water and the properties of solutions. homework the first 8 objectives are covered well in your...

Water and the Properties of Solutions

Homework• The first 8 objectives are covered well in your text. You are responsible

for making notes:

1. Distinguish between heterogeneous and homogeneous mixtures. (13-1).2. Distinguish between electrolytes and nonelectrolytes. (13-1)3. Compare the properties of suspensions, colloids, and solutions. (13-1)4. List and explain three factors that influence the rate of dissolving of a

solid in a liquid. (13-2)5. Explain terms related to solutions and solubility. (13-2)6. Explain how solutions form and use principles of polarity to determine

whether a compound will be miscible or immiscible in a particular solvent. (13-2)

7. List the three interactions that contribute to the heat of solution, and explain what causes dissolving to be exothermic or endothermic. (13-2)

8. Compare the effects of temperature and pressure on solubility. (13-2)

Water

• is an unusual substance, primarily because of polarity and hydrogen bonding.

• exists as a liquid at room temperature. It’s melting and boiling points are very high compared to substances of similar molecular weight

Substance Molar Mass Boiling Point

CH4 16.04 g/mol - 164 ºC

C3H8 44.11 g/mol - 42.1 ºC

C4H10 58.14 g/mol - 0.5 ºC

O2 32.00 g/mol - 183 ºC

CO2 44.01 g/mol - 78.5 ºC

N2 28.02 g/mol - 196 ºC

H2O 18.02 g/mol + 100 ºC

Water

• occurs in three phases normally on Earth (solid, liquid, vapour).

Water

• solid is less dense than the liquid.• ice floats.

Water

• has a high specific heat capacity.

• has a very high heat of fusion and vapourization.

• this allows water to store and move vast quantities of heat on the planet.

Water

• has a high surface tension.

• this leads to capillary action which allows vascular plants to growto great heights.

Water

• is the universal solvent

• everything dissolves in water to some degree.

• can generalize “Like dissolves Like”

Solubility

• Soluble in water– polar solutes

• sugars• alcohols

– ionic solutes• salts

– low molecular weight, non-polar molecules• O2 N2 CO2

• Not soluble in water– most hydrocarbons

• fats• oils• gasoline

Solutions

• Solutions are homogeneous mixtures of two or more pure substances.

• In a solution, the solute is dispersed uniformly throughout the solvent.

Solutions

The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.

How Does a Solution Form?

As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

How Does a Solution Form

If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal.

Types of Solutions

• SaturatedSolvent holds as much

solute as is possible at that temperature.

Dissolved solute is in dynamic equilibrium with solid solute particles.

Types of Solutions

• UnsaturatedLess than the

maximum amount of solute for that temperature is dissolved in the solvent.

Types of Solutions

• SupersaturatedSolvent holds more solute than is normally

possible at that temperature.These solutions are unstable; crystallization can

usually be stimulated by adding a “seed crystal” or scratching the side of the flask.

Factors Affecting Solubility

• Chemists use the axiom “like dissolves like”:Polar substances tend to

dissolve in polar solvents.Nonpolar substances tend

to dissolve in nonpolar solvents.

Factors Affecting Solubility

The more similar the intermolecular attractions, the more likely one substance is to be soluble in another.

Factors Affecting Solubility

Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.

Factors Affecting Solubility

• Vitamin A is soluble in nonpolar compounds (like fats).

• Vitamin C is soluble in water.

Gases in Solution

• In general, the solubility of gases in water increases with increasing mass.

• Larger molecules have stronger dispersion forces.

Gases in Solution

• The solubility of liquids and solids does not change appreciably with pressure.

• The solubility of a gas in a liquid is directly proportional to its pressure.

Henry’s Law

Sg = kPg

where

• Sg is the solubility of the gas;

• k is the Henry’s law constant for that gas in that solvent;

• Pg is the partial pressure of the gas above the liquid.

Temperature

Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

Temperature

• The opposite is true of gases:Carbonated soft drinks

are more “bubbly” if stored in the refrigerator.

Warm lakes have less O2 dissolved in them than cool lakes.

Concentration

• is a measure of the amount of solute, per unit of solution or solvent.

• is central to quantitative chemistry, since most chemical reactions happen in solution.

Molar Concentration

• is the most common measure of concentration used in the lab.

• is a measure of the number of moles of solute, per litre of solution (mol/L, mol∙L-1).

• is also called Molarity (M).

• for instance, 0.12 mol/L ≡ 0.12 M

Examples of Molarity

• Calculate the molar concentration when 0.256 mol of CuSO4 is dissolved in 1.1 L of solution.

• molar concentration = moles of solute volume of solution

• c = nv

• = 0.256 mol 1.1 L

= 0.23 mol/L (2 s.d.)

Examples of Molarity• Calculate the molar concentration when 28.2 g of KNO3 is dissolved in 750. mL of

solution.

• molar mass: KNO3

• 1 x K = 1 x 39.10 g/mol1 x N = 1 x 14.01 g/mol3 x O = 3 x 16.00 g/mol

101.11 g/mol

• moles of substance = 28.2 g = 0.279 mol 101.11 g/mol

• volume must be reported in L:• 750. mL x 1 L = 0.750 L

1000 mL• molar concentration:• c = n = 0.279 mol

V 0.750 L

• = 0.372 mol/L

Examples of Molarity

• Calculate the volume of solution made with 0.455 mol of sodium phosphate in a 2.00 M solution.

• c = nV

• V = nc

• = 0.455 mol2.00 mol/L

• = 0.228 L

Molarity Examples

• How many moles are present in 4.50 L of a 0.025 mol/L solution of magnesium nitrate?

• c = n• V• n = cV• = (0.025 mol/L)(4.50

L)• = 0.11 mol

Molarity Problems

• text, pages 941 & 942

• questions 332, 333, 335, 336, 346

Molal Concentration

• is used in industry and for measurements of colligative properties (see Ch. 14); it is useful because it does not depend on the final volume of solution.

• is a measure of the number of moles of solute, per kilogram of solvent (mol/kg, mol∙kg-1).

• is also called Molality (M).• for instance, 0.12 mol/kg ≡ 0.12 M

Molal Examples• Calculate the molal concentration when 2.56 mol of

CuSO4 is dissolved in 12.0 kg of water.

• molal concentration = moles of solute• mass of solvent

• M = n• ms

• = 2.56 mol• 12.0 kg

• = 0.213 mol/kg

Molal Examples• Calculate the molality when 19 g of Mg(NO3)2 is dissolved in 1200 g of

water.

• Mg(NO3)2

• 1 x Mg = 1 x 24.30 g/mol• 2 x N = 2 x 14.01 g/mol• 6 x O = 6 x 16.00 g/mol• 148.32 g/mol

• 1200 g x 1 kg = 1.2 kg• 1000 g

• M = n n = m• ms M

• M = m = 19 g• msM (1.2 kg)(148.32 g/mol)

• M = 0.11 mol/kg

Molality Examples• Calculate the mass of solvent used with 61.2 g of sodium phosphate in a 2.00 M solution.

• sodium phosphate = Na3PO4

• 3 x Na = 3 x 22.99 g/mol• 1 x P = 1 x 30.97 g/mol• 4 x O = 4 x 16.00 g/mol• 163.94 g/mol

• M = n• ms

• ms = n n = m• M M

• ms = m• M M• ms = 61.2 g• (2.00 mol/kg)(163.94 g/mol)

• ms = 0.187 kg

Molality Examples• What mass of lithium hydroxide is present in a 0.025 mol/kg solution

with 360. kg of water?

• lithium hydroxide = LiOH• 1 x Li = 1 x 6.94 g/mol• 1 x O = 1 x 16.00 g/mol• 1 x H = 1 x 1.01 g/mol• 23.95 g/mol• M = n n = m• ms M

• M = m• msM• m = M msM

• = (0.025 mol/kg)(360. kg)(23.95 g/mol)•• = 220 g

Molality Problems

• text, pages 941 & 942

• questions 338, 339, 340, 350, 352

Other Measures of Concentration

• Parts per million ("ppm") denotes one particle of a given substance for every 999,999 other particles. This is roughly equivalent to one drop of ink in a 150 litre (40 gallon) drum of water, or 1 g of solute in 1000 kg of solution.

• Parts per billion ("ppb") denotes one particle of a given substance for every 999,999,999 other particles. This is roughly equivalent to one drop of ink in a lane of a public swimming pool, or 1 gram of solute in 106 kg of solution (roughly Riversdale pool).

• Parts per trillion ("ppt") denotes one particle of a given substance for every 999,999,999,999 other particles. This is roughly equivalent to one drop of ink in a shipping canal lock full of water , or 1 g of solute in Blackstrap lake.

Dilution

• many substances, especially acids, are received in a concentrated form (hydrochloric acid is 12.4 mol/L). This is called the stock solution.

• to use the chemicals in the lab they are usually diluted to a concentration much less than they are received.

• the problem is to decide how much of the stock solution you need to make the solution you want.

Dilution

• For instance; you have a 12.4 mol/L stock solution of HCl and you want to make 2.00 L of 0.100 mol/L solution.

• You can calculate the number of moles in the solution:

• n = c v• = (0.100 mol/L)(2.00 L)• = 0.200 mol

Dilution• 0.200 mol represents the number of moles of HCl you

need from the stock solution. • Now you have the number of moles of stock solution and

the concentration of the stock solution, so you can calculate the volume of concentrated HCl you need to add to 2.00 L of water to make the dilute solution:

• v = n = 0.200 mol• c 12.4 mol/L

• = 0.0161 L (1000 mL/L)

• = 16.1 mL of concentrated HCl is added to water to make 2.00 L of a

0.100 mol/L solution.

Dilution

• Fortunately, there is an easier way to do this. Since the moles taken from the stock solution (ns) is the same number of moles that goes into the dilute solution (nd)

• ns = nd

• and• ns = csvs

• nd = cdvd

• so the formula for dilution is:

• csvs = cdvd

Dilution

• What volume of concentrated sulfuric acid (18 M) is needed to make 500. mL of 2.00 M dilute solution?

• CsVs = CdVd

• (18 mol/L)Vd = (2.00 mol/L)(500. mL)

• Vd = 56 mL concentrated H2SO4 is needed.

Dilution

• What is the concentration of dilute solution if 25.0 mL of glacial acetic acid (24.0 mol/L) is added to make 2.00 L of dilute solution?

• CsVs = CdVd

• (24.0 mol/L)(25.0 mL) = (Cd)(2.00 L)(1000 mL/1 L)

• Cd = 0.300 mol/L acetic acid.

Dilution

• What volume of a 1.00 mol/L stock solution of aluminum chloride is needed to make 500. mL of 3.0 x 10-4 mol/L solution?

• CsVs = CdVd

• (1.00 mol/L)Vd = (3.0 x 10-4 mol/L)(500. mL)

• Vd = 0.15 mL AlCl3 stock solution is needed.

Serial Dilution• Serial dilutions are an accurate method of making solutions of low

molar concentrations. • Since measuring small volumes of solution is prone to error, a

series of dilutions are performed in order to gradually reduce the concentration of the solution from that of the stock solution.

• Serial "ten-fold" dilutions are commonly used.

• To carry out a serial "ten-fold" dilution you would do the following:

• Add 9 ml of distilled water to each test tube. • To the first test tube, add 1 ml of the stock solution. • Mix or vortex. • Now add 1 ml of this solution to the second test tube and mix. • Repeat until you have reached your desired concentration.

Ions in Aqueous Solution• when a substance dissolves it is called a solvation reaction:

• sugar is a non-electrolyte:

C12H22O11 (s) → C12H22O11 (aq)

• soluble ionic compounds undergo dissociation in solution; they break up into their constituent ions:

NaCl (s) → Na1+(aq) + Cl1-

(aq)

(this represents both solvation and dissociation)• Where

– (s) means solid– (aq) means aqueous (dissolved in water)– (l) means liquid– (g) means gas– (ppt) means precipitate

Ions in Aqueous Solution

• to write a dissociation equation for any ionic compound you must find out the identity of the cation and anion, then write the equation, paying attention to the stoichiometry:

Al2(SO4)3 (s) contains the Al3+ and SO42- ions.

• When it dissociates you get 2 Al3+ and 3 SO42- ions:

Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO4

2-(aq)

Ions in Aqueous Solution

• this means that the concentration of the ions may be different than the calculated concentration of the substance in solution:

• If the aluminum sulfate is a 0.200 mol/L solution, then

Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO4

2-(aq)

• 0.200 mol/L 2(0.200 mol/L) 3(0.200 mol/L)• = 0.400 mol/L = 0.600 mol/L

Ions in Aqueous Solution

• 0.35 mol/L NaOH

• NaOH(s) → Na1+(aq) + OH1-

(aq)

• 0.35 M 1(0.35 M) 1(0.35 M)

• = 0.35 M = 0.35 M

Ions in Aqueous Solution

• 1.12 mol/L (NH4)2CO3

• (NH4)2CO3 (s) → 2 NH41+

(aq) + CO32-

(aq)

• 1.12 M 2(1.12 M) 1(1.12 M)

• = 2.24 M =1.12 M

Ions in Aqueous Solution

• 0.056 mol/L V3(PO4)5

• V3(PO4)5 (s) → 3 V5+(aq) + 5 PO4

3-(aq)

• 0.056 M 3(0.056 M) 5(0.056 M)

• = 0.17 M = 0.28 M

Chemical Reactions in Solution

• there are 4 indications that a chemical reaction has occurred:

1. gas production

2. energy change

3. colour change

4. precipitate formed

Chemical Reactions in Solution

• a precipitate is a solid formed from the reaction of two soluble ions in solution. For example:

• Pb(NO3)2 (aq) + 2 KI(aq) → PbI2 (s) + 2 KNO3 (aq)

• this is the chemical reaction equation for the combination of lead (II) nitrate and potassium iodide.

Chemical Reactions in Solution

• If we recognize the fact that both ionic reactants have dissociated and are in fact ions in solution we would get the overall ionic equation:

• Pb2+

(aq) + 2 NO31-

(aq) + 2 K1+(aq) + 2 I1-

(aq)

→ PbI2(s) + 2 K1+(aq) + 2 NO3

1-(aq)

• the lead (II) iodide does not break up into ions because it is a solid at the bottom of the beaker.

Chemical Reactions in Solution

• participate in the chemical reaction and are called spectator ions.

• If we eliminate the spectators we get the net ionic equation:

• Pb2+(aq) + 2 I1-

(aq) → PbI2 (s)

• if there is no precipitate formed, no chemical reaction occurs.

Chemical Reactions in Solution

• FeCl2 (aq) + K2S (aq) → FeS (s) + KCl (aq)

• Balance chemical reaction equation:• FeCl2 (aq) + K2S (aq) → FeS (s) + 2 KCl (aq)

• Overall ionic equation:• Fe2+

(aq) + 2 Cl1-(aq) + 2 K1+

(aq) + S2-(aq)

FeS(s) + 2 K1+(aq) + 2 Cl1-

(aq)

• Net Ionic Equation:• Fe2+

(aq) + S1-(aq) FeS(s)

• AlBr3 (aq) + NaOH (aq) → NaBr (aq) + Al(OH)3 (s)

• Balance chemical reaction equation:• AlBr3 (aq) + 3 NaOH (aq) → 3 NaBr (aq) + Al(OH)3 (s)

• Overall ionic equation:• Al3+

(aq) + 3 Br1-(aq) + 3 Na1+

(aq) + 3 OH1-(aq)

Al(OH)3 (s) + 3 Na1+(aq) + 3 OH1-

(aq)

• Net Ionic Equation:• Al3+

(aq) + 3 OH1-(aq) Al(OH)3 (s)

• (NH4)3PO4 (aq) + Ca(NO3)2 (aq) → NH4NO3 (aq) + Ca3(PO4)2 (s)

• Balance chemical reaction equation:• 2 (NH4)3PO4 (aq) + 3 Ca(NO3)2 (aq)

→ 6 NH4NO3 (aq) + Ca3(PO4)2 (s)

• Overall ionic equation:• 6 NH4

1+(aq) + 2 PO4

3-(aq) + 3 Ca2+

(aq) + 6 NO31-

(aq)

Ca3(PO4)2 (s)+ 6 NH41+

(aq) + 6 NO31-

(aq)

• Net Ionic Equation:• 3 Ca2+

(aq) + 2 PO43-

(aq) Ca3(PO4)2 (s)

• Aqueous solutions of silver nitrate and sodium carbonate react to produce a precipitate of silver carbonate and aqueous sodium nitrate.

• Write and balance chemical reaction equation:• 2 AgNO3 (aq) + Na2CO3 (aq) → Ag2CO3 (s) + 2 NaNO3 (aq)

• Overall ionic equation:• 2 Ag1+

(aq) + 2 NO31-

(aq) + 2 Na1+(aq) + CO3

2-(aq)

Ag2CO3 (s) + 2 Na1+(aq) + 2 NO3

1-(aq)

• Net Ionic Equation:• 2 Ag1+

(aq) + CO32-

(aq) Ag2CO3 (s)

• Aqueous solutions of barium chloride and potassium sulphate react to produce a precipitate of barium sulphate and aqueous potassium chloride

• Write and balance chemical reaction equation:• BaCl2 (aq) + K2SO4 (aq) → BaSO4 (s) + 2 KCl (aq)

• Overall ionic equation:• Ba2+

(aq) + 2 Cl1-(aq) + 2 K1+

(aq) + SO42-

(aq) BaSO4 (s) + 2 K1+

(aq) + 2 Cl1-(aq)

• Net Ionic Equation:• Ba2+

(aq) + SO42-

(aq) BaSO4 (s)

Mixing Solutions

• What happens if you mix two ionic solutions ? Using what we just learned you can calculate the concentration of the ions in solution before mixing. Take these two solutions:

• CaCl2 (aq)  →  Ca2+(aq)   +   2 Cl1-

(aq) • 0.25 mol/L       1(0.25 mol/L)     2(0.25 mol/L)        • = 0.25 mol/L           = 0.50 mol/L• • Mg3(PO4)2 (aq)   →  3 Mg2+

(aq)   +   2 PO43-

(aq) • 0.011 mol/L             3(0.011 mol/L)       2(0.011

mol/L) •                                    = 0.033 mol/L        = 0.022

mol/L

Mixing Solutions

• take 300. mL of the CaCl2 and 200. mL of the Mg3(PO4)2 and we mix them.

• Assuming that no precipitate is formed, what will the concentration of each ion in solution be after mixing ?

Mixing Solutions

• we use the dilution formula:

CsVs = CdVd

• In this case, Cs and Vs is the original concentration and volume of each ion (0.25 mol/L and 300. mL for the Ca2+, for instance) and d2 is the total volume of the mixture (300. mL + 200. mL). We are looking for Cd, the ion concentration after mixing:

Mixing Solutions

• Cd   =     CsVs                    Vd

• For the Ca2+ ion the calculation is as follows:

• [Ca2+]   =   (0.25mol/L)(300.mL)                        (300. mL + 200. mL)

• =   0.15 mol/L

Mixing Solutions

• [Cl1-]   =   (0.50 mol/L)(300. mL)                        (300. mL + 200. mL)

• =   0.30 mol/L

• [Mg2+]   =   (0.033 mol/L)(200. mL)                           (300. mL + 200. mL)

• =   0.013 mol/L

• [PO43-]   =   (0.022 mol/L)(200. mL)    

                        (300. mL + 200. mL) • =   0.0088 mol/L

Problems

1. 2.0 L of 0.40 mol/L MgSO4 mixed with 2.0 L of 0.080 mol/L KI

2. 200. mL of 0.600 mol/L AlBr3 mixed with 300. mL of 0.400 BaBr2

3. 20.0 mL of 0.50 mol/L FeCl3 mixed with 80.0 mL of 0.15 mol/L NH4Cl

Predicting Solubility

• substances dissolve in water when the attraction of the solute particles to the water molecules exceeds the attraction of the undissolved solute for itself.

• if the solute particles are held together very strongly, the attraction of the water molecules may not be enough to cause it to dissolve. Such solutes will be insoluble, or have a very low solubility.

• we can’t tell from looking at a molecule whether it will be soluble or insoluble in water. We can use empirical evidence to make predictions.

Negative Ion Positive Ion ResultAny anion Group 1 ions Soluble

Any anion Ammonium ion (NH41+) Soluble

Nitrate (NO31-) Any cation Soluble

Acetate (CH3COO1-) Ag1+ .any other cation

Not soluble .Soluble

Chloride (Cl1-), Bromide (Br1-), Iodide (I1-)

Ag1+, Pb2+, Cu1+, Hg22+ .

any other cation

Not soluble .Soluble

Sulfate (SO42-) Ca2+, Sr2+, Ba2+, Ra2+, Ag1+, Pb2+

any other cationNot soluble .Soluble

Sulfide (S2-) Group 1 & 2 ions, NH41+ .

any other cation

Soluble .Not soluble

Hydroxide (OH1-), Phosphate (PO4

3-), Carbonate (CO3

2-)

Sulfite (SO32-)

Ammonium ion and group 1 ions .

any other cation

Soluble .

Not soluble

Predicting Solubility• Na2S

– soluble, because Na is an alkali metal (group 1)

• NH4Cl– soluble, because the NH4

1+ ion is soluble with any anion

• CuNO3

– soluble, because NO31- is soluble with any cation

• AgCH3COO– insoluble

• Mg(CH3COO)2

– soluble

• PbCl4– soluble; in this compound the Pb is 4+

• PbCl2– insoluble; in this compound the Pb is 2+

Predicting Solubility

• Predict the solubility of the compounds in question 2 on page 447 of the text.

Predicting the Results of Chemical Reactions

• given any two ionic solutions you should be able to predict the products and if a precipitation reaction occurs:

• AgNO3 (aq) + NaCl(aq) → ???

• first, rewrite the reactants as separate ions in solution:

• Ag1+(aq) + NO3

1-(aq) + Na1+

(aq) + Cl1-(aq) →

• double replacement reaction; cation of the first compound combines with the anion of the second, and vice versa:

• Ag1+(aq) + Cl1-

(aq) → AgCl• Na1+

(aq) + NO31-

(aq) → NaNO3

Predicting the Results of Chemical Reactions

• use the solubility table to predict the solubility of the two products of this reaction:

• AgCl insoluble• NaNO3 soluble

• write the chemical equation for this reaction:

• AgNO3 (aq) + NaCl(aq) → AgCl(s) + NaNO3 (aq)

• write the overall ionic equation:

• Ag1+(aq) + NO3

1-(aq) + Na1+

(aq) + Cl1-(aq) → AgCl(s) + Na1+

(aq) + NO31-

(aq)

• write the net ionic equation:•• Ag1+

(aq) + Cl1-(aq) → AgCl(s)

Predicting the Results of Chemical Reactions

• determine the identity of the possible products.• find out of the products are soluble or insoluble.• if no insoluble compound is formed, simply write No Reaction in the product

place of the chemical equation:

NaCl(aq) + NH4NO3 (aq) → No Reaction

• if one (or both) of the possible products is insoluble, write a chemical equation, an overall ionic equation and a net ionic equation.

• NaBr and AgNO3

• NiCl2 and KOH• CuSO4 and K2S• CoCl2 and (NH4)3PO4

• Cu(NO3)2 and KI

NaBr and AgNO3

• Na1+ Br1- Ag1+ NO31-

• NaNO3 soluble• AgBr not soluble

• NaBr(aq) + AgNO3 (aq) → NaNO3 (aq)+ AgBr(s) • Na1+

(aq) + Br1-(aq) + Ag1+

(aq) + NO31-

(aq) →

AgBr(s) + Na1+(aq) + NO3

1-(aq)

• Ag1+(aq) + Br1-

(aq) → AgBr(s)

NiCl2 and KOH

• Ni2+ Cl1- K1+ OH1-

• Ni(OH)2 not soluble

• KCl soluble

• NiCl2 (aq) + 2 KOH(aq) → Ni(OH)2 (s) + 2 KCl (aq)

• Ni2+ (aq) + 2 Cl1-

(aq) + 2 K1+ (aq) + 2 OH1-

(aq) →

Ni(OH)2 (s) + 2 K1+ (aq) + 2 Cl1-

(aq)

• Ni2+ (aq) + 2 OH1-

(aq) → Ni(OH)2 (s)

CuSO4 and K2S

• Cu2+ SO42- K1+ S2-

• CuS not soluble• K2SO4 soluble

• CuSO4(aq) + K2S (aq) → CuS(s) + K2SO4(aq) • Cu2+

(aq) + SO42-

(aq) + 2 K1+ (aq) + S2-

(aq) →

CuS(s) + 2 K1+ (aq) + SO4

2-(aq)

• Cu2+ (aq) + S2-

(aq) → CuS(s)

CoCl2 and (NH4)3PO4

• Co2+ Cl1- NH41+ PO4

3-

• Co3(PO4)2 not soluble

• NH4Cl soluble

• 3 CoCl2(aq) + 2 (NH4)3PO4(aq) →

Co3(PO4)2 (s) + 6 NH4Cl(aq)

• 3 Co2+ (aq) + 6 Cl1-

(aq) + 6 NH41+

(aq) + 2 PO43-

(aq) →

Co3(PO4)2(s) + 6 NH41+

(aq) + 6 Cl1-(aq)

• 3 Co2+ (aq) + 2 PO4

3-(aq) → Co3(PO4)2(s)

Cu(NO3)2 and KI

• Cu2+ NO31- K1+ I1-

• KNO3 soluble

• CuI2 soluble

• Cu(NO3)2 (aq) + KI(aq) → No Reaction

Colligative Properties

• Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles.

• Among colligative properties areVapor pressure lowering Boiling point elevationMelting point depressionOsmotic pressure

Vapor Pressure

Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

Vapor Pressure

Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

Boiling Point Elevation and Freezing Point Depression

Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

Boiling Point ElevationThe change in boiling point is proportional to the molality of the solution:

Tb = Kb m

where Kb is the molal boiling point elevation constant, a property of the solvent.Tb is added to the normal

boiling point of the solvent.

Freezing Point Depression• The change in freezing

point can be found similarly:

Tf = Kf m

• Here Kf is the molal freezing point depression constant of the solvent.

Tf is subtracted from the normal freezing point of the solvent.

Boiling Point Elevation and Freezing Point Depression

Note that in both equations, T does not depend on what the solute is, but only on how many particles are dissolved.

Tb = Kb m

Tf = Kf m

Colligative Properties of Electrolytes

Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

Colligative Properties of Electrolytes

• For Instance:

Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO4

2-(aq)

0.200 mol/L 2(0.200 mol/L) 3(0.200 mol/L)

= 0.400 mol/L = 0.600 mol/L• Total ion concentration

= 0.400 mol/L + 0.600 mol/L

= 1.000 mol/L

van’t Hoff Factor

One mole of NaCl in water does not really give rise to two moles of ions.

van’t Hoff Factor

Some Na+ and Cl− reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

Osmosis

• Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles.

• In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.

Osmosis

In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the are of lower solvent concentration (higher solute concentration).

Osmotic Pressure

• The pressure required to stop osmosis, known as osmotic pressure, , is

nV

= ( )RT = MRT

where M is the molarity of the solution

If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic.

Osmosis in Blood Cells

• If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic.

• Water will flow out of the cell, and crenation results.

Osmosis in Cells

• If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic.

• Water will flow into the cell, and hemolysis results.

Molar Mass from Colligative Properties

• We can use the effects of a colligative property such as freezing point depression to determine the molar mass of a compound.

• Since Tf = Kf M• Then M = Tf ÷ Kf

• If we know the mass of solvent we can calculate the number of moles.

• If we know the mass of the solute we can calculate the molar mass.

Colloids:

Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity.

Tyndall Effect

• Colloidal suspensions can scatter rays of light.

• This phenomenon is known as the Tyndall effect.

Colloids in Biological Systems

Some molecules have a polar, hydrophilic (water-loving) end and a nonpolar, hydrophobic (water-hating) end.

Colloids in Biological Systems

Sodium stearate is one example of such a molecule.

Colloids in Biological Systems

These molecules can aid in the emulsification of fats and oils in aqueous solutions.