Chapter 11

• Suggested problems: 5, 9, 13, 15, 31, 33,

37, 39, 41, 43, 45, 53

Chapter 11

• Solutions

• Solution – solvent & solute.

• Solution – homogenous mixture – single

phase.

Chapter 11

• Types of solutions

• All nine combinations are possible:

• Solute Solvent

• Gas Gas

• Gas Liquid

• Gas Solid

• Liquid Gas

• Liquid Liquid

• Liquid Solid

• Solid Gas

• Solid Liquid

• Solid Solid

Chapter 11

• Effect of temperature on solubility

• If dissolving solute is exothermic then increase

in temperature makes solute less soluble

• If dissolving process is endothermic increasing

temperature will make solute more soluble.

• Dissolving ionic solid is usually endothermic

(takes energy to break up lattice) and so most

ionic solids are more soluble with increasing

temperature.

Chapter 11

• Dissolving a solid in a liquid is a three-step

process.

• 1) Breaking the bonds between the solid

particles - that takes energy - endothermic

• 2) Breaking the bonds between the liquid

molecules to make room for the solid particles -

that takes energy - endothermic

• 3) Making bonds between the solute and solvent

particles - that releases energy - exothermic

Chapter 11

• When a solute is soluble, steps 1 and 2

require less energy than step three

produces.

• When a solute is insoluble, the first two

steps require too much energy.

• Usually, it is the first step - breaking the

forces between solute particles - requires

too much energy.

Chapter 11

• NaCl is soluble in water because step 3,

where the Na+ and Cl- ions bind to the

water molecules, releases sufficent

energy.

Chapter 10

• ion-dipole.

Na+

H

HO δ+δ-

Chapter 11

• Hydration of NaCl(aq)

• NaCl(aq) ⇒ Na+(aq) + Cl-

Na ClNa

+

H H

O

Cl-

Chapter 11

• Oil won’t dissolve in water because,

though the oil molecules are easily

separated (step 1), there is no real

attraction for the oil molecules for the

water molecules (step 3).

Chapter 11

• Gases less soluble in liquids with

increasing temperature.

• Increasing temp increases molecular

vibrations

• Decreases intermolecular forces.

Chapter 11

• Gas-Liquid solutions

• Henry's law

• As the (partial) pressure of a gas is

increased above the liquid, the solubility of

the gas goes up.

• Carbonated beverages

Chapter 11

• Solutions

• Unsaturated: less solute dissolved than

possible

• Saturated: Maximum amount of solute

dissolved

• Supersaturated: More solute dissolved

than is “supposed” to dissolve.

Chapter 11

Molarity M = mols of solute

volume solution

% by weight (w/w) mass solute

mass solution

% by volume (v/v) volume solute

vol. solute + vol. solvent

Chapter 11

• molality m = mols solute

Kg of solvent

• mole fraction For a solution of A and B

χA = mols A

mols A + mols B

• χB = mols B

mols A + mols B

Chaper 11

• Calculate the wt of HCl in 5.00 mL of conc

HCl solution, of density = 1.19 g/mL and

37.23% HCl (w/w)

Chapter 11

• What is the M of H2SO4 solution, which

has a density of 1.2 g/mL and is 27%

(w/w)?

Chapter 11

• What is M of 16 g of CH3OH in 200 mL soln?

• M = mols solute/L solution

Chapter 11

• Find the M, m, and χacid and χwater of a solution of H2SO4 of density = 1.2 g/mL and 27.0% (w/w)

Chapter 11

• b) 1200 g solution minus 324 g H2SO4 =

876 g water

Chapter 11

• c) Mol Fraction:

Chapter 11

• Find the M, m, and χsolute and χwater of a solution of C6H12O6 of density = 2.45 g/mL and 34.5% (w/w)

Chapter 11

• b) 2450 g solution minus 845.3 g C6H12O6

= 1604.7 g water

•

Chapter 11

• c) Mol Fraction:

Chapter 11

• Those properties that depend on the

number, but not the nature, of the solute

particles

• Raoult's Law. P. 625 Accounts for vapor

pressure lowering.

Psoln = (χ solvent)(Po

pure solvent)

Chapter 11

Pure solvent Solution

Chapter 11

• The vapor pressure of water at 28oC is

28.35 Torr. Find the vapor pressure of a

solution of 68 g of C12H22O11 in 1000 g of

water at 28o C.

Chapter 11

• B.P. lowering and F.P. elevation760

ΔTf0 100 ΔTb

Chapter 11

• ΔTb = kb m and ΔTf = kf m

• Calculate the F.P. and B.P. of 2.60 g of

urea, CO(NH2)2 in 50 g of water.

Chapter 11

• Find the F.P. of 900 g ethylene glycol in 6

L water? Given: kf(H2O) = 1.86oC/m

• Ethylene glycol is (HO)CH2-CH2(OH)

Chapter 11

• 4.50 g of solute is dissolved in 125 g of

water. The solution freezes at -0.372oC.

What is the MW of the solute?

Chapter 11

• What is the MW of a solute if a solution of

0.510 g solute dissolved in 25 g of

benzene has a F.P. = 2.32oC?

Chapter 11

Chapter 11

Semi-permeablemembrane

Water moving from higher concentration(of water) to lowerconcentration of water.

Chapter 11

• ΠV = nRT or Π = MRT

• Because M = mols/V

• Π is osmotic pressure

Ocean pressure > Πbigcity

Chapter 11

• Electrolyte Solutions

• Colligative properties depend upon the

number of particles and not their nature

• So, ΔT = km is actually:

• ΔT = ikm where i is:

• actual number of particles after

dissolution

number of formula units initally dissolved

• Book calls i the van’t Hoff factor (p. 513)

Chapter 11

• Electrolyte Solutions

• Calculate the freezing point of 62 g of

H2SO4

in 500 g of water

• Remember, H2SO4 --> 2 H+ + SO42-

•

Chapter 11

• Electrolyte Solutions

• i actually usually less than the ideal value

• Due to formation of ion pairs

• That is, when ionic material dissolves

• Some cations and anions don’t separate

• Especially true when cation and anion

have high charge to size ratios

• Al2(SO4)3 --> Al3+ + SO42- --> Al3+---SO4

2-