1. What is a Solution? Solution – homogeneous mixture
Solvent – substance present in largest amount
Solutes – other substances in the solution
Aqueous solution – solution with water as the solvent

2. Various Types of Solutions

3. Solubility of Ionic Substances
Ionic substances breakup into individual cations and anions.

4. Solubility of Ionic Substances
Polar water molecules interact with the positive and negative ions of a salt.

5. Solubility of Polar Substances
Ethanol is soluble in water because of the polar OH bond.

6. Solubility of Polar Substances
Why is solid sugar soluble in water?

7. Substances Insoluble in Water
Nonpolar oil does not interact with polar water.
Water-water hydrogen bonds keep the water from mixing with the nonpolar molecules.

8. How Substances Dissolve
A “hole” must be made in the water structure for each solute particle.
The lost water-water interactions must be replaced by water-solute interactions.
“like dissolves like”

9. Concept Check
Which of the following solutes will generally not dissolve in the specified solvent? Choose the best answer. (Assume all of the compounds are in the liquid state.)
a) CCl4 mixed with water (H2O)
b) NH3 mixed with water (H2O)
c) CH3OH mixed with water (H2O)
d) N2 mixed with methane (CH4)
The correct answer is a. The general rule of solubility is “like dissolves like”. Therefore, polar solutes dissolve well in polar solvents and nonpolar solutes dissolve well in nonpolar solvents. CCl4 will not dissolve well in H2O because CCl4 is nonpolar and H2O is polar.

10. The solubility of a solute is limited.
Saturated solution – contains as much solute as will dissolve at that temperature.
Unsaturated solution – has not reached the limit of solute that will dissolve.

11. Supersaturated solution – occurs when a solution is saturated at an elevated temperature and then allowed to cool but all of the solid remains dissolved.
Contains more dissolved solid than a saturated solution at that temperature.
Unstable – adding a crystal causes precipitation.

12. Solutions are mixtures.
Amounts of substances can vary in different solutions.
Specify the amounts of solvent and solutes.
Qualitative measures of concentration
concentrated – relatively large amount of solute
dilute – relatively small amount of solute

13. Homework
Read sections 8.2, 8.3
Pages 381 #1,6,8
Page 389 #2,3,8,9

15. Exercise
What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?
6.6%
[5.5 g / (5.5 g g)] × 100 = 6.6% glucose
[5.5 g / (5.5 g g)] × 100 = 6.6%

16. Molarity (M) = moles of solute per volume of solution in liters:

17. Exercise
You have 1.00 mol of sugar in mL of solution. Calculate the concentration in units of molarity.
8.00 M
1.00 mol / (125.0 / 1000) = 8.00 M
1.00 mol / (125.0 / 1000) = 8.00 M

18. Exercise
A g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution?
1.57 M
500.0 g is equivalent to mol K3PO4 (500.0 g / g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L).
500.0 g is equivalent to mol K3PO4 (500.0 g / g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L).

19. Exercise
You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?
0.200 L
2.00 mol / 10.0 M = L
2.00 mol / 10.0 M = L

20. Exercise
Consider separate solutions of NaOH and KCl made by dissolving g of each solute in mL of solution. Calculate the concentration of each solution in units of molarity.
10.0 M NaOH
[100.0 g NaOH / g/mol] / [250.0 / 1000] = 10.0 M NaOH
5.37 M KCl
[100.0 g KCl / g/mol] / [250.0 / 1000] = 5.37 M KCl
[100.0 g NaOH / g/mol] / [250.0 / 1000] = 10.0 M NaOH
[100.0 g KCl / g/mol] / [250.0 / 1000] = 5.37 M KCl

21. Concept Check
You have two HCl solutions, labeled Solution A and Solution B. Solution A has a greater concentration than Solution B. Which of the following statements are true?
a) If you have equal volumes of both solutions, Solution B must contain more moles of HCl.
b) If you have equal moles of HCl in both solutions, Solution B must have a greater volume.
c) To obtain equal concentrations of both solutions, you must add a certain amount of water to Solution B.
d) Adding more moles of HCl to both solutions will make them less concentrated.
The correct answer is b. Solution B must have a greater volume to make it less concentrated than Solution A since Molarity = moles of solute/liters of solution.

22. For a 0.25 M CaCl2 solution: CaCl2 → Ca2+ + 2Cl–
Concentration of Ions
For a 0.25 M CaCl2 solution:
CaCl2 → Ca2+ + 2Cl–
Ca2+: 1 × 0.25 M = 0.25 M Ca2+
Cl–: 2 × 0.25 M = 0.50 M Cl–.

23. Which of the following solutions contains the greatest number of ions?
Concept Check
Which of the following solutions contains the greatest number of ions?
400.0 mL of 0.10 M NaCl.
300.0 mL of 0.10 M CaCl2.
200.0 mL of 0.10 M FeCl3.
800.0 mL of 0.10 M sucrose.
a) contains mol of ions (0.400 L × 0.10 M × 2). b) contains mol of ions (0.300 L × 0.10 M × 3). c) contains mol of ions (0.200 L × 0.10 M × 4). d) does not contain any ions because sucrose does not break up into ions. Therefore, letter b) is correct.

24. Where are we going? How do we get there? Let’s Think About It
To find the solution that contains the greatest number of moles of ions.
How do we get there?
Draw molecular level pictures showing each solution. Think about relative numbers of ions.
How many moles of each ion are in each solution?

25. the volume of the solution is the largest.
Notice
The solution with the greatest number of ions is not necessarily the one in which:
the volume of the solution is the largest.
the formula unit has the greatest number of ions.

26. A solution whose concentration is accurately known.
Standard Solution
A solution whose concentration is accurately known.

27. To Make a Standard Solution
Weigh out a sample of solute.
Transfer to a volumetric flask.
Add enough solvent to mark on flask.

28. The process of adding water to a concentrated or stock solution to achieve the molarity desired for a particular solution.
Dilution with water does not alter the numbers of moles of solute present.
Moles of solute before dilution = moles of solute after dilution
M1V1 = M2V2

29. Diluting a Solution
Transfer a measured amount of original solution to a flask containing some water.
Add water to the flask to the mark (with swirling) and mix by inverting the flask.

30. Concept Check
A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution?
Add water to the solution.
Pour some of the solution down the sink drain.
c) Add more sodium chloride to the solution.
d) Let the solution sit out in the open air for a couple of days.
e) At least two of the above would decrease the concentration of the salt solution.
For letter a), adding water to the solution will increase the total volume of solution and therefore decrease the concentration. For letter b), pouring some of the solution down the drain will not change the concentration of the salt solution remaining. For letter c), adding more sodium chloride to the solution will increase the number of moles of salt ions and therefore increase the concentration. For letter d), water will evaporate from the solution and decrease the total volume of solution and therefore increase the concentration. Therefore, since only letter a) would decrease the concentration, letter e) cannot be correct.

31. Exercise
What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a M NaOH solution?
60.0 mL
M1V1 = M2V2
(2.00 M)(V1) = (0.800 M)(150.0 mL)
The minimum volume needed is 60.0 mL. M1V1 = M2V2 (2.00 M)(V1) = (0.800 M)(150.0 mL)

32. Steps for Solving Stoichiometric Problems Involving Solutions
Write the balanced equation for the reaction. For reactions involving ions, it is best to write the net ionic equation.
Calculate the moles of reactants.
Determine which reactant is limiting.
Calculate the moles of other reactants or products, as required.
Convert to grams or other units, if required.

33. What precipitate will form? lead(II) phosphate, Pb3(PO4)2
Concept Check (Part I)
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).
What precipitate will form?
lead(II) phosphate, Pb3(PO4)2
What mass of precipitate will form?
1.1 g Pb3(PO4)2
The balanced molecular equation is: 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s).
mol Na3PO4 present to start and mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is g/mol, 1.1 g of Pb3(PO4)2 will form.

34. Where are we going? How do we get there? Let’s Think About It
To find the mass of solid Pb3(PO4)2 formed.
How do we get there?
What are the ions present in the combined solution?
What is the balanced net ionic equation for the reaction?
What are the moles of reactants present in the solution?
Which reactant is limiting?
What moles of Pb3(PO4)2 will be formed?
What mass of Pb3(PO4)2 will be formed?

35. Concept Check (Part II)
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).
What is the concentration of nitrate ions left in solution after the reaction is complete?
0.27 M
The concentration of nitrate ions left in solution after the reaction is complete is 0.27 M.
Nitrate ions are spectator ions and do not participate directly in the chemical reaction. Since there were mol of Pb(NO3)2 present to start, then mol of nitrate ions are present. The total volume in solution is 30.0 mL. Therefore the concentration of nitrate ions = mol / L = 0.27 M.

36. Where are we going? How do we get there? Let’s Think About It
To find the concentration of nitrate ions left in solution after the reaction is complete.
How do we get there?
What are the moles of nitrate ions present in the combined solution?
What is the total volume of the combined solution?

37. Concept Check (Part III)
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).
What is the concentration of phosphate ions left in solution after the reaction is complete?
0.011 M
The concentration of phosphate ions left in solution after the reaction is complete is M.
Phosphate ions directly participate in the chemical reaction to make the precipitate. There were mol of Na3PO4 present to start, therefore there was mol of phosphate ions present to start mol of phosphate ions were used up in the chemical reaction, therefore mol of phosphate ions is leftover ( – mol).
The total volume in solution is 30.0 mL. Therefore the concentration of phosphate ions = mol / L = M.

38. Where are we going? How do we get there? Let’s Think About It
To find the concentration of phosphate ions left in solution after the reaction is complete.
How do we get there?
What are the moles of phosphate ions present in the solution at the start of the reaction?
How many moles of phosphate ions were used up in the reaction to make the solid Pb3(PO4)2?
How many moles of phosphate ions are left over after the reaction is complete?
What is the total volume of the combined solution?

39. An acid-base reaction is called a neutralization reaction.
Steps to solve these problems are the same as before.
For a strong acid and base reaction:
H+(aq) + OH–(aq)  H2O(l)

40. Concept Check
For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of M sulfuric acid?
1.00 mol NaOH
The balanced equation is: H2SO4 + 2NaOH → 2H2O + Na2SO moles of sulfuric acid is present to start. Due to the 1:2 ratio in the equation, 1.00 mol of NaOH would be required to exactly react with the sulfuric acid.
1.00 mol of sodium hydroxide would be required.

41. Where are we going? How do we get there? Let’s Think About It
To find the moles of NaOH required for the reaction.
How do we get there?
What are the ions present in the combined solution? What is the reaction?
What is the balanced net ionic equation for the reaction?
What are the moles of H+ present in the solution?
How much OH– is required to react with all of the H+ present?

42. Equivalent weight – mass in grams of 1 equivalent of acid or base.
Unit of Concentration
One equivalent of acid – amount of acid that furnishes 1 mol of H+ ions.
One equivalent of base – amount of base that furnishes 1 mol of OH ions
Equivalent weight – mass in grams of 1 equivalent of acid or base.

45. To find number of equivalents:

46. Concept Check
If Ba(OH)2 is used as a base, how many equivalents of Ba(OH)2 are there in 4 mol Ba(OH)2?
a) 2
b) 4
c) 8
d) 16
The correct answer is c. Since each mole of Ba(OH)2 provides 2 moles of OH–, we have two equivalents for each mole or 8 equivalents.

47. Chapter 15 Homework Homework Reading assignment Pages 475 through 502
Homework Problems
Questions and problems 3, 5, 7, 11, 17, 19, 21, 23, 25, 27, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 57, 59, 61, 71, 73, 85, 87.
Due on
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Copyright © Cengage Learning. All rights reserved 47 47