NOTES: 16.1-16.2 – Solutions and Concentration

Solutions – a REVIEW:● SOLUTION – a homogeneous

mixture of pure substances

● the SOLVENT is the medium in which the SOLUTES are dissolved. (The solvent is usually the most abundant substance.)

● Example: Solution: Salt WaterSolute: SaltSolvent: Water

The process of dissolution is favored by:

● A decrease in the energy of the system (exothermic)

● An increase in the disorder of the system (entropy)

Liquids Dissolving in Liquids

● Liquids that are soluble in one another (“mix”) are MISCIBLE.

-“LIKE dissolves LIKE”

● POLAR liquids are generally soluble in other POLAR liquids.

● NONPOLAR liquids are generally soluble in other NONPOLAR liquids.

Liquids Dissolving in Liquids

● Liquids that are insoluble in one another (do not mix) are IMMISCIBLE.

● Example: oil and water

Factors affecting RATE of dissolution:

● Surface area / particle size Greater surface area, faster it dissolves

● Temperature Most solids dissolve faster @ higher temps

● Agitation Stirring / shaking will speed up rate of dissolution

SATURATION:

● Unsaturated solution – is able to dissolve more solute

● Saturated solution – has dissolved the maximum amount of solute

● Supersaturated solution – has dissolved excess solute (at a higher temperature); (solid crystals generally form when this solution is cooled)

SOLUBILITY

● SOLUBILITY = the AMOUNT of solute that will dissolve in a given amount of solvent

**Key difference between this and the RATE of dissolving!!

Factors Affecting SOLUBILITY:

● The nature of the solute and solvent: different substances have different solubilities

● Temperature: many solids substances become more soluble as the temp of a solvent increases;

however, gases are less soluble in liquids at higher temps.

● Pressure: Only affects the solubility of gases. As pressure increases, the solubility of gases increases.

SOLUBILITY CURVES:● shows how the

solubility of a particular substance in a particular solvent changes as temperature changes.

HENRY’S LAW:

● at a given temperature the solubility (S) of a gas is directly proportional to the pressure (P) or S1/P1 = S2/P2

Henry’s Law EXAMPLE:If the solubility of a gas in water is 0.77 g/L at 3.5 atm of

pressure, what is its solubility (in g/L) at 1.0 atm of

pressure? (the temp. is held constant at 25°C)

Henry’s Law EXAMPLE:If the solubility of a gas in water is 0.77 g/L at 3.5 atm of

pressure, what is its solubility (in g/L) at 1.0 atm of

pressure? (the temp. is held constant at 25°C)

S1/P1 = S2/P2

0.77 g/L = S2

3.5 atm 1.0 atm

S2 = 0.22 g/L

Concentration of Solution

● CONCENTRATION refers to the amount of solute dissolved in a solution.

● a DILUTE solution is one that contains a low concentration of solute

● a CONCENTRATED solution contains a high concentration of solute

MOLARITY

nsol' L

solute mol(M)Molarity

MOLARITY example #1:

What is the molarity of a 2.0 L solution

containing 4.0 mol NaCl?

MOLARITY example #1:

What is the molarity of a 2.0 L solution

containing 4.0 mol NaCl?

MOLARITY = mol solute / L solution

= 4.0 mol NaCl / 2.0 L

= 2.0 mol/L

or 2.0 M NaCl

MOLARITY example #2:

A salt solution contains 0.90 g of NaCl in

100.0 mL of solution. What is the molarity

of the solution?

MOLARITY example #2:

A salt solution contains 0.90 g of NaCl in

100.0 mL of solution. What is the molarity

of the solution?

mol NaCl = 0.90 g x 1 mol = 0.0154 mol

58.5 g

MOLARITY example #2:

A salt solution contains 0.90 g of NaCl in

100.0 mL of solution. What is the molarity

of the solution?

MOLARITY = 0.0154 mol / 0.100 L

= 0.154 mol / L

or 0.154 M NaCl

MOLARITY example #3:

How many moles of solute are present in

1.5 L of 0.24 M Na2SO4?

MOLARITY example #3:

How many moles of solute are present in

1.5 L of 0.24 M Na2SO4?

1.5 L x 0.24 mol = 0.36 mol Na2SO4

1 L

MOLARITY BY DILUTION

● When you dilute a solution, you can use this equation:

2211 VMVM

DILUTION example #1:

How many liters of 5.0 M CuSO4 would be

needed to prepare 0.1 L of 0.5 M CuSO4?

DILUTION example #1:

How many liters of 5.0 M CuSO4 would be

needed to prepare 0.1 L of 0.5 M CuSO4?

(5.0 M) (V1) = (0.5 M) (0.1 L)

V1 = 0.01 L

or 10. mL

DILUTION example #2:

How many milliliters of a stock solution of

2.00 M MgSO4 would you need to prepare

100.0 mL of 0.400 M MgSO4?

DILUTION example #2:

How many milliliters of a stock solution of

2.00 M MgSO4 would you need to prepare

100.0 mL of 0.400 M MgSO4?

(2.00 M) (V1) = (0.400 M) (100.0 mL)

V1 = 20.0 mL

Example #3: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) solid Na2SO4

b) 5.00 M Na2SO4

L 2.50

xM 665.0

nsol' L

solute mol(M)Molarity

mol 1.6625x

g236mol 1

g 1.142SONa mol 6625.1 42

Dissolve 236 g of Na2SO4 in enough water

to create 2.50 Lof solution.

Example #3: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) solid Na2SO4

b) 5.00 M Na2SO4

L) M)(2.50 665.0()M)(V (5.00 1

mL 333 L 333.0V1

2211 VMVM

Add 0.333 L of Na2SO4 to 2.17 L of water.

PERCENT SOLUTIONS:

● If both the solute and the solvent are liquids, the concentration of the solute can be expressed as a PERCENT BY VOLUME:

% (v/v) = volume of solute x 100%

solution volume

Percent by Vol. example #1

What is the percent volume of acetone in

water if 25 mL of acetone is added to 75

mL of water?

Percent by Vol. example #1

What is the percent volume of acetone in

water if 25 mL of acetone is added to 75

mL of water?

% v/v = solute vol. / solution vol.

= 25 mL / (75 + 25 mL)

= 25 mL / 100 mL

= 0.25 x 100%

= 25% acetone (v/v)

Percent by Vol. example #2:What is the percent by volume of ethanol

(C2H5OH) in the final solution when 85 mL

of ethanol is diluted to a volume of 250 mL

with water?

Percent by Vol. example #2:What is the percent by volume of ethanol

(C2H5OH) in the final solution when 85 mL

of ethanol is diluted to a volume of 250 mL

with water?

% (v/v) = 85 mL / 250 mL

= 34% ethanol (v/v)

MASS PERCENT:

● If a solid solute is dissolved into a liquid, the concentration of the solute can be expressed as a PERCENT BY MASS:

% (m/v) = mass of solute (g) x 100%

solution volume (mL)

Mass % example #1:

What is the % (m/v) of a 100 mL aqueous

solution containing 10 g of NaCl?

Mass % example #1:

What is the % (m/v) of a 100 mL aqueous

solution containing 10 g of NaCl?

% (m/v) = mass solute / vol. solution

= 10 g NaCl / 100 mL

= 0.10 x 100%

= 10% NaCl (m/v)

Mass % example #2:

What is the percent of NaCl in a solution

made by dissolving 24 g of NaCl in water

to a final volume of 174 mL?

Mass % example #2:

What is the percent of NaCl in a solution

made by dissolving 24 g of NaCl in water

to a final volume of 174 mL?

% (m/v) = mass solute / vol. solution

= 24 g NaCl / 174 mL

= 0.138 x 100%

= 13.8% NaCl (m/v)