Chapter 15: Solutions 15.1 Solubility 15.2 Solution Composition 15.3 Mass Percent 15.4 Molarity 15.7 Neutralization Reactions
Dec 31, 2015
Chapter 15: Solutions
15.1 Solubility15.2 Solution Composition15.3 Mass Percent15.4 Molarity15.7 Neutralization Reactions
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 2
Solutions
• Solutions are homogeneous mixtures.– Mixtures in which the components are uniformly intermingled
• Solvent: the substance present in the highest percentage
• Solute: the dissolved substance, which is present in lesser amount
• Aqueous solutions: solutions with water as the solvent
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 3
Solutions (cont.)
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 4
The Solution Process: Ionic Compounds
• When ionic compounds dissolve in water they dissociate into ions and become hydrated.
• When solute particles are surrounded by solvent molecules we say they are solvated.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 5
Figure 15.2: Polar water molecules interact with the positive and negative ions of a salt. These interactions replace the strong ionic forces holding the ions together in the undissolved solid, thus assisting in the dissolving process.
Solution: Solid in Liquid
• Salt in water - separate into ions
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 6
The Solution Process: Covalent Molecules
• Covalent molecules that are small and have “polar” groups tend to be soluble in water.
• The ability to H-bond with water enhances solubility.OH
H
C O
H
HH H
O
H
H
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 7
Solubility
• When one substance (solute) dissolves in another (solvent), it is said to be soluble.– Salt is soluble in water– Bromine is soluble in methylene chloride
• When one substance does not dissolve in another, it is said to be insoluble.– Oil is insoluble in water
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 8
Solubility (cont.)
• There is usually a limit to the solubility of one substance in another.– Gases are always soluble in each other– Some liquids are always mutually soluble
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 9
Solutions & Solubility
• Molecules that are similar in structure tend to form solutions: “like dissolves like”
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 10
Solutions & Solubility (cont.)
• The solubility of the solute in the solvent depends on the temperature.– Higher temp = greater solubility of solid in liquid– Lower temp = greater solubility of gas in liquid
• The solubility of gases depends on the pressure.– Higher pressure = greater solubility
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 11
Figure 15.6: An oil layer floating on water.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 12
Describing Solutions Qualitatively
• A concentrated solution has a high proportion of solute to solution.
• A dilute solution has a low proportion of solute to solution.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 13
Describing Solutions Qualitatively (cont.)
• A saturated solution has the maximum amount of solute that will dissolve in the solvent.– Depends on temp
• An unsaturated solution has less than the saturation limit.
• A supersaturated solution has more than the saturation limit.– Unstable
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 14
Describing Solutions Quantitatively (cont.)
• Solutions have variable composition.
• To describe a solution accurately, you need to describe the components and their relative amounts.
• Concentration: the amount of solute in a given amount of solution– Occasionally amount of solvent
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 15
Solution Concentration
mass percent = mass of solute X 100%
mass of solution
Molarity = moles of solute liters of solution
(mass of solute + mass of solvent)
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 16
Solution Concentration Percentage
• Mass percent = grams of solute per 100 g of solution– 5.0% NaCl has 5.0 g of NaCl in every 100 g of
solution
• Mass of solution = mass of Solute + mass of solvent
• Divide the mass of solute by the mass of solution and multiply by 100% to get mass percent.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 17
Mass Percent #15.1
mass percent = mass of solute X 100%
mass of solution
A solution is prepared by mixing 2.50 g if calcium chloride with 50.0g of water. Calculate the mass percent of calcium chloride in this solution
4.76% CaCl2mass percent = 2.50g X 100 =
2.50 + 50.0g
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 18
Mass Percent #15.2
mass percent = mass of solute X 100%
mass of solution
Concentrated hydrochloric acid solution contains 37.2% by mass HCl. What mass of HCl is contained in 35.5g of concentrated HCl?
13.2 g HCl37.2 = mass of solute X 100 =
35.5g
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 19
Solution Concentration Molarity
• Moles of solute per 1 liter of solution
• Used because it describes how many moles of solute in each liter of solution
• If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar, etc.
molarity = moles of soluteliters of solution
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 20
Molarity #15.3
Molarity = moles of solute liters of solution
Calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution
0.105 MMolarity = 0.131 mol =
1.25 L
15.6 g KBr (1 mol /119.0g) = 0.131 mol KBr
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 21
Molarity #15.4
Molarity = moles of solute liter of solution
Calculate the molarity of a solution prepared by dissolving 2.80 g of solid NaCl in enough water to make 135 mL of solution
0.355 MMolarity = 0.0479 mol =
0.135 L
2.80 g NaCl (1 mol /58.44g) = 0.0479 mol KBr
135 mL (1 L /1000 mL) = 0.135 L
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 22
Molarity & Dissociation
• The molarity of the ionic compound allows you to determine the molarity of the dissolved ions.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 23
Molarity & Dissociation (cont.)
• CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq)
• A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution
• Because each CaCl2 dissociates to one Ca+2, 1.0 M CaCl2 = 1.0 M Ca+2
• Because each CaCl2 dissociates to 2 Cl-1, 1.0 M CaCl2 = 2.0 M Cl-1
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 24
Calculating Ion Concentration from Molarity #15.5
Give the concentrations of all ions in each of the following solutions:
1. 1.20 M Na2SO4
2. 0.750 M K2CrO4
Na2SO4 Na+(aq) + SO42-(aq)
1.20 Na2SO4 2.40 Na+(aq) + 1.20 SO42-(aq)
1 12
2.40M Na+, 1.20M SO42-
K2CrO4 K+(aq) + CrO42-(aq)
0.750 K2CrO4 1.50 K+(aq) + 0.750 CrO42-(aq)
1 12
1.50M K+, 0.750M CrO42-
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 25
Dilution
• Dilution: adding solvent to decrease the concentration of a solution
• The amount of solute stays the same, but the concentration decreases.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 26
Dilution (cont.)
• Dilution Formula
M1 x V1 = M2 x V2
# Moles/L · # L = # moles– In dilution we take a certain number of moles
of solute and dilute to a bigger volume.
• Concentrations and volumes can be most units as long as they are consistent.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 27
Dilution (cont.)
M = moles of solute volume (L)
M1 x V1 = moles of solute = M2 x V2
remains constant
decreasesAdd water, therefore increases
Initial conditions Final conditions
Molarity before Dilution
Volume before Dilution
Molarity after Dilution
Volume after Dilution
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 28
Dilution #15.6 A
M1 x V1 = M2 x V2
What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution?
19 M x V1 = 0.15M x 1.0 L
V1 = 0.15M x 1.0 L = 0.0079 L = 7.9 mL
19 M
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 29
Dilution #15.6 B
M1 x V1 = M2 x V2
What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution?
19 M x V1 = 0.15M x 1.0 L
V1 = 0.15M x 1.0 L = 0.0079 L = 7.9 mL
19 M
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 30
15.7 Neutralization Reactions
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 31
Neutralization Reactions
• Acid-Base reactions are also called neutralization reactions.
• Often we use neutralization reactions to determine the concentration of an unknown acid or base.
• The procedure is called a titration. With this procedure we can add just enough acid solution to neutralize a known volume of a base solution.– Or vice-versa
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 32
Neutralization Reactions
• When a strong acid and a strong base react, the net ionic reaction is
H+(aq) + OH-(aq) H2O(l)
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 33
Neutralization Reactions # 15.7a
• What volume of 0.150 M HNO3 solution is needed to neutralize 45.0 mL of a 0.550 M KOH solution?
Answer = 165 mL
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 34
Neutralization Reactions # 15.7b
• What volume of 1.00 X 10-2 M HCL solution is needed to neutralize 35.0 mL of a 5.00 X 10-3 M Ba(OH)2 solution?
Answer = 35.0 mL