Chapter 11 Problems:

11, 15, 18, 20-23, 30, 32-35, 39, 41, 43, 45, 47, 49-51, 53, 55-57, 59-61, 63, 65, 67, 70, 71, 74, 75, 78, 81, 85, 86, 93

Chapter 11 – Properties of Solutions

Types of mixtures:homogenous – components are uniformly mixedheterogeneous – components are not uniformly mixed

Homogenous mixtures are called solutionssolvent:solute:

component in the majoritycomponent in the minority

Concentration Units:

1. % by mass = mass solute x 100mass solution

Problem: .892 g KCl are dissolved in 24.0 g water. Determine the % by mass.

2. Mole fraction (X) where Xa = mole a(mole a + mole b)

Problem: use the data above to calculate the mole fraction of KCl

Problem: What is the mole fraction of water?What is the sum of the 2 mole fractions?

3. Molarity = M = moles of solute Molar SolutionLiters of solution

4. Molality = m = moles of solute molal solutionkg of solvent

Problem: .892 g KCl is dissolved in 24.0 g water. Determine the molality of the solution.

5. Normality – N – The number of “equivalents” per li ter of solution.

H2SO4 will provide 2 mole H + ions and weighs 98.0 g/mol therefore 1 equivalent of H 2SO4 weighs 49.0 grams.

NaOH will provide 1 mole OH - ions and weighs 40.0 g/mol therefore 1 equivalent of NaOH weighs 40.0 grams.

1 equivalent of acid or base = mass of acid or base that will furnish 1 mole of H + or OH -

For Redox Reactions – The mass or quantity of oxidiz ing or reducing agent that will accept or furnish 1 mole of electrons.

MnO4- + 5e- -> Mn+2 + 4H2O

Problem: How many grams of MnO 4- is 1 equivalent?

Conversion between Molarity and Normality (acids an d bases only)

M x #H+ = N

M x #OH- = N

Since liquids expand when heated, and contract when cooled, the Molarity of a solution is not constant. Therefore, we measure Molarity at a standard temperature of 25 oC.

Density of solution must be known to convert 1 solu tion concentr ation to another.

What is the normality of:

a) 2M HClb) 3M H2SO4c) 3M Ca(OH)2

Problem: Determine the Molarity of a .396m aqueous C6H12O6 solution with a density of 1.16 g/mL.

Problem: The density of a 2.45 M aqueous CH 3OH solution is .976 g/mL. Calculate the molality.

Problem: Calculate the molality of a 34.5% by mass aqueous solution of phosphoric acid.

Problem: The density of a 3.75 M H 2SO4 solution in a battery is 1.230 g/mL. Determine its mass % and Normality.

The Process of dissolving can be broken down into 3 steps that involve energy:

Step 1 & 2: Expanding the solvent-solvent and solute-solute bonds involves breaking intermolecular bonds, which is endothermic.

Step 3 Solute/Solvent bonds form, which is exothermic.

1

2

∆ Hsolution = ∆H1 + ∆H2 + ∆H3

Problem: When CaCl 2 is dissolved, the solution becomes warm. How do process #1,2 (expanding the solute and solvent) com pare to process #3 (bonds forming between solute and solvent)?

Problem: When NH 4NO3 dissolved, the solution cools down. Compare #1,2,3

2 Factors control whether a reaction is spontaneous or not.

1. ∆H

2. Entropy – A measure of disorder, where disorder is favorable to order. Which is more favorable (disordered), pure solvent + solute, or solution?

Factors That Affect Solubility :

1. Structure of the molecule – Likes dissolve likes - -

polar substances dissolve polar substances;

non-polar substances dissolve non-polar substances.

Miscible – 2 liquids that mix in any ratioImmiscible – 2 liquids that don’t mix.

Perc, or perchloroethane, is used in the dry cleani ng industry, and is non-polar

Immiscible – 2 liquids that don’t mix

Vitamin A is essentially non-polar, thus fat soluble.

Vitamin C is quite polar, thus water soluble.

(Factors that affect Solubility)2. Temperature

A) Solid dissolved in a liquid –

Solubility – maximum amount of solute that can dissolve in a given amount of solvent.

Saturated – Rate of dissolve = rate of re-precipitation.

generally, but not always, solubility increases with increasing temperature.

B) Gas dissolved in a liquid-Generally, but not always, solubility decreases with increased temperature.

3. Pressure – Only has an effect on gases dissolved in liquids.

An increase in pressure of gas over a liquid causes an increase in gas solubility.

Henry’s Law: P = kC where k = const.P = PressureC = Concentration of gas dissolved in solution

SCUBA diving and the Bends

Problem: The solubility of N 2 in blood at 37 oC at a partial pressure of .80 atm is 5.6 x 10-4 mol/L. A SCUBA diver breathes compressed air at 13 2 feet deep with a partial pressure of 4.0 atm (N 2). The diver has 5.0 liters of blood. Calculate t he moles of N 2 gas that will dissolve in the divers blood while di ving at 132 feet deep.

Valid only for gases that dissolve, and not those t hat react.

Determine the volume of gas that will exsolve when t he diver surfaces. (1 Atm.)

The Colligative Properties (Collective Properties) – properties of solutions that don’t depend on the type of solute, just the c oncentration of the solute.

1. Boiling Point Elevation2. Freezing Point Depression3. Vapor Pressure Lowering4. Osmotic Pressure

Van’t Hoff factor: i - the relationship between mol es of solute dissolved and moles of particles in solution.

Ideal or expected i:

NaCl (s) -> Na+ + Cl- i = ?

Fe2(SO4)3 -> 2Fe+3 + 3 SO4-2 i = ?

C6H12O6 (s) -> C6H12O6 (aq) i = ?

In reality, some free ions will pair up, called ion pairs, causing actual i to be smaller than expected.

i (expected) i (observed).05 M NaCl 2 1.9.05 M FeCl3 4 3.4.05 M glucose 1 1

1. Vapor Pressure Lowering of solutions: (first co lligative property)

a) Non-Volatile solute (solute doesn’t evaporate)

The vapor pressure of a solution is lower than the pure liquid. Interference of non-volatile solute prevents a frac tion of the liquid from evaporating.

Predict what these 2 beakers will look like in a we ek or so.

Raoults Law: (For non-volatile solute)

Psolution = (X solvent ) x (Posolvent ) where P solution = vapor pressure of solution

X solvent = mole fraction of solvent

Posolvent = vapor pressure of pure solvent

Problem: Calculate the vapor pressure of a water s olution at 25 oC where 158 g of sucrose, C 12H22O11 is dissolved in 643.5 mL of water.

Problem: Do the same problem as above, using salt as the solute.

b) 2 liquids are mixed, and both are volatile (pro duce a vapor pressure)

The total vapor pressure can be calculated by:

Ptotal = PA + PB = XAPAo + XBPB

o

If a solution obeys Raults Law, it is called an ideal solution.

In reality, very few solutions are perfectly ideal.

3 Scenarios for 2 liquids mixed to make a solution:

1. 2 pure liquids are mixed and ∆Hsol’n = 0 this solution will behave ideally.Example: Benzene and Toluene

2. 2 pure liquids are mixed and ∆Hsol’n is exothermic (solution warms up). This indicates that the 2 liquids form strong bonds to each other.

Result: Calculated vapor pressure > Observed vapor pressureExample: Acetone and Water

3. 2 pure liquids are mixed and ∆Hsol’n is endothermic (solution cools). This indicates that the 2 liquids don’t bond well to eac h other.

Result: Calculated vapor pressure < Observed vap or pressureExample: Ethanol and Hexane

Problem: 5.80 grams acetone, C 3H6O is mixed with 11.9 grams of chloroform, CHCl3 at 35oC. The solutions vapor pressure is measured at 260 . torr. Is this solution ideal? (at 35 oC, Pacetone = 345 torr, P chloroform = 293 torr)

What can be said about the strength of intermolecul ar bonds in this solution?

2. Boiling Point Elevation –Boiling occurs when vapo r pressure = external pressure.

A solute lowers the vapor pressure of a solution, t herefore the solution

must be heated to a higher temperature in order for the solution to boil.

To calculate the solution boiling point:

∆TB = kB x m where k b = constantm = molality of solution

3. Freezing Point Depression – a solute lowers the fr eezing point of a pure solvent.

∆Tf = k f x m where k f = constant

Problem: Calculate the freezing point and boiling point of a solution composed of 651 grams of ethylene glycol, C 2H6O2 and 2505 grams of water.

Problem: What mass of ethylene glycol must be adde d to 10 kg water to make a solution that freezes at -10 oF?

Problem: Calculate the freezing and boiling point of a 2.0 m magnesium chloride solution.

Osmosis – Flow of a liquid through a semi-permeable membrane from an area of high to low concentration.

4. Osmotic Pressure – Solvent, but not solute can p ass through membrane. Pressure is caused by pure solvent passing through membrane faster than solvent in solution. (The pressure that stops osm osis.)

To calculate osmotic pressure:

π = MRT where π = osmotic pressure (atm)M = MolarityR = Gas Constant 0.0821 L x atm

K x molT = Kelvin Temp.

To calculate pressure from the height of a column o f solution:

torr = mm height of solution x density of so lutiondensity of Mercury

where density of Mercury is 13.6 g/mL

Problem: Convert the pressure of a 25.3 mm column of water to mm Hg:

Problem: 1.00 g of solute/liter causes a solution to rise 25.3 mm. Determine the molecular weight of this substance. Temperatur e of solution is 25 oC and the solution density is 1.00 g/mL

Osmotic Pressure is the most easily measured colli gative property. The previous problem would have had a freezing poin t depression of .00086 oC freezing point depression.

Problem: A sample of hormone weighing 0.546 grams is dissolved in 15.0 g benzene. The freezing point was determined to be 5.26oC. Calculate the molar mass of the hormone.

Problem: Calculate the osmotic pressure of a 1.2 M NaCl solution at 25 oC.

Back to electrolyte solutions:

Free ion – fully hydrated ion, completely surrounded by water molecules.

Ion Pair – at higher concentrations, some ions won’t be free ions, but rather, will form an ion pair. Ion pairs decrease conductivity and decrease the effect of the colligative property.

Higher charged ions are more likely to form ion pairs

Problem: The observed osmotic pressure of a 0.10 M Fe(NH4)2(SO4)2solution at 25 oC is 10.8 atm. Compare the experimental vs. expect ed van’tHoff factor, i.

Desalination of sea water:

Distill Freeze Reverse Osmosis

Colloidal Solution – Suspension of relatively large uncharged particles in a solution. These particles don’t settle, unlik e most uncharged particles.

Although the particle overall is neutral, the outsi de of the particle has a charge that neighboring molecules will be repelled by.

This charged particle suspends itself indefinitely in a solution.

Tyndall Effect – often used to detect suspended particles in a solution.

Milk, a colloid, and K2Cr2O7, a pure solution