5 Steps to a 5: AP Chemistry 2017 (2016)
Review the Knowledge You Need to Score High
Solutions and Colligative Properties
IN THIS CHAPTER
Summary: A solution is a homogeneous mixture composed of a solvent and one or more solutes. The solvent is the substance that acts as the dissolving medium and is normally present in the greatest amount. Commonly the solvent is a liquid, but it doesn’t have to be. Our atmosphere is a solution with nitrogen as the solvent; it is the gas present in the largest amount (79%). Many times, you will be dealing with a solution in which water is the solvent, an aqueous solution . The solute is the substance that the solvent dissolves and is normally present in the smaller amount. You may have more than one solute in a solution. For example, if you dissolved table salt (sodium chloride) and table sugar (sucrose) in water, you would have one solvent (water) and two solutes (sodium chloride and sucrose).
Some substances will dissolve in a particular solvent and others will not. There is a general rule in chemistry that states that “like dissolves like .” This general statement may serve as an answer in the multiple-choice questions, but does not serve as an explanation in the free-response questions. This simply means that polar substances (salts, alcohols, etc.) will dissolve in polar solvents such as water, and nonpolar solutes, such as iodine, will dissolve in nonpolar solvents such as carbon tetrachloride. The solubility of a particular solute is normally expressed in terms of grams solute per 100 mL of solvent (g/mL) at a specified temperature. The temperature must be specified because the solubility of a particular substance will vary with the temperature. Normally, the solubility of solids dissolving in liquids increases with increasing temperature, while the reverse is true for gases dissolving in liquids.
A solution in which one has dissolved the maximum amount of solute per given amount of solvent at a given temperature is called a saturated solution . An unsaturated solution has less than the maximum amount of solute dissolved. Sometimes, if the temperature, purity of the solute and solvent, and other factors are just right, you might be able to dissolve more than the maximum amount of solute, resulting in a supersaturated solution . Supersaturated solutions are unstable, and sooner or later separation of the excess solute will occur, until a saturated solution and separated solute remain.
The formation of a solution depends on many factors, such as the nature of the solvent, the nature of the solute, the temperature, and the pressure. Some of these factors were addressed in the Reactions and Periodicity chapter. In general, the solubility of a solid or liquid will increase with temperature and be unaffected by pressure changes. The solubility of a gas will decrease with increasing temperature and will increase with increasing partial pressure of the gas (Henry’s law ).
Keywords and Equations
π = osmotic pressure
i = van’t Hoff factor
K f = molal freezing-point depression constant
K b = molal boiling-point elevation constant
K f for H2 O = 1.86 K kg mol−1
K b for H2 O = 0.512 K kg mol−1
ΔT f = iKf molality
ΔT b = iKb molality
π = iMRT
molarity, M = moles solute per liter solution
molality, m = moles solute per kilogram solvent
There are many ways of expressing the relative amounts of solute(s) and solvent in a solution. The terms saturated , unsaturated , and supersaturated give a qualitative measure, as do the terms dilute and concentrated . The term dilute refers to a solution that has a relatively small amount of solute in comparison to the amount of solvent. Concentrated , on the other hand, refers to a solution that has a relatively large amount of solute in comparison to the solvent. However, these terms are very subjective. If you dissolve 0.1 g of sucrose per liter of water, that solution would probably be considered dilute; 100 g of sucrose per liter would probably be considered concentrated. But, what about 25 g per liter––dilute or concentrated? In order to communicate effectively, chemists use quantitative ways of expressing the concentration of solutions. Several concentration units are useful, including percentage, molarity, and molality.
One common way of expressing the relative amount of solute and solvent is through percentage, amount-per-hundred. Percentage can be expressed in three ways:
Mass (Sometimes Called Weight) Percentage
The mass percentage of a solution is the mass of the solute divided by the mass of the solution, multiplied by 100% to get percentage. The mass is commonly measured in grams.
mass % = (grams of solute/grams solution) × 100%
For example, a solution is prepared by dissolving 25.2 g of sodium chloride in 250.0 g of water. Calculate the mass percent of the solution.
A common error is forgetting to add the solute and solvent masses together in the denominator.
When solutions of this type are prepared, the solute and solvent are weighed out separately and then mixed together to form a solution. The final volume of the solution is unknown.
The mass/volume percent of a solution is the mass of the solute divided by the volume of the solution, multiplied by 100% to yield percentage. The volume of the solution is generally expressed in milliliters.
mass/volume % = (grams solute/volume of solution) × 100%
When mass/volume solutions are prepared, the grams of the solute are weighed out, dissolved, and diluted to the required volume.
For example, a solution is prepared by mixing 125.0 g of benzene with 250.0 g of toluene. The density of benzene is 0.8765 g/mL, and the density of toluene is 0.8669 g/mL. Determine the mass/volume percentage of the solution. Assume that the volumes are additive.
First, determine the volume of the solution.
Notice that it is not necessary to know the chemical formula of either constituent. A common error is forgetting to add the solute and solvent volumes together.
The third case is one in which both the solute and solvent are liquids. The volume percent of the solution is the volume of the solute divided by the volume of the solution, multiplied by 100% to generate the percentage.
volume % = (volume solute/volume solution) × 100%
When volume percent solutions are prepared, the mL of the solute are diluted with solvent to the required volume.
For example, determine the volume percentage of carbon tetrachloride in a solution prepared by dissolving 100.0 mL of carbon tetrachloride and 100.0 mL of methylene chloride in 750.0 mL of chloroform. Assume the volumes are additive.
A common error is not to add all the volumes together to get the volume of the solution.
If the solute is ethyl alcohol and the solvent is water, then another concentration term is used, proof. The proof of an aqueous ethyl alcohol solution is twice the volume percent. A 45.0 volume % ethyl alcohol solution would be 90.0 proof.
Percentage concentration is common in everyday life (3% hydrogen peroxide, 5% acetic acid, commonly called vinegar, etc.). The concentration unit most commonly used by chemists is molarity. Molarity (M ) is the number of moles of solute per liter of solution.
M = moles solute/liter solution
In preparing a molar solution, the correct number of moles of solute (commonly converted to grams using the molar mass) is dissolved and diluted to the required volume.
Determine the molarity of sodium sulfate in a solution produced by dissolving 15.2 g of Na2 SO4 in sufficient water to produce 750.0 mL of solution.
The most common error is not being careful with the units. Grams must be converted to moles, and milliliters must be converted to liters.
Another way to prepare a molar solution is by dilution of a more concentrated solution to a more dilute one by adding solvent. The following equation can be used:
(M before )(V before ) = (M after )(V after )
In the preceding equation, before refers to before dilution and after refers to after dilution.
Let’s see how to apply this relationship. Determine the final concentration when 500.0 mL of water is added to 400.0 mL of a 0.1111 M solution of HC1. Assume the volumes are additive.
The most common error is forgetting to add the two volumes.
Sometimes the varying volumes of a solution’s liquid component(s) due to changes in temperature present a problem. Many times volumes are not additive, but mass is additive. The chemist then resorts to defining concentration in terms of the molality. Molality (m ) is defined as the moles of solute per kilogram of solvent.
m = moles solute/kilograms solvent
Notice that this equation uses kilograms of solvent, not solution. The other concentration units use mass or volume of the entire solution. Molal solutions use only the mass of the solvent . For dilute aqueous solutions, the molarity and the molality will be close to the same numerical value.
For example, ethylene glycol (C2 H6 O2 ) is used in antifreeze. Determine the molality of ethylene glycol in a solution prepared by adding 62.1 g of ethylene glycol to 100.0 g of water.
The most common error is to use the total grams in the denominator instead of just the grams of solvent.
Electrolytes and Nonelectrolytes
An electrolyte is a substance that, when dissolved in a solvent or melted, conducts an electrical current. A nonelectrolyte does not conduct a current when dissolved. The conduction of the electrical current is usually determined using a light bulb connected to a power source and two electrodes. The electrodes are placed in the aqueous solution or melt, and if a conducting medium is present, such as ions, the light bulb will light, indicating the substance is an electrolyte.
The ions that conduct the electrical current can result from a couple of sources. They may result from the dissociation of an ionically bonded substance (a salt). If sodium chloride (NaCl) is dissolved in water, it dissociates into the sodium cation (Na+ ) and the chloride anion (Cl− ). But, certain covalently bonded substances may also produce ions if dissolved in water, a process called ionization. For example, acids, both inorganic and organic, will produce ions when dissolved in water. Some acids, such as hydrochloric acid (HCl), will essentially completely ionize. Others, such as acetic acid (CH3 COOH), will only partially ionize. They establish an equilibrium with the ions and the unionized species (see Chapter 15 for more on chemical equilibrium).
Species such as HCl that completely ionize in water are called strong electrolytes , and those that only partially ionize are called weak electrolytes . Most soluble salts also fall into the strong electrolyte category.
Some of the properties of solutions depend on the chemical and physical nature of the individual solute. The blue color of a copper(II) sulfate solution and the sweetness of a sucrose solution are related to the properties of those solutes. However, some solution properties simply depend on the number of solute particles, not the type of solute. These properties are called colligative properties and include:
⊠ vapor pressure lowering
⊠ freezing-point depression
⊠ boiling-point elevation
⊠ osmotic pressure
Vapor Pressure Lowering
If a liquid is placed in a sealed container, molecules will evaporate from the surface of the liquid and eventually establish a gas phase over the liquid that is in equilibrium with the liquid phase. The pressure generated by this gas is the vapor pressure of the liquid. Vapor pressure is temperature-dependent; the higher the temperature, the higher the vapor pressure. If the liquid is made a solvent by adding a nonvolatile solute, the vapor pressure of the resulting solution is always less than that of the pure liquid. The vapor pressure has been lowered by the addition of the solute; the amount of lowering is proportional to the number of solute particles added and is thus a colligative property.
Solute particles are evenly distributed throughout a solution, even at the surface. Thus, there are fewer solvent particles at the gas–liquid interface where evaporation takes place. Fewer solvent particles escape into the gas phase, and so the vapor pressure is lower. The higher the concentration of solute particles, the less solvent is at the interface and the lower the vapor pressure. This relationship is referred to as Raoult’s law .
The freezing point of a solution of a nonvolatile solute is always lower than the freezing point of the pure solvent. It is the number of solute particles that determines the amount of the lowering of the freezing point. The amount of lowering of the freezing point is proportional to the molality of the solute and is given by the equation
ΔTf = iKf molality
where ΔTf is the number of degrees that the freezing point has been lowered (the difference in the freezing point of the pure solvent and the solution); Kf is the freezing-point depression constant (a constant of the individual solvent); the molality is the molality of the solute; and i is the van’t Hoff factor—the ratio of the number of moles of particles released into solution per mole of solute dissolved. For a nonelectrolyte, such as sucrose, the van’t Hoff factor would be 1. For an electrolyte, such as sodium chloride, you must take into consideration that if 1 mol of NaCl dissolves, 2 mol of particles would result (1 mol Na+ , 1 mol Cl− ). Therefore, the van’t Hoff factor should be 2. However, because sometimes there is a pairing of ions in solution, the observed van’t Hoff factor is slightly less (for example, it is 1.9 for a 0.05 m NaCl solution). The more dilute the solution, the closer the observed van’t Hoff factor should be to the expected factor. If you can calculate the molality of the solution, you can also calculate the freezing point of the solution.
Let’s learn to apply the preceding equation. Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.
The most common mistake is to forget to subtract the ΔT value from the normal freezing point.
The freezing-point depression technique is also commonly used to calculate the molar mass of a solute.
For example, a solution is prepared by dissolving 0.490 g of an unknown compound in 50.00 mL of water. The freezing point of the solution is –0.201°C. Assuming the compound is a nonelectrolyte, what is the molecular mass of the compound? Use 1.00 g/mL as the density of water.
m = ΔT /Kf = 0.201 K/(1.86 K kg mol−1 ) = 0.108 mol/kg
50.00 mL (1.00 g/mL) (1 kg/1,000 g) = 0.0500 kg
(0.108 mol/kg) (0.0500 kg) = 0.00540 mol
0.490 g/0.00540 mol = 90.7 g/mol
Many students make the mistake of stopping before they complete this problem.
Just as the freezing point of a solution of a nonvolatile solute is always lower than that of the pure solvent, the boiling point of a solution is always higher than the solvent’s. Again, only the number of solute particles affects the boiling point. The mathematical relationship is similar to the one for the freezing-point depression above and is
ΔT b = iK b molality
where ΔT b is the number of degrees the boiling point has been elevated (the difference between the boiling point of the pure solvent and the solution); K b is the boiling-point elevation constant; the molality is the molality of the solute; and i is the van’t Hoff factor. You can calculate a solution’s boiling point if you know the molality of the solution. If you know the amount of the boiling-point elevation and the molality of the solution, you can calculate the value of the van’t Hoff factor, i .
For example, determine the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g water. (K b = 0.512 K kg mol−1 )
A 1.00 molal aqueous solution of trichloroacetic acid (CCl3 COOH) is heated to the boiling point. The solution has a boiling point of 100.18°C.
Determine the van’t Hoff factor for trichloroacetic acid (Kb for water = 0.512 K kg mol−1 ).
ΔT = (101.18 – 100.00) = 0.18°C = 0.18 K
i = ΔT /K b m = 0.18 K/(0.512 K kg mol−1 )(1.00 mol kg−1 ) = 0.35
A common mistake is the assumption that the van’t Hoff factor must be a whole number. This is true only for strong electrolytes at very low concentrations.
If you were to place a solution and a pure solvent in the same container but separate them by a semipermeable membrane (which allows the passage of some molecules, but not all particles) you would observe that the level of the solvent side would decrease while the solution side would increase. This indicates that the solvent molecules are passing through the semipermeable membrane, a process called osmosis . Eventually the system would reach equilibrium, and the difference in levels would remain constant. The difference in the two levels is related to the osmotic pressure . In fact, one could exert a pressure on the solution side exceeding the osmotic pressure, and solvent molecules could be forced back through the semipermeable membrane into the solvent side. This process is called reverse osmosis and is the basis of the desalination of seawater for drinking purposes. These processes are shown in Figure 13.1 .
Figure 13.1 Osmotic pressure.
The osmotic pressure is a colligative property and mathematically can be represented as π = (nRT /V ) i, where π is the osmotic pressure in atmospheres; n is the number of moles of solute; R is the ideal gas constant 0.0821 L · atm/K · mol; T is the Kelvin temperature; V is the volume of the solution; and i is the van’t Hoff factor. Measurements of the osmotic pressure can be used to calculate the molar mass of a solute. This is especially useful in determining the molar mass of large molecules such as proteins.
For example, a solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25.0°C. Assuming the fragment is a nonelectrolyte, determine the molar mass of the gene fragment.
Rearrange π = (nRT /V ) i to n = π V /RT (i = 1 for a nonelectrolyte)
If you watch a glass of muddy water, you will see particles in the water settling out. This is a heterogeneous mixture where the particles are large (in excess of 1,000 nm), and it is called a suspension . In contrast, dissolving sodium chloride in water results in a true homogeneous solution , with solute particles less than 1 nm in diameter. True solutions do not settle out because of the very small particle size. But, there are mixtures whose solute diameters fall in between solutions and suspensions. These are called colloids and have solute particles in the range of 1 to 1,000 nm diameter. Table 13.1 shows some representative colloids.
Many times it is difficult to distinguish a colloid from a true solution. The most common method is to shine a light through the mixture under investigation. A light shone through a true solution is invisible, but a light shown through a colloid is visible because the light reflects off the larger colloid particles. This is called the Tyndall effect .
Table 13.1 Common Colloid Types
Experimental procedures for solutions involve concentration units. Keeping close track of the units may simplify the problem.
Concentration problems are concerned with the definitions of the various units. It is possible to calculate the mass and/or volume of the solvent and solute by taking the difference between the final and initial measurements. The density, if not given, is calculated, not measured. It is important to recognize the difference between the values that must be measured and those that can be calculated. Moles are also calculated, not measured.
Do not forget that nearly all the concentration units use the total for the solution in the denominator. For these units it is important to remember to combine the quantities for the solvent and all solutes present.
Common Mistakes to Avoid
- In molarity problems, be sure to use liters ofsolution .
- Make sure your units cancel, leaving you with the units desired in your final answer.
- Round off your final numerical answers to the correct number of significant figures.
- Remember, most molecular compounds––compounds containing only nonmetals––do not ionize in solution. Acids are the most common exceptions.
Use these questions to review the content of this chapter and practice for the AP Chemistry exam. First are 14 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry exam. Following those are two multipart free-response questions like the ones in Section II of the exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.
Answer the following questions in 20 minutes. You may not use a calculator. You may use the periodic table and the equation sheet at the back of this book.
1 . A solution is prepared by dissolving 1.25 g of an unknown substance in 100.0 mL of water. Which procedure from the following list could be used to determine whether the solute is an electrolyte?
(A) Measure the specific heat of the solution.
(B) Measure the volume of the solution.
(C) Measure the freezing point of the solution.
(D) Determine the specific heat of the solution.
2 . What is the final K+ concentration in a solution made by mixing 300.0 mL of 1.0 M KNO3 and 700.0 mL of 2.0 M K3 PO4 ?
(A) 4.5 M
(B) 5.0 M
(C) 3.0 M
(D) 2.0 M
3 . Strontium sulfate, SrSO4 , will precipitate when a solution of sodium sulfate is added to a strontium nitrate solution. What will be the strontium ion, Sr2+ , concentration remaining after 30.0 mL of 0.10 M Na2 SO4 solution are added to 70.0 mL of 0.20 M Sr(NO3 )2 solution?
(A) 0.14 M
(B) 0.15 M
(C) 0.11 M
(D) 0.20 M
4 . Which of the following is a strong electrolyte when it is mixed with water?
(C) C2 H5 OH
(D) CH3 COOH
5 . A solution with a total chloride ion, Cl− , concentration of 1.0 M is needed. Initially, the solution is 0.30 M in MgCl2 . How many moles of solid CaCl2 must be added to 400 mL of the MgCl2 solution to achieve the desired concentration of chloride ion?
6 . Assuming the volumes are additive, what is the final H+ (aq) concentration produced by adding 30.0 mL of 0.50 M HNO3 to 70.0 mL of 1.00 M HCl?
(A) 0.75 M
(B) 1.50 M
(C) 1.25 M
(D) 0.85 M
7 . To prepare 3.0 L of a 0.20 molar K3 PO4 solution (molecular weight 212 g/mol), a student should follow which of the following procedures?
(A) The student should weigh 42 g of solute and add sufficient water to obtain a final volume of 3.0 L.
(B) The student should weigh 42 g of solute and add 3.0 Kg of water.
(C) The student should weigh 130 g of solute and add sufficient water to obtain a final volume of 3.0 L.
(D) The student should weigh 42 g of solute and add 3.0 L of water.
8 . How many grams of MgSO4 (molecular weight 120.4 g/mol) are in 100.0 mL of a 5.0 molar solution?
(A) 600 g
(B) 5.0 g
(C) 12 g
(D) 60.0 g
9 . How many milliliters of concentrated nitric acid (16.0 molar HNO3 ) are needed to prepare 0.500 L of 6.0 molar HNO3 ?
(A) 0.19 mL
(B) 250 mL
(C) 375 mL
(D) 190 mL
10 . A solution has 10 grams of urea in 100 grams of solution. Which item(s) from the following list are needed to calculate the molarity of this solution?
(A) the density of the solution and the molecular weight of urea
(B) the density of the solution and the molecular weight of urea
(C) the density of the solvent and the density of the solution
(D) the molecular weight of urea and the density of the solvent
11 . Which of the following aqueous solutions would have the greatest conductivity?
(A) 0.2 M NaOH
(B) 0.2 M RbCl
(C) 0.2 M K3 PO4
(D) 0.2 M HNO2
12 . How many milliliters of water must be added to 50.0 mL of 10.0 M HNO3 to prepare 4.00 M HNO3 , assuming that the volumes of nitric acid and water are additive?
(A) 50.0 mL
(B) 125 mL
(C) 500 mL
(D) 75.0 mL
13 . The best method to isolate pure MgSO4 from an aqueous solution of MgSO4 is:
(A) evaporate the solution to dryness
(B) titrate the solution
(C) electrolyze the solution
(D) use paper chromatography
14 . Pick the conditions that would yield the highest concentration of N2 (g) in water.
(A) partial pressure of gas = 1.0 atm; temperature of water = 25°C
(B) partial pressure of gas = 0.50 atm; temperature of water = 55°C
(C) partial pressure of gas = 2.0 atm; temperature of water = 25°C
(D) partial pressure of gas = 2.0 atm; temperature of water = 85°C
Answers and Explanations for the Multiple-Choice Questions
1 . C —If the solute is an electrolyte, the solution will conduct electricity and the van’t Hoff factor, i , will be greater than 1. The choices do not include any conductivity measurements; therefore, the van’t Hoff factor would need to be determined. This determination is done by measuring the osmotic pressure, the boiling-point elevation, or the freezing-point depression. The freezing point depression may be found by measuring the freezing point of the solution.
2 . A —The potassium ion contribution from the KNO3 is:
(300.0 mL)(1.0 mol KNO3 /1,000 mL)(1 mol K+ /1 mol KNO3 ) = 0.300 mol K+
The potassium ion contribution from K3 PO4 is:
(700.0 mL)(2.0 mol K3 PO4 /1,000 mL)(3 mol K+ /1 mol K3 PO4 ) = 4.20 mol K+
The total potassium is 4.50 mol in a total volume of 1.000 L. Thus, the potassium concentration is 4.50 M .
3 . C —The reaction is:
Sr2+ (aq) + SO4 2– (aq) → SrSO4 (s)
The strontium nitrate solution contains:
(70.0 mL)(0.20 mol Sr(NO3 )2 /1,000 mL) × (1 mol Sr2+ /1 mol Sr(NO3 )2 ) = 0.014 mol Sr2+
The sodium sulfate solution contains:
(30.0 mL)(0.10 mol Na2 SO4 /1,000 mL)(1 mol SO4 2– /1 mol Na2 SO4 ) = 0.0030 mol SO4 2–
The strontium and sulfate ions react in a 1:1 ratio, so 0.0030 mol of sulfate ion will combine with 0.0030 mol of strontium ion, leaving 0.011 mol of strontium in a total volume of 100.0 mL. The final strontium ion concentration is:
4 . B —A (nitrous acid) and D (acetic acid) are weak acids. Weak acids and bases are weak electrolytes. C (ethanol) is a nonelectrolyte. Potassium nitrate (B) is a water-soluble ionic compound, which is normally a strong electrolyte.
5 . B —The number of moles of chloride ion needed is: (400 mL)(1.0 mol Cl– /1,000 mL) = 0.40 mol Cl–
The initial number of moles of chloride ion in the solution is:
(400 mL)(0.30 mol MgCl2 /1,000 mL)(2 mol Cl– /mol MgCl2 ) = 0.24 mol Cl–
The number of moles needed = [(0.40 – 0.24) mol Cl– ](1 mol CaCl2 /2 mol Cl– ) = 0.080 mol
6 . D —Both of the acids are strong acids and yield 1 mol of H+ each. Calculate the number of moles of H+ produced by each of the acids. Divide the total number of moles by the final volume. (30.0 mL)(0.50 mol H+ /1,000 mL) + (70.0 mL) × (1.00 mol H+ /1,000 mL) = 0.085 mol H+
7 . C —To produce a molar solution of any type, the final volume must be the desired volume. This eliminates answer D. B involves mass of water instead of volume. A calculation of the required mass will allow a decision between A and C.
(3.0 L)(0.20 mol K3 PO4 /L)(212 g K3 PO4 /
1 mol K3 PO4 ) = 130 g K3 PO4
8 . D —(5.0 mol MgSO4 /1,000 mL)(100.0 mL) × (120.4 g MgSO4 /mol MgSO4 ) = 60.0 g MgSO4
9 . D —This is a dilution problem. V before = (M after )(V after )/(M before )
(6.0 M HNO3 )(0.500 L)(1,000 mL/1 L)/
(16.0 M HNO3 ) = 190 mL
10 . A —To calculate the molarity, the moles of urea and the volume of the solution are necessary. The density of the solution and the mass of the solution give the volume of the solution (it may be necessary to convert to liters). The mass of urea and the molecular weight of urea give the moles of urea.
11 . C —The strong electrolyte with the greatest concentration of ions is the best conductor. D is a weak electrolyte, not a strong electrolyte. The number of ions for the strong electrolytes may be found by simply counting the ions: A – 2, B – 2, C – 4. The best conductor has the greatest value when the molarity is multiplied by the number of ions.
12 . D —This is a dilution problem. V after = (M before V before )/(M after )
(10.0 M HNO3 × 50.0 mL)/(4.0 M HNO3 ) = 125 mL
The final volume is 125 mL. Since the original volume was 50.0 mL, an additional 75.0 mL must be added.
13 . A —Solutions cannot be separated by titrations or filtering. Electrolysis of the solution would produce hydrogen and oxygen gas. Chromatography might achieve a minimal separation.
14 . C— The solubility of a gas is increased by increasing the partial pressure of the gas and by lowering the temperature.
You have 5 minutes to answer the following two-part question. You may use a calculator and the tables in the back of the book.
Five beakers each containing 100.0 mL of an aqueous solution are on a lab bench. The solutions are all at 25°C. Solution 1 contains 0.20 M KNO3 . Solution 2 contains 0.10 M BaCl2 . Solution 3 contains 0.15 M C2 H4 (OH)2 . Solution 4 contains 0.20 M (NH4 )2 SO4 . Solution 5 contains 0.25 M KMnO4 .
(a) Which solution has the lowest pH? Explain.
(b) Which solution would be the poorest conductor of electricity? Explain.
You have 15 minutes to answer the following four-part question. You may use a calculator and the tables in the back of the book.
Five beakers are placed in a row on a countertop. Each beaker is half filled with a 0.20 M aqueous solution. The solutes, in order, are: (1) potassium sulfate, K2 SO4 , (2) methyl alcohol, CH3 OH, (3) sodium carbonate, Na2 CO3 , (4) ammonium chromate, (NH4 )2 CrO4 , and (5) barium chloride, BaCl2 . The solutions are all at 25°C. Answer the following questions with respect to the five solutions listed above.
(a) Which solution will form a precipitate when ammonium chromate is added to it?
(b) Which solution is the most basic? Explain.
(c) Which solution would be the poorest conductor of electricity? Explain.
(d) Which solution is colored?
Answers and Explanations for the Free-Response Questions
(a) Solution 4, because the ammonium ion is a weak acid
You get 1 point for picking solution 4 and 1 point for saying the ammonium ion is a weak acid, or that it undergoes hydrolysis.
(b) Solution 3, because the solute is a nonelectrolyte
Give yourself 1 point for picking solution 3 and 1 point for saying it is a nonelectrolyte or that it does not ionize.
(a) The ammonium ion, from the ammonium chromate, will not form a precipitate since most ammonium compounds are water soluble. Therefore, the precipitate must contain the chromate ion combined with a cation from one of the solutions. Solution (2) is a nonelectrolyte; therefore, there are no cations present to combine with the chromate ion. The potassium and sodium ions, from solutions (1) and (3), give soluble salts like the ammonium ion. This only leaves solution (5) barium chloride that will give a precipitate. The formula of the precipitate is BaCrO4 .
You get 1 point for picking the correct solution.
(b) Solution (3) sodium carbonate is the most basic. Since the carbonate ion is the conjugate base of a weak acid, it will undergo significant hydrolysis to produce a basic solution.
You get 1 point for picking the correct solution and 1 point for the correct explanation.
(c) Methyl alcohol is a nonelectrolyte, so its solutions do not conduct electricity. The remaining solutions contain ionic salts, which in general are electrolytes in solution.
You get 1 point for picking the correct solution and 1 point for the correct formula for the explanation.
(d) Solution (4) ammonium chromate is yellow. Most solutions containing a transition metal ion are colored.
You get 1 point for picking the correct solution.
Total your points for the different parts. There are 6 points possible.
- A solution is a homogeneous mixture composed of a solvent and one or more solutes. A solute is a substance that dissolves in the solvent and is normally present in smaller amount.
- The general rule of solubility is “like dissolves like.” This means that polar solvents dissolve polar solutes and nonpolar solvents dissolve nonpolar solutes. Remember, however, that simply quoting this rule will not be sufficient as an explanation in the free-response section.
- A saturated solution is one in which the maximum amount of solute is dissolved for a given amount of solvent at a given temperature. Any solution with less than the maximum solute is called unsaturated. A solution with greater than maximum solute is supersaturated (an unstable state).
- For the chemist the most useful unit of concentration is molarity (M), which is the moles of solute per liter of solution. Know how to work molarity problems. Be careful not to confuse molarity, M or [ ], with moles, n or mol.
- Electrolytes conduct an electrical current when melted or dissolved in a solvent, whereas nonelectrolytes do not.
- A colloid is a mixture in which the solute particle size is intermediate between a true solution and a suspension. If a light is shone through a colloid, the light beam is visible. This is the Tyndall effect.