Solution - a homogeneous mixture of two or more chemical substances

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Solution - a homogeneous mixture of two or more chemical substances. Solvent - the major component of a solution ( or sometimes the only liquid component). Solute - a minor component of a solution (the stuff which is dissolved in the solution). - PowerPoint PPT Presentation

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Solution - a homogeneous mixture of two or more chemical substances

Solvent - the major component of a solution ( or sometimes the only liquid component)

Solute - a minor component of a solution (the stuff which is dissolved in the solution)

Solutions: Concentration Expressions

Molarity: Moles of solute per liter of solution

Molality: Moles of solute per Kg of solvent

Mole Fraction: Moles of solute divided by totalmoles of solute and solvent.

Thermodynamics of Solubility

What is H° of solution?

Thermodynamics of Solubility

NaCl(s) Na+(g) + Cl-(g) Ho1 = 786 kJ/mol

H2O(l) + Na+(g) + Cl-(g) Na+(aq) + Cl-(aq)

Hohyd = Ho

2 + Ho3 = -783 kJ/mol

H° = +3 kJ/mol

So why do salts with H positive dissolve?

Entropy G = H - TS

NaCl S° = 46 J/Kmol

But some S° values are negative!

M

O

O

O O

O

O

H

H

H

H

HH

H

H

HH

H H

H

O H

H

OH

XH

OH

H O

H

The higher the charge on the ion and the smallerthe size of the ion, the greater the interaction. Thisleads to a greater ordering of the water.

Hydration of a salt - MX

The solvation of the ions leads to anordering of the water solution. Thus andecrease in the entropy of the water.

Chemical species that interact favorablywith water are call hydrophilic.

Chemical species that do not interact favorablywith water are called hydrophobic.

Non polar molecules such as hexane donot dissolve in water.

Oil floats on the top of water.

The enthalpy for dissolving hexane in wateris actually negative. So why doesn’t it dissolve?

HO

H

H

OH H

OH

HO

H

H

OHH

OH

H

CH H

H

When a non polar molecule is dissolved in water the water molecules form a highly ordered clatharate structure

The structure hereis a two dimensionalrepresentation of thetrue structure whichthree dimensional.

This ordering of the watermolecules leads to a negative S.

This is the basis of the hydrophobic effect

Solubilities of alcohols in water

Methanol infinite

Ethanol infinite

1-propanol infinite

1-butanol 80 g/L

1-pentanol 22 g/L

1-hexanol 5 g/L

SO O

ONa

Surfactants

The amount of gas dissolved in a liquid solutionis directly proportional to the pressure.

Partial Pressure = P = kHHenry’s Law

kH (atm) for water

CH4 4.13 x 102

O2 4.34 x 104

N2 8.57 x 104

In the atmosphere how much O2 will dissolve in H2O?

P = kHkH O2 4.34 x 104 atm

P(O2) in atm .21atm

.21atm = 4.34 x 104 atm x

= 4.84x 10-6 mol fraction of O2

1000 g H2O/18 g/mol = 55.6 mol H2O

4.84x 10-6 x 55.6 = 2.69 x 10-4 mol O2

2.69 x 10-4 mol x 32 g/mol = .0086 g O2

.0086 g O2 / liter H2O

at normal atmospheric pressure

In the actual atmosphere

.21 x 32.0 g / 22.4 = .30 g O2 / liter air

Aren’t you glad you are not a fish!

What will happen if you have two beakers in a closed system? One contains 1 liter of 1M NaCl,the other 1 liter of pure H2O.

Raoult’s Law

Vapor pressure ofsolution with a nonvolatile solute.

Pobs = solvPosolv

Raoult’s Law for a mixture of two volatile liquids.

Ptotal = PA + PB = APoA + BPo

B

Raoult’s says that the presence of a solutewill lower the vapor pressure of a solution ascompared to the pure solvent.

What will this do to the freezing point andboiling point of the solution?

Freezing Point Depression

T = Kf msolute

Boiling Point Elevation

T = Kb msolute

Calculate the boiling point of a solution containing6.16 g of sucrose, C12H22O11, dissolved in 30.g of H2O.

T = Kb msolute

Kb = 0.51 C°kg/mol m.wt. sucrose 342 g/mol

mol = 6.16 g/(342g/mol) = .018 mol

T = 0.51 C°kg/mol x .60 mol/kg =.31 C°

Boiling point of solution = 100.31 C°

m = .018 mol / .030 kg = .60 mol/kg

A solution of .90g of glucose in 20.0g of water hasa freezing point of -0.465 C°. What is mwt. of glucose?

Kf = 1.86 C°kg/mol

m = T/Kf = .465 C° / (1.86 C°kg/mol) =.25 m

.25 mol/kg = X mol/.020kg

X mol of glucose = .005 mol

.90 g/ .005 mol = mwt. / 1mol = 180 g / mol

Osmosis

Osmotic Pressure isthe pressure that just stops the osmosis.

= MRT

= MRT

= MRT M = / RT

A 2.2 gram sample of polyethylene was disolved intoluene to give 100ml solution. The osmoticpressure at 25°C was measured to be 1.10 x 10-2atm.What is the mwt. of the polyethylene?

M = 1.10 x 10-2 / (.082 x 298) = 4.5 x 10-4 M

.10 L x 4.5 x 10-4 mol/L = 4.5 x 10-5 mol

2.2 g / 4.5 x 10-5 mol = mwt = 49,000 g/mol

49,000/ 28 = 1750 units CH2CH2n

ReverseOsmosis

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