1

1

Chapter 8 Solutions

8.4 Percent Concentration

2

The concentration of a solution §  Is the amount of solute dissolved in a specific amount

of solution. amount of solute amount of solution

Concentration

3

Mass percent (%m/m) concentration is the •  Percent by mass of solute in a solution.

mass percent (%m/m) = g of solute x 100 g of solute + g of solvent •  Amount in g of solute in 100 g of solution. mass percent = g of solute x 100

100 g of solution

Mass Percent Concentration

4

Mass of Solution

8.00 g KCl

50.00 g KCl solution

Add water to give 50.00 g solution

2

5

The calculation of mass percent (%m/m) requires the §  Grams of solute (g KCl) and §  Grams of solution (g KCl solution).

g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g

8.00 g KCl (solute) x 100 = 16.0% (m/m)

50.00 g KCl solution

Calculating Mass Percent

6

A solution is prepared by mixing 15.0 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution.

1) 15.0% (m/m) Na2CO3

2) 6.38% (m/m) Na2CO3 3) 6.00% (m/m) Na2CO3

Learning Check

7

3) 6.00% (m/m) Na2CO3 STEP 1 mass solute = 15.0 g Na2CO3

mass solution = 15.0 g + 235 g = 250. g STEP 2 Use g solute/ g solution ratio STEP 3 mass %(m/m) = g solute x 100

g solution STEP 4 Set up problem mass %(m/m) = 15.0 g Na2CO3 x 100 = 6.00% Na2CO3 250. g solution

Solution

8

The volume percent (%v/v) is §  Percent volume (mL) of solute (liquid) to volume (mL)

of solution. §  Volume % (v/v) = mL of solute x 100

mL of solution §  Solute (mL) in 100 mL of solution.

volume % (v/v) = mL of solute 100 mL of solution

Volume Percent

3

9

The mass/volume percent (%m/v) is §  Percent mass (g) of solute to volume (mL) of solution. §  mass/volume % (m/v) = g of solute x 100

mL of solution §  Solute (g) in 100 mL of solution.

mass/volume % (m/v) = g of solute 100 mL of solution

Mass/Volume Percent

10

Percent Conversion Factors

§  Two conversion factors can be written for each type of % value.

TABLE 8.9

5% (m/v) glucose There are 5 g of glucose 5 g glucose and 100 mL solution in 100 mL of solution. 100 mL solution 5 g glucose

11

Write two conversion factors for each solution: A. 8.50%(m/m) NaOH

B. 5.75%(v/v) ethanol

C. 4.8 %(m/v) HCl

Learning Check

12

A. 8.50 g NaOH and 100 g solution 100 g solution 8.50 g NaOH

B. 5.75 mL alcohol and 100 mL solution 100 mL solution 5.75 mL alcohol C. 4.8 g HCl and 100 mL HCl 100 mL solution 4.8 g HCl

Solution

4

13

How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1 Given: 225 g solution; 10.0% (m/m) NaCl

Need: g of NaCl STEP 2 g solution g NaCl STEP 3 Write the 10.0 %(m/m) as conversion factors.

10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl

STEP 4 Set up to cancel g solution. 225 g solution x 10.0 g NaCl = 22.5 g NaCl

100 g solution

Using Percent Concentration (m/m) Factors

14

How many grams of NaOH are needed to prepare 75.0 g of 14.0%(m/m) NaOH solution?

1) 10.5 g NaOH 2) 75.0 g NaOH 3) 536 g NaOH

Learning Check

15

1) 10.5 g NaOH

75.0 g solution x 14.0 g NaOH = 10.5 g NaOH 100 g solution 14.0 % (m/m) factor

Solution

16

How many milliliters of a 5.75 % (v/v) ethanol solution can be prepared from 2.25 mL ethanol?

1) 2.56 mL 2) 12.9 mL 3) 39.1 mL

Learning Check

5

17

3) 39.1 mL 2.25 mL ethanol x 100 mL solution

5.75 mL ethanol 5.75 %(v/v) inverted

= 39.1 mL solution

Solution

18

How many mL of a 4.20%(m/v) will contain 3.15 g KCl? STEP 1 Given: 3.15 g KCl(solute); 4.20% (m/v) KCl

Need: mL of KCl solution STEP 2 Plan: g KCl mL KCl solution STEP 3 Write conversion factors.

4.20 g KCl and 100 mL solution 100 mL solution 4.20 g KCl

STEP 4 Set up the problem 3.15 g KCl x 100 mL KCl solution = 75.0 mL KCl

4.20 g KCl

Using Percent Concentration(m/v) Factors

19

Learning Check

How many grams of NaOH are needed to prepare 125 mL of a 8.80%(m/v) NaOH solution?

20

Solution

How many grams of NaOH are needed to prepare 125 mL of a 8.80%(m/v) NaOH solution? 125 mL solution x 8.80 g NaOH = 11.0 g NaOH 100 mL solution

6

21

Chapter 8 Solutions

8.5 Molarity and Dilution

22

Molarity (M)

Molarity (M)

§  Is a concentration term for solutions.

§  Gives the moles of solute in 1 L solution.

§  = moles of solute liter of solution

23

Preparing a 1.0 Molar Solution

A 1.00 M NaCl solution is prepared §  By weighing out 58.5 g NaCl (1.00 mole) and §  Adding water to make 1.00 liter of solution.

24

What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution

Need molarity (mole/L) STEP 2 Plan g NaOH mole NaOH molarity

Calculation of Molarity

7

25

STEP 3 Conversion factors 1 mole NaOH = 40.0 g

1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH

STEP 4 Calculate molarity.

6.00 g NaOH x 1 mole NaOH = 0.150 mole 40.0 g NaOH

0.150 mole = 0.300 mole = 0.300 M NaOH 0.500 L 1 L

Calculation of Molarity (cont.)

26

What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?

1) 0.557 M 2) 1.44 M 3) 1.71 M

Learning Check

27

3) 1.71 M 46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3 84.0 g NaHCO3

0.557 mole NaHCO3 = 1.71 M NaHCO3 0.325 L

Solution

28

What is the molarity of 225 mL of a KNO3 solution containing 34.8 g KNO3? 1) 0.344 M 2) 1.53 M 3) 15.5 M

Learning Check

8

29

2) 1.53 M 34.8 g KNO3 x 1 mole KNO3 = 0.344 mole KNO3 101.1 g KNO3

M = mole = 0.344 mole KNO3 = 1.53 M L 0.225 L In one setup

34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M 101.1 g KNO3 0.225 L

Solution

30

Molarity Conversion Factors

The units of molarity are used to write conversion factors for calculations with solutions.

TABLE 8.10

31

Calculations Using Molarity

How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl

Need Grams of KCl STEP 2 Plan L KCl moles KCl g KCl

32

Calculations Using Molarity

STEP 3 Conversion factors 1 mole KCl = 74.6 g 1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl 1 L KCl = 0.720 mole KCl 1 L and 0.720 mole KCl 0.720 mole KCl 1 L STEP 4 Calculate g KCl 0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g KCl 1 L 1 mole KCl

9

33

How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution?

1) 20.0 g AlCl3 2) 16.7g AlCl3

3) 2.50 g AlCl3

Learning Check

34

Solution

3) 2.50 g AlCl3 0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3 1 L 1 mole

35

How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3? 1) 12.0 mL 2) 83.3 mL 3) 190. mL

Learning Check

36

24.0 g HNO3 x 1 mole HNO3 x 1000 mL =

63.0 g HNO3 2.00 moles HNO3 Molarity factor inverted

= 190. mL HNO3

Solution

10

37

Dilution

In a dilution, §  Water is added. §  Volume increases. §  Concentration decreases.

38

Initial and Diluted Solutions

In the initial and diluted solution, §  The moles of solute are the same. §  The concentrations and volumes are related

by the following equations: For percent concentration C1V1 = C2V2

initial diluted

For molarity M1V1 = M2V2

initial diluted

39

Dilution Calculations with Percent

What volume of a 2.00 %(m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0%(m/v) HCl solution? Prepare a table:

C1= 14.0 %(m/v) V1 = 25.0 mL C2= 2.00%(m/v) V2 = ?

Solve dilution equation for unknown and enter values: C1V1 = C2V2

V2 = V1C1 = (25.0 mL)(14.0%) = 175 mL

C2 2.00%

40

Learning Check

What is the percent (%m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

11

41

Solution

What is the percent (%m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? Prepare a table:

C1= 9.00 %(m/v) V1 = 10.0 mL C2= ? V2 = 60.0 mL

Solve dilution equation for unknown and enter values: C1V1 = C2V2

C2 = C1 V1 = (10.0 mL)(9.00%) = 1.50 %(m/v)

V2 60.0 mL

42

Dilution Calculations

What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO3 to 0.540 L? Prepare a table:

M1= 0.600 M V1 = 0.180 L M2= ? V2 = 0.540 L

Solve dilution equation for unknown and enter values: M1V1 = M2V2

M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M

V2 0.540 L

43

Learning Check

What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? 1) 27.0 mL 2) 60.0 mL 3) 90.0 mL

44

Solution

What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? Prepare a table:

M1= 1.80 M V1 = 15.0 mL M2= 0.300 M V2 = ?

Solve dilution equation for V2 and enter values: M1V1 = M2V2

V2 = M1V1 = (1.80 M)(15.0 mL) = 90.0 mL

M2 0.300 M

12

45

Molarity in Chemical Reactions

In a chemical reaction, §  The volume and molarity of a solution are used to

determine the moles of a reactant or product. molarity ( mole ) x volume (L) = moles

1 L §  If molarity (mole/L) and moles are given, the volume

(L) can be determined moles x 1 L = volume (L)

moles

46

Using Molarity of Reactants

How many mL of 3.00 M HCl are needed to completely react with 4.85 g CaCO3? 2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l) STEP 1 Given 3.00 M HCl; 4.85 g CaCO3

Need volume in mL STEP 2 Plan g CaCO3 mole CaCO3 mole HCl mL HCl

47

Using Molarity of Reactants (cont.)

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l) STEP 3 Equalitites 1 mole CaCO3 = 100.1 g; 1 mole CaCO3 = 2 mole HCl 1000 mL HCl = 3.00 mole HCl

STEP 4 Set Up 4.85 g CaCO3 x 1 mole CaCO3 x 2 mole HCl x 1000 mL HCl

100.1 g CaCO3 1 mole CaCO3 3.00 mole HCl = 32.3 mL HCl required

48

Learning Check

How many mL of a 0.150 M Na2S solution are needed to completely react 18.5 mL of 0.225 M NiCl2 solution? NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq) 1) 4.16 mL 2) 6.24 mL 3) 27.8 mL

13

49

Solution

3) 27.8 mL 0.0185 L x 0.225 mole NiCl2 x 1 mole Na2S x 1000 mL 1 L 1 mole NiCl2 0.150 mole Na2S

= 27.8 mL Na2S solution

50

Learning Check

If 22.8 mL of 0.100 M MgCl2 is needed to completely react 15.0 mL of AgNO3 solution, what is the molarity of the AgNO3 solution? MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq) 1) 0.0760 M 2) 0.152 M 3) 0.304 M

51

Solution

3) 0.304 M AgNO3 0.0228 L x 0.100 mole MgCl2 x 2 moles AgNO3 x 1 1 L 1 mole MgCl2 0.0150 L = 0.304 mole/L = 0.304 M AgNO3

52

Learning Check

How many liters of H2 gas at STP are produced when Zn react with 125 mL of 6.00 M HCl? Zn(s) + 2HCl(aq) ZnCl2 (aq) + H2(g) 1) 4.20 L H2 2) 8.40 L H2 3) 16.8 L H2

14

53

Solution

2) 8.40 L H2 gas 0.125 L x 6.00 moles HCl x 1 mole H2 x 22.4 L 1 L 2 moles HCl 1 mole H2

= 8.40 L H2 gas

54

Chapter 8 Solutions

8.6 Properties of Solutions

55

Solutions

Solutions §  Contain small particles (ions or molecules).

§  Are transparent.

§  Do not separate.

§  Cannot be filtered.

§  Do not scatter light.

56

Colloids

Colloids

§  Have medium size particles.

§  Cannot be filtered.

§  Can be separated by semipermeable membranes.

§  Scatter light (Tyndall effect).

15

57

Examples of Colloids

TABLE 8.11

58

Suspensions

Suspensions §  Have very large particles.

§  Settle out.

§  Can be filtered.

§  Must be stirred to stay suspended.

§  Examples include blood platelets, muddy water, and Calamine lotion.

59

Solutions, Colloids, and Suspensions

60

Summary

TABLE 8.12

16

61

Learning Check

A mixture that has solute particles that do not settle out, but are too large to pass through a semipermeable membrane is called a 1) solution 2) colloid 3) suspension

62

Solution

A mixture that has solute particles that do not settle out, but are too large to pass through a semipermeable membrane is called a 2) colloid

63

Osmosis

In osmosis, water (solvent) flows from the lower solute concentration into the higher solute concentration. The level of the solution with the higher concentration rises. The concentrations of the two solutions become equal with time.

64

Osmosis

A semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens?

semipermeable membrane

10% starch 4% starch H2O

17

65

Water flow Equalizes

§  The 10% starch solution is diluted by the flow of water out of the 4% and its volume increases.

§  The 4% solution loses water and its volume decreases.

§  Eventually, the water flow between the two becomes equal.

7% starch 7% starch

H2O 66

Osmotic Pressure

Osmotic pressure is

§  Produced by the solute particles dissolved in a solution.

§  Equal to the pressure that would prevent the flow of additional water into the more concentrated solution.

§  Greater as the number of dissolved particles in the solution increases.

67

Learning Check

A semipermeable membrane separates a 10% starch solution (A) from a 5% starch solution (B). If starch is a colloid, fill in the blanks in the statements below.

1. Solution ____ has the greater osmotic pressure. 2. Water initially flows from ___ into ___. 3. The level of solution ____will be lower.

68

Solution

A semipermeable membrane separates a 10% starch solution (A) from a 5% starch solution (B). If starch is a colloid, fill in the blanks in the statements below.

1. Solution A has the greater osmotic pressure.

2. Water initially flows from B into A.

3. The level of solution B will be lower.

18

69

Osmotic Pressure of the Blood

Red blood cells §  Have cell walls that are semipermeable

membranes. §  Maintain an osmotic pressure that cannot

change or damage occurs. §  Must maintain an equal flow of water between

the red blood cell and its surrounding environment.

70

Isotonic Solutions

An isotonic solution §  Exerts the same osmotic

pressure as red blood cells.

§  Is known as a “physiological solution”.

§  Of 5.0% glucose or 0.90% NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells.

71

Hypotonic Solutions

A hypotonic solution §  Has a lower osmotic

pressure than red blood cells.

§  Has a lower concentration than physiological solutions.

§  Causes water to flow into red blood cells.

§  Causes hemolysis: RBCs swell and may burst.

72

Hypertonic Solutions

A hypertonic solution §  Has a higher osmotic

pressure than RBCs. §  Has a higher

concentration than physiological solutions.

§  Causes water to flow out of RBCs.

§  Causes crenation: RBCs shrinks in size.

19

73

Dialysis

In dialysis, §  Solvent and small solute

particles pass through an artificial membrane.

§  Large particles are retained inside.

§  Waste particles such as urea from blood are removed using hemodialysis (artificial kidney).

74

Learning Check

Indicate if each of the following solutions is 1) isotonic 2) hypotonic 3) hypertonic A.____ 2% NaCl solution B.____ 1% glucose solution C.____ 0.5% NaCl solution D.____ 5% glucose solution

75

Solution

Indicate if each of the following solutions is 1) isotonic 2) hypotonic 3) hypertonic

A._3_ 2% NaCl solution B._2_ 1% glucose solution C._2_ 0.5% NaCl solution D._1_ 5% glucose solution

76

Learning Check

When placed in each of the following, indicate if a red blood cell will

1) not change 2) hemolyze 3) crenate

A.____ 5% glucose solution

B.____ 1% glucose solution

C.____ 0.5% NaCl solution

D.____ 2% NaCl solution

20

77

Solution

When placed in each of the following, indicate if a red blood cell will

1) not change 2) hemolyze 3) crenate

A._1_ 5% glucose solution

B._2_ 1% glucose solution

C._2_ 0.5% NaCl solution

D._3_ 2% NaCl solution

78

Hemodialysis

§  When kidneys fail, an artificial kidney uses hemodialysis to remove waste particles such as urea from blood.

79

Learning Check

Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if

any, will be found in the water outside the bag?

A. 10% KCl solution

B. 5% starch solution

C. 5% NaCl and 5% starch solutions

80

Solution

Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if

any, will be found in the water outside the bag?

A. 10% KCl solution KCl ( K+, Cl−)

B. 5% starch solution None; starch is retained.

C. 5% NaCl and 5% starch solutions

NaCl (Na+, Cl−), but starch is retained.