with
  • 础驳狈翱鈧�
  • 笔产(狈翱鈧�)鈧�
  • 叠补颁濒鈧�
Bromide (Br鈦�) and carbonate (CO鈧兟测伝) ions indicate their presence. Since these ions create insoluble compounds, observing precipitates cues us to their identity. In contrast, nitrate (NO鈧冣伝) does not form a precipitate, helping rule out its presence in the tested compound.

Hence, testing which reactions result in precipitation provides the necessary clues for identifying the unknown anion.
Chemical Solubility
Understanding chemical solubility is key to predicting which compounds will form precipitates. Solubility rules tell us which ionic compounds dissolve in water and which do not. Solubility is influenced by factors like lattice energy of the solid and the hydration energy of the ions.

For example, most nitrate ( NO鈧冣伝) compounds are soluble, so they don't typically form precipitates. Bromides ( Br鈦�), however, are only slightly soluble with lead and silver, and most carbonates (like CO鈧兟测伝) are insoluble, forming a solid in solution when mixed with ions like Ag鈦�, Pb虏鈦�, or Ba虏鈦�.

This knowledge allows chemists to predict precipitation reactions and identify unknown ions by observing results in a controlled setting.

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Most popular questions from this chapter

A solution of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is mixed with a solution of \(200.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NiSO}_{4}\). (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

In each of the following pairs, indicate which has the higher concentration of I \({ }^{-}\)ion: (a) \(0.10 \mathrm{M} \mathrm{BaI}_{2}\) or \(0.25 \mathrm{M} \mathrm{KI}\) solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KI}\) solution or \(200 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{ZnI}\) solution, (c) \(3.2 \mathrm{M}\) HI solution or a solution made by dissolving \(145 \mathrm{~g}\) of \(\mathrm{NaI}\) in water to make \(150 \mathrm{~mL}\) of solution.

A person suffering from hyponatremia has a sodium ion concentration in the blood of \(0.118 \mathrm{M}\) and a total blood volume of \(4.6 \mathrm{~L}\). What mass of sodium chloride would need to be added to the blood to bring the sodium ion concentration up to \(0.138 \mathrm{M}\), assuming no change in blood volume?

State whether each of the following statements is true or false. Justify your answer in each case. (a) Sulfuric acid is a monoprotic acid. (b) \(\mathrm{HCl}\) is a weak acid. (c) Methanol is a base.

A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If \(0.2815 \mathrm{~g}\) of barium sulfate was obtained, what was the mass percentage of barium in the sample?
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