SolutionsChapter 16

Copyright © The McGraw-Hill Companies, Inc. 

16.1

A solution is a homogenous mixture of 2 or more substancesThe solute what is being dissolved… usually the one with the smaller amount(s)The solvent is the substance present in the larger amount, doing the dissolving

A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

Sodium acetate crystals rapidly form when a seed crystal isadded to a supersaturated solution of sodium acetate.

16.1

Three types of interactions in the solution process:• solvent-solvent interaction• solute-solute interaction• solvent-solute interaction

Hsoln = H1 + H2 + H3

solid solubility usually increases with temperature,however, gas solubility generally decreases…

16.1

Temperature and Solubility

“like dissolves like”

Two substances with similar intermolecular forces are likely to be soluble in each other.

• non-polar molecules are soluble in non-polar solvents

CCl4 in C6H6

• polar molecules are soluble in polar solvents

C2H5OH in H2O

• ionic compounds are more soluble in polar solvents

NaCl in H2O or NH3 (l)

16.1

Which of the following in each pair is likely to be the more soluble in hexane, C6H14 ?

(A) cyclohexane (C6H12 ) or glucose (C6H12O6 )

(B) propionic acid (CH3CH2COOH) or sodium propionate (CH3CH2COONa)

(C)Hydrochloric acid (HCl) or ethyl chloride (CH3CH2Cl)

The Cleansing Action of Soap

A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.

Colloid versus solution

• collodial particles are much larger than solute molecules

• collodial suspension is not as homogeneous as a solution

12.8

Pressure and Solubility of Gases

12.5

The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

S1 = S2

P1 P2

S1 is the solubility of a gas at one pressure (P1)

low P

low c

high P

high c

S2 is the solubility of a gas at one pressure (P2)

Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

Percent by Mass (or volume)

% by mass = x 100%mass of solutemass of solute + mass of solvent

= x 100%mass of solutemass of solution

16.2

Mole Fraction (X)

XA = moles of A

sum of moles of all components

% by mass

• What is the mass percentage of Iodine, I2, in a solution containing 0.0650 mol I2 in 120.0g of CCl4?

Percent by volume

• What is the percent by volume of vinegar if 25mL of pure vinegar (acetic acid) are diluted by 150mL of water?

Mole fraction

• What is the mole fraction of NaOH in water if 15.00g NaOH are dissolved in 1000.0g of water?

Concentration Units Continued

M =moles of solute

liters of solution

Molarity (M) *note the capital letter

Molality (m) *note the lower case

m =moles of solute

mass of solvent (kg)

16.2

Molarity

• What is the molarity of a solution formed by dissolving 1.80 mol of KCl in 1.600 L of water?

To dilute concentrations already in molarity,

use this equation: M1V1 = M2V2

• Ex: How can I make 500.0 mL of a 1.0 M HCl solution using 12 M stock solution?M1= 12 M

V1= ?

M2= 1.0 M

V2 = 500.0 mL

V1 = M2V2/M1 = (1.0M)(500.0mL)/(12M)

V1 = 42mL

* I will need 42 mL stock and 458 mL water (500mL- 42mL)

What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?

m =moles of solute

mass of solvent (kg)M =

moles of solute

liters of solution

Assume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)

mass of solvent = mass of solution – mass of solute

= 927 g sol’n – 270 g eth = 657 g solvent = 0.657 kg solvent

m =moles of solute

mass of solvent (kg)=

5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

16.2

Boiling-Point Elevation

Tb = Tb – T b0

Tb > T b0 Tb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

Tb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m) for a given solvent 12.6

Freezing-Point Depression

Tf = T f – Tf0

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m) for a given solvent

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

Tf = Kf m

m =moles of solute

mass of solvent (kg)= 2.41 m=

3.202 kg solvent

478 g x 1 mol62.01 g

Kf water = 1.86 0C/m

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C