5/5/15

1

1

Chapter 8A

Solutions

2

CHAPTER OUTLINE���

§  Type of Solutions §  Electrolytes & Non-electrolytes §  Equivalents of Electrolytes §  Solubility & Saturation §  Soluble & Insoluble Salts §  Formation of a Solid §  Precipitation Reactions

5/5/15

2

3

TYPE OF���SOLUTIONS���

q  A solution is a homogeneous mixture of two substances:

Solute:

Solvent:

substance being dissolved

present in smaller amount

substance doing the dissolving

present in larger amount

Solutes and solvents may be of any form of matter: solid, liquid or gas.

4

TYPE OF���SOLUTIONS���

Type Example Solute Solvent

Gas in gas Air Oxygen Nitrogen

Soda water CO2 Water Gas in liq.

Vinegar Acetic acid Water Liq. in liq.

Seawater Salt Water Solid in liq.

5/5/15

3

5

TYPE OF���SOLUTIONS���

Type Example Solute Solvent

Dental amalgam Mercury Silver

Brass Zinc Copper

Liquid in solid

Solid in solid

6

SOLUBILITY���

q  Solutions form between solute and solvent molecules because of similarities between them.

Like dissolves Like

Ionic solids dissolve in water because the charged ions (polar) are attracted to the polar water molecules.

Non-polar molecules such as oil and grease dissolve in non-polar solvents such as kerosene.

5/5/15

4

7

SOLUBILITY���

Water (polar) CH2Cl2 (non-polar)

I2 (non-polar)

Ni(NO3 )2 (polar)

8

ELECTROLYTES &���NON-ELECTROLYTES���

q  Solutions can be characterized by their ability to conduct an electric current.

q  Solutions containing ions are conductors of electricity and those that contain molecules are non-conductors. q  Substances that dissolve in water to form ions are called electrolytes.

q  The ions formed from these substances conduct electric current in solution, and can be tested using a conductivity apparatus.

5/5/15

5

9

STRONG���ELECTROLYTES���

q  Electrolytes are further classified as strong electrolytes and weak electrolytes.

q  In water, a strong electrolyte exists only as ions.

q  Strong electrolytes make the light bulb on the conductivity apparatus glow brightly.

q  Ionic substances such as NaCl are strong electrolytes.

NaCl (s) Na+ (aq) + Cl- (aq)

Only ions present

after solution

10

WEAK���ELECTROLYTES���

q  Solutions containing weak electrolytes contain only a few ions.

q  These solutions make the light bulb on the conductivity apparatus glow dimly.

q  Weak acids and bases that dissolve in water and produce few ions are weak electrolytes.

HF (aq) H+ (aq) + F- (aq)

Few ions present

after solution

5/5/15

6

11

NON-���ELECTROLYTES���

q  Substances that do not form any ions in solution are called non-electrolytes.

q  With these solutions the bulb on the conductivity apparatus does not glow.

q  Covalent molecules that dissolve in water but do not form ions, such as sugar, are non-electrolytes.

C12H22O11 (s) C12H22O11 (aq)

No ions present

after solution

12

ELECTROLYTES &���NON-ELECTROLYTES���

5/5/15

7

13

Example 1:���Identify the predominant particles in each of the following solutions and write the equation for the formation of the solution:

NH4Br Strong electrolyte (only ions)

NH4Br (s) NH4+ (aq) + Br- (aq)

14

Example 1:���Identify the predominant particles in each of the following solutions and write the equation for the formation of the solution:

CH4 N2O Non-electrolyte (only molecules)

CH4N2O (s) CH4N2O (aq)

5/5/15

8

15

Example 1:���Identify the predominant particles in each of the following solutions and write the equation for the formation of the solution:

HClO Weak electrolyte (few ions)

HClO (aq) H+ (aq) + ClO- (aq)

16

EQUIVALENTS OF���ELECTROLYTES���

q  Body fluids typically contain a mixture of several electrolytes, such as Na+

, Cl–, K+ and Ca2+. q  Each individual ion is measured in terms of an

equivalent (Eq), which is the amount of that ion equal to 1 mole of positive or negative electrical charge.

q  For example, 1 mole of Na+ ions and 1 mole of Cl– ions are each 1 equivalent (or 1000 mEq) because they each contain 1 mole of charge.

5/5/15

9

17

EQUIVALENTS OF���ELECTROLYTES���

q  An ion with a charge of 2+ or 2– contains 2 equivalents per mole.

q  Some examples of ions and their equivalents are shown below:

Ion Electrical Charge

No. of Equivalents in 1 Mole

Na+ 1+ 1 Eq Ca2+ 2+ 2 Eq Fe3+ 3+ 3 Eq Cl– 1– 1 Eq

SO42– 2– 2 Eq

18

EQUIVALENTS OF���ELECTROLYTES���

q  In body, the charge of the positive ion is always balanced by the charge of the negative ion.

q  For example, a solution containing 25 mEq/L of Na+ and 4 mEq/L of K+ must have 29 mEq/L of Cl– to balance.

q  Shown next are examples of some common intravenous solutions and their ion concentrations.

5/5/15

10

19

EQUIVALENTS OF���ELECTROLYTES���

20

Example 1:���Indicate the number of equivalents in each of the following: 2 mol K+ 2 mol x

1 Eq 1 mol = 2 Eq

0.5 mol Mg2+ 0.5 mol x 2 Eq 1 mol

= 1 Eq

3 mol CO32- 3 mol x

2 Eq 1 mol

= 6 Eq

5/5/15

11

21

Example 2:���A typical concentration for Ca2+ in blood is 8.8 mEq/L. How many moles of Ca2+ are present in 0.50 L of blood?

0.50 L x

Liter blood mEq moles

Ca2+

8.8 mEq 1 L

1 mol 2 Eq

x1 Eq

103 mEq x = 0.0022 mol

22

Example 3:���An IV solution contains 155 mEq/L of Cl–. If a patient received 1250 mL of the IV solution, how many moles of chloride were given to him?

1250 mL x 1 L

103 mL 1 mol 1 Eq

x155 mEq

1 L x

= 0.194 mol

1 Eq 103 mEq

x

5/5/15

12

23

Example 4:���A sample of Ringer’s solution contains the following concentrations (mEq/L) of cations: Na+ 147, K+ 4, and Ca2+ 4. If Cl– is the only anion in the solution, what is the concentration of Cl– in mEq/L?

Total cation mEq/L = 147 + 4 + 4 = 155 mEq/L

Total cation mEq/L = Total anion mEq/L

Total Cl- mEq/L = 155 mEq/L

24

SOLUBILITY���

q  Solubility refers to the maximum amount of solute that can be dissolved in a given amount of solvent.

q  Many factors affect the solubility of a solute in a solution.

Type of solute

Type of solvent Temperature

Solubility is measured in grams of solute per 100 grams of solvent at a given temperature.

5/5/15

13

25

SATURATION���

q  A solution that does not contain the maximum amount of solute in it, at a given temperature, is called an unsaturated solution.

q  A solution that contains the maximum amount of solute in it, at a given temperature, is called a saturated solution.

Un-dissolved solid in solution

26

SOLUBILITY���

q  Solubility of most solids in water increases as temperature increases.

q  Using a solubility chart, the solubility of a solute at a given temperature can be determined.

q  For example, KNO3 has a solubility of 80 g/100 g H2O (80%) at 40 °C.

5/5/15

14

27

SOLUBILITY���OF GASES���

q  Solubility of gases in water decreases as temperature increases.

q  At higher temperatures more gas molecules have the energy to escape from solution.

q  Henry’s law states that the solubility of a gas is directly proportional to the pressure above the liquid.

q  For example, a can of soda is carbonated at high pressures in order to increase the solubility of CO2. Once the can is opened, the pressure is reduced and the excess gas escapes from the solution.

28

SOLUBLE &���INSOLUBLE SALTS���

q  Many ionic solids dissolve in water and are called soluble.

q  However, some ionic salts do not dissolve in water and do not form ions in solution. These salts are called insoluble salts and remain solid in solution.

q  Chemists use a set of solubility rules to predict whether a salt is soluble or insoluble.

q  These rules are summarized next.

5/5/15

15

29

SOLUBILITY���RULES���

S O L U B L E

NO3- No exceptions

Na+, K+

NH4+ No exceptions

Cl-, Br-, I- Except those containing Ag+ , Pb2+

SO42-

Except those containing Ba2+ , Pb2+, Ca2+

30

SOLUBILITY���RULES���

I N S O L U B L E

S2-, CO32-

PO43-

Except those containing Na+ , K+, NH4

+

OH- Except those containing Na+ , K+, Ca2+, NH4

+

5/5/15

16

31

Example 1:���Use the solubility rules to determine if each of the following salts are soluble or insoluble:

K3PO4 soluble

CaCO3 insoluble (most carbonates are insoluble)

(all salts of K are soluble)

32

Example 2:���Using the solubility chart, determine if each of the following solutions is saturated or unsaturated at 20°C: 25 g NaCl in 100 g water

Solubility of NaCl at 20 °C is 40%

Solution is unsaturated

5/5/15

17

33

Example 2:���Using the solubility chart, determine if each of the following solutions is saturated or unsaturated at 20°C: 11 g NaNO3 in 25 g water

Solubility of NaNO3 at 20 °C is 85%

Solution is unsaturated

34

Example 2:���Using the solubility chart, determine if each of the following solutions is saturated or unsaturated at 20°C: 400. g of glucose in 125 g water

Solubility of glucose at 20 °C is 80%

Solution is saturated

5/5/15

18

35

FORMATION���OF A SOLID���

q  Solubility rules can be used to predict whether a solid, called a precipitate, can be formed when two solutions of ionic compounds are mixed.

q  A solid is formed when two ions of an insoluble salt come in contact with one another.

q  For example, when a solution of K2CrO4 is mixed with a solution of Ba(NO3)2 a yellow insoluble salt BaCrO4 is produced.

36

AQUEOUS���REACTIONS���

K2CrO4 (aq) + Ba(NO3)2 (aq) BaCrO4 (s) + 2 KNO3 (aq)

Solid BaCrO4 forms

precipitate

5/5/15

19

37

PRECIPITATION���REACTIONS���

q  Double replacement reactions in which a precipitate is formed are called precipitation reactions.

q  To predict a precipitate, follow the steps outlined next.

38

PRECIPITATION���REACTIONS���

1.  Write the reactant ions that form after dissolution. 2.  Write the product combinations possible when reactant

ions combine. 3.  Use solubility rules to determine if any of the products are

insoluble. 4.  If a precipitate forms, write the formula for the solid.

Write other ions that form soluble salts as ions. If no precipitate forms, write “NO REACTION” after the arrow.

5.  Cancel ions that appear the same on both sides of the equation (spectator ions), to form Net Ionic Equation.

5/5/15

20

39

The reaction of K2CrO4 and Ba(NO3)2 can be predicted as shown below

2 K+ +CrO42- Step 1: + Ba2+ + 2 NO3

- → ???

Step 2: 2 K+ +CrO42- +Ba2+ + 2 NO3

- → BaCrO4(?) + 2K+NO3- (?)

Step 3:

Step 4:

spectator ions Step 5: Ba2+ + CrO42- → BaCrO4 (s)

precipitate

PRECIPITATION���REACTIONS���

2 K+ +CrO42- +Ba2+ + 2 NO3

- → BaCrO4(s) + 2K+NO3- (aq)

2 K+ +CrO42- +Ba2+ + 2 NO3

- → BaCrO4(s) + 2K+ +2NO3-

Net Ionic Equation

40

Predict the products, if any, for the reaction of AgNO3 and NaCl and write the net ionic equation.

Example 1:���

Ag+ +NO3- Step 1: + Na+ + Cl- → ???

Step 2: Ag+ +NO3- +Na+ +Cl- → Ag+Cl- (?) + Na+ NO3

- (?)

Step 3: Ag+ +NO3- +Na+ +Cl- → Ag+Cl- (s) + Na+ NO3

- (aq)

Step 4: Ag+ +NO3- +Na+ +Cl- → AgCl (s) + Na+ + NO3

- (aq)

spectator ions Step 5: Ag+ +Cl- → AgCl (s)

Net Ionic Equation

precipitate spectator ions

5/5/15

21

41

Predict the products, if any, for the reaction of Na2SO4 and Pb(NO3)2 and write the net ionic equation.

Example 2:���

2 Na+ +SO42- Step 1: + Pb2+ + 2 NO3

- → ???

Step 2: 2Na+ +SO42- +Pb2+ +2NO3

- → 2Na+NO3-(?) + Pb2+SO4

2-(?)

Step 3:

Step 4:

Step 5: Pb2+ +SO42- → PbSO4 (s)

precipitate

2Na+ +SO42- +Pb2+ +2NO3

- → 2Na+NO3-(aq) + Pb2+SO4

2-(s)

2Na+ +SO42- +Pb2+ +2NO3

- → 2Na+ + 2NO3-(aq) + PbSO4

(s)

(NIE)

42

Predict the products, if any, for the reaction of PbCl2 and KI and write the net ionic equation.

Example 3:���

Pb2+ + 2 Cl- Step 1: + K+ + I- → ???

Step 2: Pb2+ + 2 Cl- + K+ + I- → Pb2+I2-(?) + K+Cl-(?)

Step 3:

Step 4:

Step 5: Pb2+ + 2 I- → PbI2 (s) precipitate

Pb2+ + 2 Cl- + 2 K+ +2 I- → Pb2+I2-(s) + 2 K+Cl-(aq)

Pb2+ + 2 Cl- + 2 K+ +2 I- → PbI2 (s) + 2 K+ + 2 Cl-

(NIE)

5/5/15

22

43

Predict the products, if any, for the reaction of NH4Cl and KNO3 and write the net ionic equation.

Example 4:���

NH4+ + Cl- Step 1: + K+ + NO3

- → ???

Step 2: NH4+ + Cl- + K+ + NO3

- → NH4+NO3

-(?) + K+Cl-(?)

Step 3:

Step 4:

NH4+ + Cl- + K+ + NO3

- → NH4+NO3

-(aq) + K+Cl-(aq)

NH4+ + Cl- + K+ + NO3

- → NH4+ + NO3

- + K+ + Cl-

→ No Reaction

44

THE END