solution:

a homogeneous mixture

solute: substance that gets dissolved

solvent: substance that does the dissolving

tincture: sol’n in which alcohol is solvent

aqueous: water is solvent

water is an excellent solvent because of its polarity

salt dissolving in water

hydrochloric acid dissolving in water

δ+ HCl δ-δ+ HCl δ-

“like dissolves like”

polar mixes w/ polar; nonpolar w/nonpolar

miscible: two liquids that mixex: water and alcohol

immiscible: liquids that don’t mix

ex: water and oil

types of compounds that dissolve in water

I. Ionic (metal w/nonmetal) ex: NaCl, KI

II. Acids (H+ w/ anion) ex: HCl, H2SO4

III. polar covalent ex: NH3, H2O2

cmpds w/ -OH groups ex: sugars, alcohols

ethanol: glucose: C6H12O6C2H5OH

compounds that don’t dissolve in water

nonpolar covalent cmpds

ex: hydrocarbons like C4H10

symmetrical molecules like CCl4

Conductivity of Solutions

Electrolyte:

cmpd that dissociates into ions; conducts electricity

must be ionic cmpd or an acid ex: CuSO4, HNO3

nonelectrolyte: doesn’t dissociate; won’tconduct

covalent cmpds that are not acids

ex: sugar, C12H22O11

alcohols, C2H5OH

Sample problems: determine formula,classify as soluble or insoluble, electrolyte or non

acetic acid HC2H3O2 sol. elec.

calcium chloride CaCl2 sol. elec.

Hexane, C6H14 insol. nonelec

silver nitrate

methanol

hydrochloric acid

butane

potassium iodide

AgNO3 sol, elec

CH3OH sol, nonelec

HCl sol, elec

KI sol, elec

C4H10 insol, nonelec

saturated, unsaturated, supersaturated

saturated sol’n:

contains the maximum amt of dissolved solute;equilibrium btwn dissolved and undissolved solute

unsaturated sol’n:contains less than max. amt of solute

supersaturated sol’n

contains more than the normal amt of solute

crystallizes completely by adding a “seed”crystal of the solute.

solubility: concentration of a saturatedsol’n

ex: solubility of NaCl at 25°C is 36g/100 mL

Effect of temperature on solubilitysolid solutes:solubility increases as temp increases

Gases and solubility

Effect of temp decreasing temp (water)increases solubility of gas

Effect of pressure Increasing pressure(over the water) increasessolubility

cold, high pressure

solubility curves

What is the solubilityof ammonium chlorideat 70°C ?

61 g/100 mLHow many gramsof potassium nitratewill dissolve in 100 mL water at 50°C?

83 g

How many gramsKNO3 will saturate250 mL water at 50°?

207.5 g

100 mL of saturated potassiumchlorate sol’n is cooled from80°C to 30°C? How many gramsof solute will crystallize?

at 80°: 42 g

at 30°: 12 g

30 gHow many grams staydissolved? 12 g

a sol’n contains 20 g sodium chloride dissolved in 100 mLwater at 25°C; is it saturated,unsaturated, or supersaturated?

unsaturated

Solution Concentration

dilute: contains little solute

concentrated: contains a lot of solute

ways to indicate concentration

percent: mass of solutetotal mass sol’n

100

ppm: parts per million ppb parts per billion

Molarity (M):moles solute

Liter of sol’n

Calculating Molarity M= MOL L

What is the molarity of a sol’n made by dissolving0.25 mole of sucrose in enough water to make 100.0 mL of sol’n ?

M= 0.25 mol 0.1 L

2.5 M

What is the molarity of a sol’n that contains 30.0 grams of sodium hydroxide dissolved in 0.50 liter of sol’n?

30.0 g

g

mol1

40

0.75 mol

0.50 L1.5 M

How many moles of HCl are contained in 5.0 litersof a 6.0 M HCl solution?

M= MOL L

6.0 M X mol

5.0 LX = 30 mol

How many grams of potassium sulfate are neededto dissolve in water to make 250 mL of a 2.0-molarsolution? K2SO4

2.0 MX mol

0.25 LX = 0.5 mol

0.5 mol

1 mol

174.3 g87 g

Calculate the molarity of a solution that has 12.5 g ofglucose, C6H12O6, dissolved in 500 mL of solution.

12.5 g

g

mol1

1800.06944 mol

0.5 L0.14 M

How many moles and grams of potassium iodide would be needed to prepare 0.75 liter of a 0.15 Msolution?

0.15 M = x mol 0.75 L

X = 0.1125 mol

0.1125 mol

mol1

g16718.8 g

Diluting a Solution M1V1 M2V2

concentrated diluted

If 100.0 mL of 6.0 M NaOH solution is diluted withwater to 800.0 mL, what is the molarity of the diluted solution?

(6.0 M) (100.0 mL) (800.0 mL)M2

800 mL800 mL

M2 = 0.75 M

(12.0 M) V1 (1.0 L)(3.0 M)

12.0 M12.0 M

What volume of 12.0 M HCl solution is needed to make 1.0 liter of 3.0 M solution of HCl ?

25.0 mL of concentrated acetic acid (17 molar) arepipetted into a flask. Water is added to the 1.0 literMark. Calculate the molarity of the final solution.

V1 = 0.25 L

(17 M) (25.0 mL) (1000 mL)M2

1000 mL1000 mL

M2 = 0.43 M

Colligative PropertiesProperties that depend on the concentrationof a solution

1. Lowering of vapor pressure

a solution has a lower V.P. than the puresolvent

pure water

water +a solute

the more concentrated the greaterthe effect

2. Boiling Point Elevation

solutions have a higherB.P. than the pure solvent

more conc., higher the B.P.

100°C

B> 100°C

3. Freezing Point Depression

solutions have a lowerF.P. than the pure solvent

0°C

< 0°C

Electrolytes vs. Nonelectrolytes

electrolytes dissociate into ions;

produces a higher concentration of dissolved solute particles

NaCl Na+ + Cl-

1 mol 2 moles ions

Nonelectrolytes don’t dissociate

C12H22O11 (s) C12H22O11 (aq)

1 mol 1 mol

Electrolytes have the greater

effect onC.P. ‘s

examples:Which of these 1-molar solutions has thelowest freezing point?

AlCl3

KI

C12H22O11

Na2CO3

Al3+ + Cl-3

K+ + I-

C12H22O11

Na+ + CO3

2-2

4 ions

2 ions

1 molecule

3 ions

Which of these 1-molal solutions has thehighest boiling point?

CuSO4 LiBr Mg(NO3)2 C6H12O6

2 2 3 1

Calculating Freezing & Boiling Point

ΔT = kf m ichange in F.P.

molality

freezing pt.constant

m = mol/kg solv.

ionization factor

ΔT = kb m i

kf = 1.86°C/mkb = 0.512°C/m

Calculate the freezing & boiling point ofa 1.5-molal solution of sodium hydroxide.

F.P.

ΔT = kf m i

ΔT = (1.86 °C/m) (1.5 m) (2)

NaOH Na+ + OH-

ΔT = 5.58 °CF.P. = -5.6 °C

B.P.

ΔT = (0.512 °C/m) (1.5 m) (2)ΔT = 1.536 °C

B.P. = 101.5 °C

Calculate the freezing and boiling point of a 2.0 maqueous solution of aluminum chloride.

AlCl3 Al+ + 3 Cl- ΔT = kf m i

ΔT = kbm i

ΔT = (1.86 °C/m) (2.0 m) (4)

ΔT = (0.512 °C/m) (2.0 m) (4)

ΔT = 14.88 °CF.P. = -15 °C

ΔT = 4.096 °C

B.P. = 104 °C