Unit 10:Solutions

Chemistry

Solution Definitions

solution: a homogeneous mixture

-- e.g., --

alloy: a solid solution of metals

-- e.g.,

solvent: the substance that dissolves the solute

evenly mixed at the particle level salt water

bronze = Cu + Snbrass = Cu + Zn

e.g., water e.g., salt

soluble: “will dissolve in”

miscible: refers to two liquids that mix evenly in all proportions

-- e.g., food coloring and water

Factors Affecting the Rate of Dissolution

1. temperature

2. particle size

3. mixing

4. nature of solvent or solute

With more mixing,rate

As temp. , rate

As size , rate

We can’t controlthis factor

Classes of Solutionsaqueous solution: solvent = water

amalgam:e.g.,

tincture:

e.g.,

organic solution:

solvent = alcohol

dental amalgam for cavities

tincture of iodine (for cuts)

solvent = mercury

water = “the universal solvent”

solventcontains ________

Organic solvents includebenzene, toluene, hexane, etc.

carbon

Non-Solution Definitions

insoluble: “will NOT dissolve in”e.g.,

immiscible: refers to two liquids thatwill NOT form a solution

e.g.,

suspension: appears uniform while being stirred, but settles over time

e.g.,

sand and water

water and oil

liquid medications, Italian dressing

Molecular Polaritynonpolar molecules:

-- e– are shared equally -- tend to be symmetric e.g.,

e.g.,

polar molecules:-- e– NOT shared equally

“Like dissolves like.”

fats and oils

H–C–H H

H–C–H H–C–H H–C–H

H

water H H

O

polar + polar = solution nonpolar + nonpolar = solution

polar + nonpolar = suspension (won’t mix evenly)

Anabolic steroids and HGH arefat-soluble, synthetic hormones.

Using Solubility Principles

Chemicals used by body obey solubility principles.

-- water-soluble vitamins: e.g.,

-- fat-soluble vitamins: e.g.,

vitamin C

vitamins A & D

Dry cleaning employs nonpolar liquids

Using Solubility Principles (cont.)

-- polar liquids damage wool, silk

-- also, dry clean for stubborn stains(ink, rust, grease)

-- tetrachloroethylene was in longtime use

C=CCl

ClCl

Cl

emulsifying agent (emulsifier):

molecules w/both a polar AND a nonpolar end --

-- allows polar and nonpolar substances to mix

e.g., soap lecithin eggs

MODEL OF A SOAP MOLECULE

NONPOLARHYDROCARBON

TAILPOLARHEAD

Na+

detergent

soap vs. detergent-- --made from animal

and vegetable fats made from petroleum

-- works better in hardwater

Hard water contains minerals w/ions like Ca2+, Mg2+,and Fe3+ that replace Na+ at polar end of soapmolecule. Soap is changed into an insolubleprecipitate (i.e., soap scum).

NONPOLARHYDROCARBON

TAILPOLAR HEAD

Na+

micelle: a liquid droplet covered w/soap or detergent molecules

oil

H2OH2O

H2O H2O

Solubility

Temp. (oC)

Solubility(g/100 g

H2O)

KNO3 (s)

KCl (s)

HCl (g)

SOLUBILITYCURVE

unsaturated: sol’n could dissolve more solute;

saturated: sol’n dissolves “just right” amt. of solute;

supersaturated: sol’n has “too much” solute dissolved in it;

how much solutedissolves in a givenamt. of solvent at agiven temp.

below the line

on the line

above the line

sudden stresscauses thismuch ppt

Solids dissolved Gases dissolved in liquids in liquids

To

Sol.

To

Sol.

As To ,solubility ___

As To ,solubility ___

[O2]

supersaturated

NaCl

0 10 20 30 40 50 60 70 80 90 100

Sol

ubili

ty (g

ram

s of

sol

ute/

100

g H

2O)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

KClO3SO2

gases

solids

150

Using an available solubility curve, classify as unsaturated, saturated,or supersaturated.

80 g NaNO3 @ 30oC

45 g KCl @ 60oC

50 g NH3 @ 10oC

70 g NH4Cl @ 70oC

per 1

00 g

H2O

unsaturated

saturated

unsaturated

NaCl

0 10 20 30 40 50 60 70 80 90 100

Solubility vs. Temperature for Solids

Sol

ubili

ty (g

ram

s of

sol

ute/

100

g H

2O)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

KClO3SO2

gases

solids

150Per 500 g H2O,120 g KNO3 @ 40oC

saturation point@ 40oC for 100 g H2O

= 70 g KNO3

So saturation pt.@ 40oC for 500 g H2O

= 5 x 70 g= 350 g

120 g < 350 g

unsaturated

(Unsaturated, saturated, or supersaturated?)

NaCl

0 10 20 30 40 50 60 70 80 90 100

Sol

ubili

ty (g

ram

s of

sol

ute/

100

g H

2O)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

KClO3SO2

gasessolids

150

Describe each situation below.(A) Per 100 g H2O, 100 g NaNO3 @ 50oC.

(B) Cool sol’n (A) very slowly to 10oC.

(C) Quench sol’n (A) in an ice bath to 10oC.

unsaturated;all solute dissolves;clear sol’n.

supersaturated;extra solute remainsin sol’n; still clear

saturated; extra solute (20 g)can’t remain in sol’n and becomes visible

Solubility Curve

NaCl

0 10 20 30 40 50 60 70 80 90 100

Sol

ubili

ty (g

ram

s of

sol

ute/

100

g H

2O)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

KClO3SO2

gases

solids

150

Glassware – Precision and Cost

beaker vs. volumetric flask 1000 mL + 5% 1000 mL + 0.30 mL

When filled to1000 mL line,

how much liquidis present?

5% of 1000 mL = 50 mL min:max:R

ange

:

WE DON’T KNOW.

min:max:R

ange

:

imprecise; cheap precise; expensive

950 mL 1050 mL

999.70 mL 1000.30 mL

** Measure to part of meniscus w/zero slope.

water ingrad. cyl.

mercury ingrad. cyl.

measure tobottom

measure totop

Concentration…a measure of solute-to-solvent ratio

concentrated dilute

Add water to dilute a sol’n;boil water off to concentrate it.

“lots of solute” “not much solute”

“watery”“not much solvent”

Selected units of concentration

A. mass % = mass of solute x 100mass of sol’n

parts per million (ppm) = mass of solute x 106

mass of sol’n B.

also, ppb and ppt

-- commonly used for minerals or contaminants in water supplies

molarity (M) = moles of soluteL of sol’n

C.

-- used most often in this class Lmol M

mol

L M

(Use 109 or 1012 here)

Na+

How many mol solute are req’d to make1.35 L of 2.50 M sol’n?

A. What mass sodium hydroxide is this?

B. What mass magnesium phosphate is this?

mol

L Mmol = M L = 2.50 M (1.35 L )= 3.38 mol

mol 1g 40.03.38 mol = 135 g NaOH

OH– NaOH

Mg2+

mol 1g 262.93.38 mol = 889 g Mg3(PO4)2

PO43– Mg3(PO4)2

Find molarity if 58.6 g barium hydroxide areDissolved in 5.65 L sol’n.

You have 10.8 g potassium nitrate. How many mLof sol’n will make this a 0.14 M sol’n?

mol

L M

Lmol M

g 171.3mol 158.6 g

Ba2+ OH–

Ba(OH)2

= 0.342 mol

L 5.65mol 0.342 = 0.061 M Ba(OH)2

K+ NO3–

KNO3

g 101.1mol 110.8 g = 0.1068 mol

Mmol L

M 0.14mol 0.1068 = 0.763 L

L 1mL 1000

(convert to mL)

= 763 mL

Molarity andStoichiometry

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

What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2?

Strategy: (1)

(2)

Find mol KI needed to yield 89 g PbI2.

Based on (1), find volume of 4.0 M KI sol’n.

22

PbI2 KI V of gasesat STP

V of sol’ns

L Mpart.

vol.

mass mass

vol.

part.

mol L M

mol

Strategy: (1)

(2)

Find mol KI neededto yield 89 g PbI2. Based on (1), find volumeof 4.0 M KI sol’n.

2

2

PbI g 461PbI mol 1

2PbI mol 1KI mol 2

89 g PbI2 = 0.39 mol KI

Mmol L

KI M 4.0KI mol 0.39

= 0.098 L of 4.0 M KI

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

What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2?

22

PbI2 KI

How many mL of a 0.500 MCuSO4 sol’n will react w/excessAl to produce 11.0 g Cu?

CuSO4(aq) + Al(s) Cu(s)

Cu g 63.5Cu mol 1

Cu mol 3CuSO mol 3 411.0 g Cu

= 0.173 mol CuSO4

Mmol L

4

4

CuSO M 0.500CuSO mol 0.173

= 346 mL of 0.500 M CuSO4

+ Al2(SO4)3(aq) 3 3 2

= 0.346 L

L 1mL 1000

Cu CuSO4Cu CuSO4

MC VC = MD VD

Dilutions of Solutions

Acids (and sometimes bases) are purchased inconcentrated form (“concentrate”) and are easilydiluted to any desiredconcentration.

**Safety Tip:

When diluting, add acid(or base) to water.

Dilution Equation: C = conc.D = dilute

14.8

Conc. H3PO4 is 14.8 M. What volume of concentrateis req’d to make 25.00 L of 0.500 M H3PO4?

MC VC = MD VD

(VC) = 0.500 (25)

How would you mix the above sol’n?

1. Measure out _____ L of conc. H3PO4. 2. In separate container, obtain ____ L of cold H2O.

3. In fume hood, slowly pour H3PO4 into cold H2O.

4. Add enough H2O until 25.00 L of sol’n is obtained.

0.845

14.8 14.8

VC = 0.845 L

~20

Cost Analysis with Dilutions

2.5 L of 12 M HCl(i.e., “concentrate”)

Cost: $25.71

0.500 L of0.15 M HCl

Cost: $6.35

How many 0.500 L-samples of 0.15 m HCl can bemade from the bottle of concentrate?

12 M

MC VC = MD VD

(2.5 L) = 0.150 MVD = 200 L

400 samples @ $6.35 ea. = $2,540

Moral:Buy the concentrateand mix it yourself to

any desired concentration.

(Expensive water!)

(VD)

You have 75 mL of conc. HF (28.9 M); you need15.0 L of 0.100 M HF. Do you have enough to dothe experiment?

28.9 M

MC VC = MD VD

(0.075 L) = 0.100 M (15 L)

Yes;we’re OK.

Calc. how much conc. you need…

28.9 M (VC) = 0.100 M (15 L)

VC = 0.052 L = 52 mL needed

Dissociation occurs when neutral combinations ofparticles separate into ions while in aqueous solution.

In general, _____ yield hydrogen (H+) ions in aque-ous solution; _____ yield hydroxide (OH–) ions.

acidsbases

sodium hydroxide NaOH hydrochloric acid HCl

sodium chloride NaCl Na+ + Cl–

Na+ + OH–

H+ + Cl–

sulfuric acid H2SO4 2 H+ + SO42–

acetic acid CH3COOH CH3COO– + H+

NOT in water:

in aq. sol’n:

“Strong” or “weak” is a property of the substance.We can’t change one into the other.

NOT in water:

in aq. sol’n:

1000 0 0

1 999 999

NaCl Na+ + Cl–

1000 0 0

980 20 20

CH3COOH CH3COO– + H+ Weak electrolytes exhibit little dissociation.

Strong electrolytes exhibit nearly 100% dissociation.

electrolytes: solutes that dissociate in sol’n

-- conduct elec. current because of free-moving ions -- e.g.,

-- are crucial for many cellular processes -- obtained in a healthy diet

--

acids,bases,most ionic compounds

For sustained exercise ora bout of the flu, sports drinks ensure adequateelectrolytes.

nonelectrolytes: solutes that DO NOT dissociate

--

-- e.g., any type of sugar

DO NOT conduct elec. current (not enough ions)

Colligative Properties

properties that depend on the conc. of a sol’n

Compared tosolvent’s…

…normal freezing point (NFP)

…normal boiling point (NBP)

a sol’n w/thatsolvent has a…

…lower FP (freezing point depression)

…higher BP (boiling point elevation)

water + more salt

water + a little salt

water

BPFP

0oC (NFP)

Applications of Colligative Properties (NOTE: Data is fictitious.)

1. salting roads in winter

100oC (NBP)

–11oC 103oC

–18oC 105oC

50% water + 50% AF

water + a little AF

water

BPFP

0oC (NFP)

Applications of Colligative Properties (cont.)

2. Antifreeze (AF) (a.k.a., “coolant”)

100oC (NBP)

–10oC 110oC

–35oC 130oC

Applications of Colligative Properties (cont.)

3. law enforcement

C

B

A

finishes melting

at…

starts melting

at…white powder penalty, if

convicted

120oC 150oC comm.service

130oC 140oC

134oC 136oC

2 yrs.

20 yrs.

Calculations withColligative Properties

The freezing point depression andboiling point elevation are given by: DTx = Kx m i

DTx = FP depression or BP elevation

Kx = Kf (molal FP depression constant) or Kb (molal BP elevation constant)

-- they depend on the solvent -- for water: Kf = 1.86oC/m, Kb = 0.52oC/m

Adding a nonvolatile solute to a solvent decreasesthe solution’s freezing point (FP) and increases itsboiling point (BP).

m = molality of solute solvent of kg

solute of mol m

i = van’t Hoff factor (accounts for # of particles in solution)

-- i = __ for nonelectrolytes

In aq. soln., assume that…

-- i = __ for KBr, NaCl, etc.

-- i = __ for CaCl2, etc.Jacobus Henricus van’t Hoff

(1852 – 1911)

In reality, the van’t Hoff factor isn’t alwaysan integer. Use the guidelines unlessgiven information to the contrary.

1

2

3

360 g BaCl2

Find the FP and BP of a soln. containing 360 g bariumchloride and 2.50 kg of water.

g 208.7

mol 1= 1.725 mol BaCl2

kg 2.50 0.690 m

DTx = Kx m iKf = 1.86oC/mKb = 0.52oC/m

DTf = Kf m i = 1.86(0.690)(3)

i = 3

= 3.85oC

FP = –3.85oC

DTb = Kb m i = 0.52(0.690)(3) = 1.08oC

BP = 101.08oC