1 © 2006 Brooks/Cole - Thomson Solutions Why does a raw egg swell or shrink when placed in...

1

© 2006 Brooks/Cole - Thomson

Solutions

Why does a raw egg swell or shrink when placed in different solutions?

2


Some Definitions

A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase.

One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

3


Solutions can be classified as saturated or unsaturated.

A saturated solution contains the maximum quantity of solute that dissolves at that temperature.

Definitions

SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.

4


Dissolving An Ionic Solid

Active Figure 14.9

5


Energetics of the Solution

Process

Figure 14.8

6


Energetics of the Solution

Process

If the enthalpy of formation of the solution is more negative that that of the solvent and solute, the enthalpy of solution is negative.

The solution process is exothermic!

7


SupersaturatedSodium

Acetate

• One application of a supersaturated solution is the sodium acetate “heat pack.”

• Sodium acetate has an ENDOthermic heat of solution.

8


Supersaturated Sodium Acetate

Sodium acetate has an ENDOthermic heat of solution.

NaCH3CO2 (s) + heat ----> Na+(aq) + CH3CO2

-(aq)

Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC.

Na+(aq) + CH3CO2-(aq) ---> NaCH3CO2 (s) + heat

9


Colligative PropertiesOn adding a solute to a solvent, the properties

of the solvent are modified.

• Vapor pressure decreases

• Melting point decreases

• Boiling point increases

• Osmosis is possible (osmotic pressure)

These changes are called COLLIGATIVE PROPERTIES.

They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

10


An IDEAL SOLUTION is one where the properties depend only on the concentration of solute.

Need concentration units to tell us the number of solute particles per solvent particle.

The unit “molarity” does not do this!

Concentration Units

11


Concentration UnitsMOLE FRACTION, X

For a mixture of A, B, and C

XA mol fraction A = mol A

mol A + mol B + mol C

m of solute = mol solute

kilograms solvent

WEIGHT % = grams solute per 100 g solution(also ppm, ppb, which is like %, which is parts per hundred)

MOLALITY, m

12


Calculating Concentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight % of ethylene glycol.

13



250. g H2O = 13.9 mol

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.

Xglycol = 1.00 mol glycol

1.00 mol glycol + 13.9 mol H 2O

X glycol = 0.0672

Calculate mole fraction

14



Calculate molality

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of H2O. Calculate X, m, and % of glycol.

conc (molality) = 1.00 mol glycol0.250 kg H 2O

4.00 molal

%glycol = 62.1 g

62.1 g + 250. g x 100% = 19.9%

Calculate weight %

15


Dissolving Gases & Henry’s Law

Sg Gas solubility (mol/L) = kH • Pgas

kH for O2 = 1.66 x 10-6 M/mmHg

When Pgas drops, solubility drops.

16


Understanding Colligative Properties

To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

17


Psolvent = Xsolvent • Posolvent

Understanding Colligative Properties

Vapor pressure of H2O over a solution

depends on the number of H2O molecules

per solute molecule.

Psolvent proportional to Xsolvent

VP of solvent over solution = (Mol frac solvent)•(VP pure solvent)

RAOULT’S LAW

18


PA = XA • PoA

Raoult’s LawAn ideal solution is one that obeys Raoult’s

law.

Because mole fraction of solvent, XA, is always

less than 1, then PA is always less than PoA.

The vapor pressure of solvent over a solution is

always LOWERED!

19


Vapor Pressur

e Lowerin

g

Figure 14.14

20


Raoult’s LawAssume the solution containing 62.1 g of

glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg)

Solution

Xglycol = 0.0672 and so Xwater = ?

Because Xglycol + Xwater = 1

Xwater = 1.000 - 0.0672 = 0.9328

Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)

Pwater = 29.7 mm Hg

21


Elevation of Boiling Point

Elevation in BP = ∆TBP =

KBP•m

(where KBP is characteristic of solvent)

V P so lv enta fter a ddingso lute

V P Pure so lv ent

BP pureso lv ent

BP so lutio n

1 a tm

P

T

The boiling point of a solution is higher than that of the pure solvent.

22


Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g

of water. What is the BP of the solution?

KBP = +0.512 oC/molal for water (see Table 14.3).

Solution

1. Calculate solution molality = 4.00 m

2. ∆TBP = KBP • m

∆TBP = +0.512 oC/molal (4.00 molal)

∆TBP = +2.05 oC

BP = 102.05 oC

23


Change in Freezing Point

The freezing point of a solution is LOWER than that of the pure

solvent.

FP depression = ∆TFP = KFP•m

Pure water Ethylene glycol/water solution

24


Lowering the Freezing Point

Water with and without antifreeze When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

25


How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?.

Solution

Calc. required molality (*This one is a little

different because the solute is ionic, not

molecular.)

∆TFP = KFP • m

-10.00 oC = (-1.86 oC/molal) • Conc

Conc = 5.38 molal

Freezing Point Depression

26


How much NaCl must be dissolved in 4.00 kg of water to lower

FP to -10.00 oC?.

Solution

Conc req’d = 5.38 molal

This means we need 5.38 mol of dissolved particles per kg of solvent.

*Recognize that m represents the total concentration of all dissolved particles.

Recall that 1 mol NaCl(aq)--> 1 mol Na+(aq) + 1 mol Cl-(aq)


27


How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?.

SolutionConc req’d = 5.38 molalWe need 5.38 mol of dissolved particles per

kg of solvent. Each mol of NaCl contributes 2 mol of dissolved particles.

NaCl(aq) --> Na+(aq) + Cl-(aq)To get 5.38 mol/kg of particles we need

5.38 mol / 2 = 2.69 mol NaCl / kg2.69 mol NaCl / kg ---> 157 g NaCl / kg(157 g NaCl / kg)•(4.00 kg) = 629 g NaCl


28


Boiling Point Elevation and Freezing Point

Depression ∆T = K•m•iA generally useful equation

i = van’t Hoff factor = number of particles produced per formula unit.

Compound Theoretical Value of i

glycol 1

NaCl 2

CaCl2 3

29


Osmosis

Dissolving the shell in vinegar

Egg in corn syrupEgg in pure water

30


Osmosis

The semipermeable membrane allows only the movement of solvent molecules.

Solvent Solution

Semipermeable membrane

Solvent molecules move from pure solvent to solution in an attempt to make both have the

same concentration of solute.

Driving force is entropy

31


Process of Osmosis

32


Osmosis at the Particulate Level

Figure 14.17

33


Osmosis at the Particulate Level

Figure 14.17

34


Osmosis

• Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.

35


Osmosis and Living Cells

36


Reverse OsmosisWater Desalination

Water desalination plant in Tampa

37


Osmosis Calculating a Molar Mass

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.

Solution

(a) Calc. ∏ in atmospheres

∏ = 10.0 mmHg • (1 atm / 760 mmHg)

= 0.0132 atm

(b) Calculate molarity

38


Osmosis Calculating a Molar Mass

Conc = 5.39 x 10-4 mol/L

(c) Calc. molar mass

Molar mass = 35.0 g / 5.39 x 10-4 mol/L

Molar mass = 65,100 g/mol

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.

Solution

(b)Calc. molarity from ∏ = iMRT

Conc = 0.0132 atm

(0.0821 L • atm/K • mol)(298K)

1 © 2006 Brooks/Cole - Thomson Solutions Why does a raw egg swell or shrink when placed in...

Documents

Transcript of 1 © 2006 Brooks/Cole - Thomson Solutions Why does a raw egg swell or shrink when placed in...