crsch04-concept

Chapter 4

Types of Chemical

Reactions and Solution

Stoichiometry

QUESTION

Which statement below best describes the connection between water and electrolytes?

1. Polar water molecules enable weak electrolytes to dissociate completely to brightly light a conductivity apparatus.

2. Strong electrolytes do not need to become attached to polar water molecules to conduct electricity.

3. Non-electrolytes do not interact with polar water.

4. In strong electrolytes conductivity can arise from dissociation or ionization.

Copyright © Houghton Mifflin Company. All rights reserved.

CRS Question, 4 –2

ANSWER

Choice 4 is correct. Strong acids such as HCl are not ionic until water interacts with HCl molecules to facilitate ionization. Ionic salts dissociate into separate ions.

Section 4.2: The Nature of Aqueous Solutions: Strong and Weak

Electrolytes



QUESTION

If 250.0 mL of 0.200 M NaCl were evaporated without the loss of any solute, how many grams of NaCl would remain?

1. 2.92 g

2. 11.6 g

3. 2,900 g

4. I do not know how to use these values to determine the answer.



ANSWER

Choice 1 provides the correct number of grams. When volume, in liters, is multiplied by molarity, moles of solute can be determined.

Once this is known it can be converted to grams using the solute’s molar mass.

Section 4.3: The Composition of Solutions



QUESTION

The acid in car batteries is H

2

SO

4

(sulfuric acid). Besides water, what reaction product would likely form when sulfuric acid reacts with calcium hydroxide? Would the product likely be a strong or weak electrolyte?

1. Ca(SO

4

)

2

2. CaSO

4

3. CaSO

4

4. Ca(SO

4

)

2 weak weak strong strong



ANSWER

Choice 2 indicates the correct formula and rating for electrolyte behavior. The anion of the acid reacts with the cation of the base.

Ca has a +2 ionic charge so one sulfate ion (with –2 charge) is needed to balance the calcium charge. CaSO

4 is not very soluble.

Section 4.2: The Nature of Aqueous Solutions: Strong and Weak

Electrolytes

Section 4.5: Precipitation Reactions



QUESTION

A 0.14 M solution of NaCl provides the same concentration of NaCl typically found in blood serum. If you had a 0.50 M solution of

NaCl but wanted to produce 200.0 mL of 0.14 M NaCl, how many mL of the concentrated solution would be needed to make the dilution?

1. 28 mL

2. 56 mL

3. 14 mL

4. 200 mL



ANSWER

Choice 2 provides the correct number of mL of the more concentrated solution. The first step is to determine the number of moles of NaCl needed in the final solution (0.2000 L  0.14 mol/L

= 0.028 moles), then determine the volume of the concentrated solution that contains that many moles. (0.028 moles = V  0.50 mol/L)

Section 4.3: The Composition of Solutions



QUESTION



QUESTION (continued)

Of the following compounds which, when placed in distilled water, would give nearly the same conductivity result as pure water alone?

1. CuSO

4

2. NH

4

NO

3. AgCl

3

4. FeCl

3



ANSWER

Choice 3 represents an insoluble compound. The proper use of the solubility rules chart reveals that most chlorides dissociate in water.

AgCl does not appreciably do so.

Section 4.5: Precipitation Reactions



QUESTION



QUESTION (continued)

If you began a reaction with the following ions in solution (all would be written with an aq subscript), how would you represent the proper final net ionic equation?

6Na + + 2PO

4

3– + 3Fe 2+ + 6NO

3

– 

1. 3Na

2. 6Na

3. 3Na + + PO

4. 2PO

+

+

+ PO

4

3–

4

3–

+ 2PO

4

3–

4

3–

+ Fe 2+ + 2NO

+ 3Fe 2+ + 6NO

+ Fe 2+ + 2NO

+ 3Fe 2+  Fe

3

(PO

4

)

3

–

3

–

 No Reaction

2

3

–

 Fe

( s )

 Fe

3

(PO

3

(PO

4

)

2

4

)

2

( s )+ 6NaNO

3

( s )+ 6 Na + + 6 NO

3

–



ANSWER

Choice 4 correctly depicts the reacting ions and the resulting solid product. Use the chart to determine if an insoluble solid will be produced by the combination of ions given. If so, write that and balance the atoms. If soluble ions from strong electrolytes are present, they may cancel out from the left and right side of the equation.

Section 4.6: Describing Reactions in Solution



QUESTION

Given the insoluble compound Al

2

(CO

3

)

3 predict the ions and coefficients that would be necessary to complete the following net ionic equation:

__ _____ + __ _____  Al

2

(CO

3

)

3

1. 2 AlCl

3

+ 3 Na

2. 3 Al 3+ + 2 CO

3. 2 Al 3+ + 3 CO

3

2

2–

3

2–

CO

3 also include 6 NaCl on right

4. 2 Al 3+ 6 Cl – + 3 CO

3

2– + 6 Na +



ANSWER

Choice 3 correctly depicts the completed net ionic equation. Only those soluble reacting ions that form the compound need to be shown. The other ions could be considered spectator ions.

Section 4.6: Describing Reactions in Solution



QUESTION

Suppose a 250.0 mL water sample has a lead nitrate Pb(NO

3

)

2 concentration of 0.10 M . If you wanted to precipitate all the lead from the water you could add a solution containing NaCl to get

PbCl

2 to precipitate; thus removing the lead. How many mL of

0.50 M NaCl would you have to add to your sample to precipitate all the Pb 2+ ?

1. 50. mL

2. 1.0  10 2 mL

3. 2.0  10 2 mL

4. I know this should be possible, but I do not get any of those answers.



ANSWER

Choice 2 provides the consideration of stoichiometry and solution concentrations. First, consider that multiplying the volume  M of lead nitrate will provide the number of moles of Pb 2+ that need to be precipitated. Next, note that each lead ion will need two chloride anions. To keep that 1:2 mole ratio, twice as many moles of Cl – than Pb 2+ are required. Once the Cl – anions are known, then the volume containing that number of moles can be calculated.

Moles = V  M

Section 4.7: Stoichiometry of Precipitation Reactions



QUESTION

In order to facilitate x-ray examinations of some internal organs, patients drink a solution containing nearly insoluble BaSO

4

. Good thing this is insoluble, otherwise we would be exposed to toxic barium ions! How many moles of BaSO mixing 250.0 mL of 0.200 M Ba(OH)

2

H

2

SO

4

? What is the limiting reagent?

4 could be precipitated from with 125.0 mL of 0.100 M

1. 0.0500 moles; barium hydroxide is the limiting reagent.

2. 0.0500 moles; sulfuric acid is the limiting reagent.

3. 0.0125 moles; barium hydroxide is the limiting reagent.

4. 0.0125 moles; sulfuric acid is the limiting reagent.



ANSWER

Choice 4 properly shows both the number of moles and identifies the limiting reagent. Volume (in liters)  M = moles. This needs to be done for both reagents. Then the smallest can be used to predict the maximum yield of product (because this reaction has 1:1 stoichiometry).

Section 4.7: Stoichiometry of Precipitation Reactions



QUESTION

Phosphoric acid is often found in soft drinks. How many mL of

0.100 M NaOH would be required to neutralize 300.0 mL of

0.0010 M phosphoric acid?

1. 30. mL

2. 3.0  10 3 mL

3. 9.0 mL

4. 10. mL



ANSWER

Choice 3 properly considers the concentrations and stoichiometry.

Phosphoric acid is H

3

PO

4 so each mole requires 3 moles of NaOH to neutralize. For phosphoric acid: V  M = moles of phosphoric acid initially. Three times this shows how many moles of NaOH are needed. Then the moles of NaOH can be used to determine the volume needed from a 0.100 M solution.

Section 4.8: Acid–Base Reactions



QUESTION

Benzoic acid (HC

7

H

5

O

2

; molar mass = 122 g/mol) can be used to make the important food preservative sodium benzoate. Benzoic acid and sodium hydroxide react in a 1:1 mol ratio. Suppose a food scientist had 1.00 gram of an impure sample of benzoic acid.

When dissolved, the sample required 42.0 mL of 0.185 M NaOH solution to neutralize. What percent of the impure sample is benzoic acid?

1. 94.8%

2. 0.77%

3. 27.7%

4. 77.0%



ANSWER

Choice 1 represents the percent of the 1.00 gram sample that is benzoic acid. Multiplying the volume (in liters) of the NaOH solution used by its molarity yields the number of moles of NaOH required for neutralization. Since stoichiometry is 1:1 this is also the number of moles of benzoic acid in the sample. This can be converted to grams and divided by the mass of the sample (times

100) to provide the % of the sample that is actually benzoic acid.

Section 4.8: Acid–Base Reactions



QUESTION

In a redox reaction, oxidation and reduction must both occur.

Which statement provides an accurate premise of redox chemistry?

1. The substance that is oxidized must be the oxidizing agent.

2. The substance that is oxidized must gain electrons.

3. The substance that is oxidized must have a higher oxidation number afterwards.

4. The substance that is oxidized must combine with oxygen.



ANSWER

Choice 3 displays the proper connection between oxidation and oxidation number. During oxidation negatively charged electrons leave the atom thereby increasing the oxidation number.

Section 4.9: Oxidation–Reduction Reactions



QUESTION

One reason gold is so prized is because it seldom reacts. However, a mixture of nitric acid and hydrochloric acid (called aqua regia; i.e. king water) will, through a redox reaction, dissolve gold. When the following redox equation is properly balanced, how many H + ions will appear on the left side of the equation?

Au( s ) + NO

3

– ( aq ) + Cl – ( aq )  AuCl

4

– ( aq ) + NO

2

( g )

1. 0

2. 2

3. 4

4. 6



ANSWER

Choice 4 reflects the correct H + ion count for the balanced equation.

6H + ( aq ) + Au( s ) + 3NO

3

( aq ) + 4Cl ( aq ) 

AuCl

4

( aq ) + 3NO

2

( g ) + 3H

2

O( l )

Section 4.10: Balancing Oxidation–Reduction Equations



QUESTION

Oxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO

4

– for oxalate.

could be used to analyze samples

MnO

4

– + C

2

O

4

2–  MnO

2

+ CO

3

2– (basic solution)

When properly balanced, how many OH – ions are present?

1. 1

2. 2

3. 3

4. 4



ANSWER

Choice 4 is correct. The redox equation could be balanced according to the steps for an acid solution, but then OH – must be added to neutralize the H + ions. When done properly 4 OH – ions will be present in the basic equation.

4OH + 2MnO

4

+ 3C

2

O

4

2 2MnO

2

+ 6CO

3

2+ 2H

2

O

Section 4.10: Balancing Oxidation–Reduction Equations



crsch04-concept

Chapter 4

Types of Chemical

Reactions and Solution

Stoichiometry

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Types of Chemical

Reactions and Solution

Stoichiometry

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