CH 10 Solutions

Principles of SolubilityNature of the solute and solvent

Like likes like.

Electrolytes will dissolve in a polar solvent (like water)

Non-electrolytes will dissolve more easily in non polar or slightly polar solvents (like benzene, or toluene)

CH 10 Solutions

Non electrolytes (molecular substances) dissolving in water.

This ‘unusual’ case will involve Hydrogen bonding or otherwise polar molecules.

Ex: alcohols (-OH

group)Methanol,

Ethanol Hydrogen peroxide

H2O2

Solubilities of Alcohols in Water

Formula Name Solubility in Water (g/100 g)

CH3OH methanol infinitely soluble

CH3CH2OH ethanol infinitely soluble

CH3(CH2)2OH propanol infinitely soluble

CH3(CH2)3OH butanol 9

CH3(CH2)4OH pentanol 2.7

CH3(CH2)5OH hexanol 0.6

CH3(CH2)6OH heptanol 0.18

CH3(CH2)7OH octanol 0.054

CH3(CH2)9OH decanol insoluble in water

“greasy end” is larger proportion in large alcohols.

CH 10 Solutions

DDT no more mosquitoes!

Pesticide used widely in the 60’s

Soluble in non polar solvents.

Concentrates in tissues of fish, birds, and other wildlife. (biological ½ life of 8 years)

Thins eggshells.CH 10 Solutions

CH 10 Solutions

Principles of SolubilityEffect of Pressure/Henry’s Law

For gases only Solubility is directly proportional

to pressure.

Soda is bottled under a pressure of 4atm.

This drives more of the CO2 into solution,

when the cap is released, the pressure drops to 1 atm, and CO2

bubbles and rapidly escapes solution.

Cg =k Pg Concentration in solution = constant x Pressure of gas

CH 10 Solutions

Principles of Solubility Effect of Temperature

Solid + Water Solution Usually + ΔH heat in

Increased T favors an endothermic process

So: Solubility of a solid usually increases with increased temperature

Gas + Water Soln. Usually –ΔH heat out

Increased T is detrimental for an exothermic process

• So: Solubility of a gas usually decreases with higher temperature.

CH 10 Solutions

Concentration Unitsproblem type 1. Molarity (M)

Molarity = moles of solute/

liters of solution

Units are moles/liter

Tricks:

change mass to moles

change mL to Liters

Typical Problems Making a volume of

solution from the solid. Creation

Preparing a volume of a solution from another (stock) solution. Dilution

CH 10 Solutions

ProblemsMolarity (M)

Prepare 35.0 mL of 0.200 M Aluminum nitrate from a solid sample.

1. How many moles of Al-nitrate would be in the solution?

2. Measure out that many grams and then add water to make 35.0 mL

Prepare 35.0 mL of 0.20 M Aluminum Nitrate solution from a 0.50M stock solution.

How many moles do you need? Remember

M1V1=M2V2

McVc=MdVd

Meausure out the volume of stock you need, then dilute

Molar mass (gfm) of aluminum nitrate is 213.03 g/n

CH 10 Solutions

Problem type 2.

Mole Fraction (X) Similar to percent

composition…you are interested in part of the whole

Moles of substance/ total moles

Typical problems will have you determine the mole fraction

from % mass. determine the mole fraction

of solute in solution

CH 10 Solutions

ProblemsMole Fraction (X)

If you dissolve 12.0 g of methanol in 100.0 g of water.

What is the mole fraction that is methanol?

1. Determine the moles of water

2. Determine the moles of methanol

3. Add together for the total moles, then take part/whole

Methane is CH4

Methanol is CH3OH

CH 10 Solutions

Problem type 3. Mass % Solute

Simply determine the percent composition by mass.

Trick is to change the decimal to a percent

(x100 + “%”)

Solute is the solid dissolved in

solution

CH 10 Solutions

Problem Type 4. Molality (m)

This is similar to molarity, but uses mass (kilograms) of the solvent, rather than liters of solvent.

Typical problems will have you determine the molality from other concentration units.

Molarity (M) = # moles solute/ Liter Solvent

CH 10 Solutions

Problems : Molality (m)

What is the molality of methanol if 12g of methanol are dissolved into 100g of water?

1. Find Moles of solute (methanol)

2. Find kilograms of solvent

3. Divide

if the solvent is water, M and m are numerically equal

12g/32.0 g/n = 0.375 n

100g/ 1000g/kg = 0.100kg

= 3.75m

CH 10 Solutions

Conversion between units-decide on a fixed amount of solution

When the original concentration [ ] is:

Start with:

Mass Percent 100 g solution

Molarity (M) 1.00L of solution

Molality (m) 1000g of solvent (1Kg)

Mole Fraction (X) 1 mol (solute+solvent)

CH 10 Solutions

Part III: Colligative PropertiesVapor Pressure Lowering

VPL = X2Pº1

VPL= (vapor pressure lowering)

X2= mole fraction of the solute

Pº1= vapor pressure of the PURE solvent

Each liquid has a vapor pressure.

The pressure to escape the surface into the gas phase.

Dissolved particles get in the way and block the escape, thereby reducing the vapor pressure.

The higher the concentration (X2), the more the VP is lowered.

CH 10 Solutions

Raoult’s Law

CH 10 Solutions

Part III: Colligative Properties*Boiling Point Elevation ΔTb

*Freezing Point Depression ΔTf

ΔTb = kbm

ΔTf = kfmThe change in temperature

from the normal boiling/freezing points is a constant (k) times the molality(m).

The higher the concentration of ‘contaminant’, the greater the change in freezing/boiling points

Boiling Elevation:

The particles lower the vapor pressure.

Kb for water is 0.52 °C/m

Freezing Depression:

The particles obstruct the formation of the solid crystal. There must be a lower temperature to form solid structure

Kf for water is 1.86 °C/m

CH 10 Solutions

Problems: Colligative Propertiesfreezing point depression

ΔTf = kfm Change in freezing

temp is (Kf) x (molality)

Find the freezing point of a solution containing 20.0 g of ethanol in 50.0g of water.

What is the molality of the solution?

Grams to moles Moles / Kg of

Solvent Molality x Kf

Kf for water is 1.86 °C/m

CH 10 Solutions

Problems: Colligative Propertiesboiling point elevation

ΔTf = kbm Change in boiling temp is (Kb) x (molality)

Find the boiling point of a solution containing 20.0 g of ethylene glycol in 50.0g of water.

What is the molality of the solution?

Grams to moles Moles / Kg of

Solvent Molality x Kb

Kb for water is 0.52 °C/m

CH 10 Solutions

Colligative PropertiesOsmotic Pressure (π)

π = M R T

M = molarityR= .0821 L * atm/n *K

T = Temperature

Water moves through a semi-permeable membrane from an area of high vapor pressure to low vapor pressure.

CH 10 Solutions

Problems:Using freezing point depression to find molar mass

A solution is made using .0100g of a substance in 1.00g of water.

The freezing point depresses 1.00 °C

What is the molar mass?

Solve for molarity