Chapter 15 Notes
15-1 What are Solutions?
CHAPTER 15 SOLUTIONS
SOLUTIONS are homogeneous mixtures that contain a solute and a solvent.
SOLUTE is the material dissolved in a solution.
SOLVENT-the material dissolving the solute to make the solution.
Solutions, solvents, and solutes may be solid, liquid, or gaseous.
A substance that dissolves in a solvent is said to be soluble in that solvent.
A substance that does not dissolve in a solvent is said to be insoluble in that solvent.
Two liquids that are soluble in each other are said to be miscible.
Liquids that are not soluble in each other are immiscible.
A soluble substance is able to dissolve in a solvent because attractive forces between solvent and solute particles are strong enough to overcome the attractive forces holding the solute particles together. Solvent particles surround the solute particles to form a solution in a process called solvation. Solvation in water is also called hydration.
In general, likes dissolve likes; that is, polar substances tend to be soluble in polar solvents, and nonpolar substances tend to be soluble in nonpolar solvents. When an ionic compound dissolves in water, a polar solvent, the attraction between the water dipoles and the ions cause the ions to become solvated. The ions break away from the surface of the ionic solid and move into solution. The overall energy change that occurs during solution formation is called the heat of solution.
There are 3 ways to increase the rate at which a solute dissolves. (Essentially, you are increasing the collisions between the solute and solvent particles, which increase the rate at which the solute dissolves.)
1. Agitating the mixture
2. Increasing the surface area of the solute
3. Increasing the temperature of the solvent. (Increasing the temp. causes collisions to become more frequent and to have greater energy.)
The solubility of a solute is the maximum amount of the solute that will dissolve in a given amount of solvent at a specified temperature and pressure. Usually expressed in grams of solute per 100 grams of solvent. (g solute/100 g solvent)
A solution that contains the
maximum amount of dissolved solute for a given amount of solvent at a specific
temperature and pressure is called a saturated solution.
Unsaturated solution is one that contains less dissolved solute for a given temperature and pressure than does a saturated solution.
Supersaturated solution contains more dissolved solute than does a saturated solution at the same temperature. Supersaturated solutions are unstable, and the excess solute they contain often precipitates out of solution if the solution is disturbed.
1. Temperature-most solids are more soluble at higher temps. than at lower ones.
Use the graph in your textbook (pg. 458 Figure 15-7) to answer the following questions.
1. Determine the approximate solubility of potassium chloride (KCl) at 20oC.
2. How many grams of KCl would dissolve in 300 grams of water at 20oC.
3. How many more grams of KCl could be dissolved if the temperature of the water was raised to 40oC?
ANSWER
Locate the curve representing KCl on the graph. At 20oC , the solubility of KCl is approximately 33 g per 100 g water. Using this solubility, calculate the amount of KCl that will dissolve in 300 g of water at 20oC.
33 g KCl/100 g H2O X 300 g water = 99 g KCl
According to the graph, the solubility of KCl at 40oC is 40 g KCl per 100 g water. Find the amount of KCl that will dissolve in 300 g of water at 40oC.
40 g KCl/100 g water X 300 g water = 120 g KCl
The amount of additional KCl that dissolves at the higher temperature in 300 g of water is the difference between the 2 amounts.
120 g KCl - 99 g KCl = 21 g KCl
2.
Pressure-the solubility of a gas increases as its external pressure
(its pressure above the solution) increases. This fact is expressed as Henry’s Law.
Henry’s Law-states that at a given temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid, or
S1/ P1
= S2/P2
Where S1 is the solubility of a gas at pressure P1, and S2 is the solubility of the gas at P2.
Example Problem
Using Henry’s Law
If 0.24 grams of a gas dissolves in 1.0 L of water at 1.5 atm of pressure how much of the gas will dissolve if the pressure is raised to 6.0 atm? Assume the temperature is held constant.
S1= 0.24 g/l P1=1.5 atm P2=6.0 atm. Use Henry’s Law to calculate S2.
Answer
Rearrange Henry’s Law to solve for S2, then substitute the known values into the equation and solve.
S2=S1(P2/P1) = (0.24 g/L) (6.0 atm/1.5 atm) = 0.96 g/L
Practice Problems
1. A gas has a solubility of 0.086 g/L at a pressure of 3.5 atm. At what pressure would its solubility be 2.3 g/L?
2. The solubility of a gas changes from 0.95 g/L to 0.72 g/L. If the initial pressure was 2.8 atm, what is the final pressure?
Answers 1. 94 atm 2. 2.1 atm
15-2 Solution Concentration
The concentration of a solution is a measure of the amount of solute dissolved in a given amount of solvent or solution.
A concentrated solution contains a large amount of solute.
A dilute solution contains a small amount of solute.
It is useful to describe solution concentrations quantitatively, in terms of relative masses or volumes. One method is percent by mass.
Percent by mass is the ratio of the solute's mass to the solution's mass, expressed as a percent.
Percent by mass = Mass of Solute/mass of solution X 100
Example Problem
Determining Percent by Mass.
What is the percent by mass of potassium nitrate in a solution made by mixing 5.4 g of this substance with 260.0 ml of water?
The mass in grams of water is numerically equal to the volume of water in milliliters, because 1 ml of water has a mass of 1g. WE LEARNED THIS FIRST SEMESTER. THE DENSITY OF WATER IS 1g/ml. REMEMBER??? Thus, the mass of the solvent (water) is 260.0 g. The mass of the solution is the sum of the masses of solute and solvent.
Mass of solution = solute + solvent = 5.4 g + 260.0 g = 265.4 g.
Substitute the mass of solution and the given mass of solute into the equation for percent by mass and solve.
Percent by mass = Mass of Solute/mass of solution X 100
Percent by mass = 5.4 g/265.4 g X 100 = 2.0 %
Practice Problems
3. What is the percent by mss of sodium carbonate in a water solution containing 0.497 g of NaCO3 in 58.3 g of solution?
4. The percent by mass of magnesium chloride in a water solution is 1.47 %. How many grams of solute are dissolved in each 500.0 g of solution?
5. What is the mass of the solvent in Practice Problem 4?
Answers: 3. 0.852 % 4. 7.35 g 5. 492.65 g
Percent by volume is a method of expressing the concentration of solute in a solution made by mixing two liquids.
Percent by volume = volume of solute/volume of solution X 100
Example
100 ml of a solution is made by mixing water with 50 ml of acetone . The percent by volume = 50ml/100 ml X 100 = 50 %
Practice Problems
6. What is the percent by volume of isopropyl alcohol in a solution made by mixing 75 ml of the alcohol with enough water to make 288 ml of solution?
7. What volume of acetone was used to make 3.11 L of a water solution if the percent acetone by volume is 27.9 %?
Answers: 6. 26 % 7. 0.868 L
Molarity
A common way to express solution concentration.
Molarity (M) = moles of solute/liter(s) of solution
Example
Calculating Molarity and Preparing Molar Solutions
a. What is the molarity of an aqueous solution that contains 14.2 g NaCl dissolved in 2364 ml of solution?
b. How many grams of NaCl are contained in 1.00 L of this solution?
Answers
a. Use molar mass to convert grams of solute to moles of solute.
(14.2 g NaCl)(1 mole NaCl/58.5 g NaCl) = 0.243 mol NaCl ----------the 58.5 g NaCl came from the Periodic Table. mass
of Na plus mass of Cl.
The solution's volume must be expressed in liters.
(2364 ml)(1 L /100 ml) = 2.364 L
Substitute the known values and solve.
Molarity (M) = moles of solute/liter(s) of solution = 0.243 mol NaCl/2.364 L solution = 0.103 mol/L = 0.103 M
b. Given the definition of molarity, it is now known that there is 0.103 mol of NaCl per liter of solution. To find the number of grams this represents, multiply by the molar mass of NaCl.
0.103 mol NaCl/1L solution X 58.5 g NaCl/ 1 mol NaCl = 6.03 g NaCl/ 1 L solution
Practice Problems
8. A solution is made by dissolving 17.0 g of lithium iodide (LiI) in enough water to make 387 ml of solution. What is the molarity of the solution?
9. Calculate the molarity of a water solution of calcium chloride (CaCl2) , given that 5.04 L of the solution contains 612 g of calcium chloride.
Answers: 8. 0.328 M 9. 1.09 M
Diluting Molar Solutions
Suppose you want to dilute a stock solution of known concentration to make a given quantity of solution of lower concentration? You would need to know the volume of stock solution to use. To find that information, you can use the equation:
M1V1=M2V2
where M1 and V1 are the molarity and volume, respectively, of the stock solution, and M2 and V2 are the molarity and volume, respectively, of the dilute solution. The equation can be solved for any of the four variables, given the other three.
Example
Diluting a Solution
What volume, in milliliters, of a 1.15 M stock solution of potassium nitrate is needed to make 0.75 L of 0.578 M potassium nitrate?
Answer:
M1V1=M2V2
Rearrange the equation to solve for the volume of stock solution, V1, and substitute the known values into the equation.
V1 = V2(M2/M1) = (0.75 L0(0.578 M/1.15L) = 0.377 L
Convert to milliliters:
(0.377 L) (1000ml/1L) = 377 ml
Thus, in making the 0.578 M solution, 377 ml of the 1.15 M stock solution should be diluted with enough water to make 0.75 L of solution.
Practice Problems
10. Suppose you wish to make 0.879 L of 0.250 M silver nitrate by diluting a stock solution of 0.675 M silver nitrate. How many milliliters of the stock solution would you need to use?
11. If 55.0 ml of a 2.45 M stock solution of sucrose is diluted with water to make 168 ml of sucrose solution, what is the molarity of the final solution?
Answers: 10. 326 ml 11. 0.802 M
Molality (m)
Molality of a solution is equal to the number of moles of solute per kilogram of solvent.
Molality (m) = moles solute/kg solvent
Example
Calculating Molality
What is the molality of a solution that contains 16.3 g of potassium chloride dissolved in 845 g of water?
Answer:
Convert the mass of solute to moles.
(16.3 g KCl)(1 mol KCl/74.6 g KCl) = 0.218 mol KCl -----NOTE--74.6 g KCl came from using Periodic Table to find molar mass.
The solvent mass, 845 g, must be expressed in kilograms.
(845 g H2O)(1 kg H2O/1000 g H2O) = 0.845 kg H2O
Substitute the known values into the equation for molality and solve.
Molality (m) = moles solute/kg solvent
= 0.218 mol KCl/0.845 kg H2O = 0.258 mol KCl/kg water = 0.258 m
Practice Problems
12. What is the molality of the solution formed by mixing 104 g of silver nitrate (AgNO3) with 1.75 kg of water?
13. Suppose that 5.25 g of sulfur (S8) is dissolved in 682 g of the liquid solvent carbon disulfide (CS2). What is the molality of the sulfur solution?
Answers: 12. 0.350 m 13. 0.0300m