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Chemistry. Unit 10: Solutions. e.g., water. e.g., salt. Solution Definitions. solution : a homogeneous mixture. --. evenly mixed at the particle level. -- e.g.,. salt water. alloy : a solid solution of metals. -- e.g.,. bronze = Cu + Sn brass = Cu + Zn.

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Unit 10: Solutions

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Unit 10 solutions

Chemistry

Unit 10:

Solutions


Unit 10 solutions

e.g., water

e.g., salt

Solution Definitions

solution: a homogeneous mixture

--

evenly mixed at the particle level

-- e.g.,

salt water

alloy: a solid solution of metals

-- e.g.,

bronze = Cu + Sn

brass = Cu + Zn

solvent: the substance that dissolves the solute


Unit 10 solutions

soluble: “will dissolve in”

miscible: refers to two liquids that mix

evenly in all proportions

-- e.g.,

food coloring and water


Unit 10 solutions

As temp. , rate

As size ,

rate

Factors Affecting the Rate of Dissolution

1. temperature

2. particle size

3. mixing

With more mixing,

rate

We can’t control

this factor

4. nature of solvent or solute


Unit 10 solutions

Classes of Solutions

aqueous solution:

solvent = water

water = “the universal solvent”

amalgam:

solvent = mercury

e.g.,

dental amalgam for cavities

tincture:

solvent = alcohol

e.g.,

tincture of iodine (for cuts)

organic solution:

solvent

contains ________

carbon

Organic solvents include

benzene, toluene, hexane, etc.


Unit 10 solutions

Non-Solution Definitions

insoluble: “will NOT dissolve in”

e.g.,

sand and water

immiscible: refers to two liquids that

will NOT form a solution

e.g.,

water and oil

suspension: appears uniform while

being stirred, but

settles over time

e.g.,

liquid medications,

Italian dressing


Unit 10 solutions

H

H–C–H

H–C–H

H–C–H

H–C–H

H

H

H

O

Molecular Polarity

nonpolar molecules:

-- e– are shared equally

-- tend to be symmetric

fats and oils

e.g.,

polar molecules:

-- e– NOT shared equally

water

e.g.,

“Like dissolves like.”

polar + polar = solution

nonpolar + nonpolar = solution

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


Unit 10 solutions

Using Solubility Principles

Chemicals used by body obey solubility principles.

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

vitamin C

vitamins A & D

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

Anabolic steroids and HGH are

fat-soluble, synthetic hormones.


Unit 10 solutions

Cl

Cl

C=C

Cl

Cl

Using Solubility Principles (cont.)

Dry cleaning employs nonpolar liquids

-- polar liquids damage wool, silk

-- also, dry clean for stubborn stains

(ink, rust, grease)

-- tetrachloroethylene was in longtime use


Unit 10 solutions

lecithin

e.g., soap

eggs

detergent

MODEL OF A SOAP MOLECULE

Na+

NONPOLAR

HYDROCARBON

TAIL

POLAR

HEAD

emulsifying agent (emulsifier):

molecules w/both a polar AND a nonpolar end

--

--

allows polar and nonpolar substances to mix


Unit 10 solutions

Na+

NONPOLAR

HYDROCARBON

TAIL

POLAR HEAD

H2O

H2O

oil

H2O

H2O

soap vs. detergent

-- --

made from petroleum

made from animal

and vegetable fats

--

works better in hard

water

Hard water contains minerals w/ions like Ca2+, Mg2+,

and Fe3+ that replace Na+ at polar end of soap

molecule. Soap is changed into an insoluble

precipitate (i.e., soap scum).

micelle: a liquid droplet covered

w/soap or detergent molecules


Unit 10 solutions

SOLUBILITY

CURVE

sudden stress

causes this

much ppt

KNO3 (s)

KCl (s)

Solubility

(g/100 g H2O)

HCl (g)

Temp. (oC)

Solubility

how much solute

dissolves in a given

amt. of solvent at a

given temp.

unsaturated: sol’n could dissolve more

solute;

below the line

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

of solute;

on the line

supersaturated: sol’n has “too much” solute

dissolved in it;

above the line


Unit 10 solutions

Sol.

Sol.

As To ,

solubility ___

As To ,

solubility ___

To

To

Solids dissolved Gases dissolved

in liquids in liquids

[O2]


Unit 10 solutions

gases

solids

per 100 g H2O

150

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

KI

140

130

120

NaNO3

110

100

80 g NaNO3 @ 30oC

KNO3

90

unsaturated

80

HCl

NH4Cl

70

45 g KCl @ 60oC

Solubility (grams of solute/100 g H2O)

KCl

NH3

60

saturated

50

50 g NH3 @ 10oC

40

30

NaCl

unsaturated

20

70 g NH4Cl @ 70oC

KClO3

10

SO2

supersaturated

0 10 20 30 40 50 60 70 80 90 100


Unit 10 solutions

Solubility vs. Temperature for Solids

gases

solids

(Unsaturated, saturated, or supersaturated?)

150

KI

140

Per 500 g H2O,

120 g KNO3 @ 40oC

130

120

NaNO3

110

saturation point

@ 40oC for 100 g H2O

= 70 g KNO3

100

KNO3

90

80

HCl

NH4Cl

70

Solubility (grams of solute/100 g H2O)

So saturation pt.

@ 40oC for 500 g H2O

= 5 x 70 g

KCl

NH3

60

50

40

30

NaCl

= 350 g

20

120 g < 350 g

KClO3

10

SO2

unsaturated

0 10 20 30 40 50 60 70 80 90 100


Unit 10 solutions

gases

solids

Describe each situation below.

150

KI

140

(A) Per 100 g H2O,

100 g NaNO3 @ 50oC.

130

120

unsaturated;

all solute dissolves;

clear sol’n.

NaNO3

110

100

KNO3

90

(B) Cool sol’n (A) very

slowly to 10oC.

80

HCl

NH4Cl

70

Solubility (grams of solute/100 g H2O)

supersaturated;

extra solute remains

in sol’n; still clear

KCl

NH3

60

50

40

30

NaCl

(C) Quench sol’n (A) in

an ice bath to 10oC.

20

KClO3

10

SO2

saturated; extra solute (20 g)

can’t remain in sol’n and becomes visible

0 10 20 30 40 50 60 70 80 90 100


Solubility curve

gases

solids

Solubility Curve

150

KI

140

130

120

NaNO3

110

100

KNO3

90

80

HCl

NH4Cl

70

Solubility (grams of solute/100 g H2O)

KCl

NH3

60

50

40

30

NaCl

20

KClO3

10

SO2

0 10 20 30 40 50 60 70 80 90 100


Unit 10 solutions

beaker vs.volumetric flask

1000 mL + 5% 1000 mL + 0.30 mL

min:

min:

Range:

Range:

max:

max:

Glassware – Precision and Cost

When filled to

1000 mL line,

how much liquid

is present?

WE DON’T KNOW.

5% of 1000 mL

= 50 mL

999.70 mL

950 mL

1050 mL

1000.30 mL

precise; expensive

imprecise; cheap


Unit 10 solutions

water in

grad. cyl.

mercury in

grad. cyl.

measure to

bottom

measure to

top

** Measure to part of meniscusw/zero slope.


Unit 10 solutions

Concentration…a measure of solute-to-solvent ratio

concentrated

dilute

“lots of solute”

“not much solute”

“not much solvent”

“watery”

Add water to dilute a sol’n;

boil water off to concentrate it.


Unit 10 solutions

Selected units of concentration

A.

mass % = mass of solute x 100

mass of sol’n

B.

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

mass of sol’n

also, ppb and ppt

(Use 109 or 1012 here)

mol

-- commonly used for minerals or

contaminants in water supplies

M

L

molarity (M) = moles of solute

L of sol’n

C.

-- used most often in this class


Unit 10 solutions

mol

M

L

How many mol solute are req’d to make

1.35 L of 2.50 M sol’n?

mol = M L

= 2.50 M (1.35 L )

= 3.38 mol

A. What mass sodium hydroxide is this?

Na+

OH–

NaOH

3.38 mol

= 135 g NaOH

B. What mass magnesium phosphate is this?

Mg2+

PO43–

Mg3(PO4)2

3.38 mol

= 889 g Mg3(PO4)2


Unit 10 solutions

mol

M

L

(convert to mL)

Find molarity if 58.6 g barium hydroxide are

Dissolved in 5.65 L sol’n.

Ba2+

OH–

Ba(OH)2

58.6 g

= 0.342 mol

= 0.061 M Ba(OH)2

You have 10.8 g potassium nitrate. How many mL

of sol’n will make this a 0.14 M sol’n?

K+

NO3–

KNO3

10.8 g

= 0.1068 mol

= 0.763 L

= 763 mL


Unit 10 solutions

Molarity and

Stoichiometry

V of gases

at STP

PbI2

KI

V of sol’ns

2

2

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?

Find mol KI needed to yield 89 g PbI2.

Strategy: (1)

(2)

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

mass

mass

vol.

vol.

mol

mol

M

L

M

L

part.

part.


Unit 10 solutions

2

2

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.

PbI2

KI

89 g PbI2

= 0.39 mol KI

= 0.098 L of 4.0 M KI


Unit 10 solutions

How many mL of a 0.500 M

CuSO4 sol’n will react w/excess

Al to produce 11.0 g Cu?

Cu

CuSO4

3

2

3

CuSO4(aq) + Al(s) 

Cu(s)

+ Al2(SO4)3(aq)

11.0 g Cu

Cu

CuSO4

= 0.173 mol CuSO4

= 0.346 L

= 346 mL of 0.500 M CuSO4


Unit 10 solutions

Dilutions of Solutions

Acids (and sometimes bases) are purchased in

concentrated form (“concentrate”) and are easily

diluted to any desired

concentration.

**Safety Tip:

When diluting, add acid

(or base) to water.

C = conc.

D = dilute

MC VC = MD VD

Dilution Equation:


Unit 10 solutions

14.8

14.8

Conc. H3PO4 is 14.8 M. What volume of concentrate

is req’d to make 25.00 L of 0.500 M H3PO4?

MC VC = MD VD

(VC)

=

14.8

0.500

(25)

VC = 0.845

L

How would you mix the above sol’n?

0.845

1. Measure out _____ L of conc. H3PO4.

~20

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.


Unit 10 solutions

400 samples

Cost Analysis with Dilutions

2.5 L of 12 M HCl

(i.e., “concentrate”)

0.500 L of

0.15 M HCl

Cost: $25.71

Cost: $6.35

How many 0.500 L-samples of 0.15 m HCl can be

made from the bottle of concentrate?

MC VC = MD VD

(Expensive water!)

(2.5 L)

=

12 M

0.150 M

(VD)

Moral:

Buy the concentrate

and mix it yourself to

any desired concentration.

VD =

200 L

@ $6.35 ea. =

$2,540


Unit 10 solutions

You have 75 mL of conc. HF (28.9 M); you need

15.0 L of 0.100 M HF. Do you have enough to do

the experiment?

MC VC = MD VD

(0.075 L)

28.9 M

0.100 M

(15 L)

=

Calc. how much conc. you need…

(VC)

28.9 M

0.100 M

(15 L)

=

VC = 0.052

L

= 52 mL needed

Yes;

we’re OK.


Unit 10 solutions

Dissociation occurs when neutral combinations of

particles separate into ions while in aqueous solution.

sodium chlorideNaCl 

Na+ + Cl–

sodium hydroxide NaOH 

Na+ + OH–

H+ + Cl–

hydrochloric acid HCl 

sulfuric acid H2SO4

2 H+ + SO42–

acetic acid CH3COOH 

CH3COO– + H+

acids

In general, _____ yield hydrogen (H+) ions in aque-

ous solution; _____ yield hydroxide (OH–) ions.

bases


Unit 10 solutions

Strong electrolytes exhibit nearly 100% dissociation.

NaCl Na+ + Cl–

Weak electrolytes exhibit little dissociation.

CH3COOH CH3COO– + H+

1000

0

0

NOT in water:

in aq. sol’n:

1

999

999

1000

0

0

NOT in water:

in aq. sol’n:

980

20

20

“Strong” or “weak” is a property of the substance.

We can’t change one into the other.


Unit 10 solutions

electrolytes: solutes that dissociate in sol’n

-- conduct elec. current because

of free-moving ions

-- e.g.,

acids,

bases,

most ionic compounds

-- are crucial for many

cellular processes

-- obtained in a healthy diet

--

For sustained exercise or

a bout of the flu, sports

drinks ensure adequate

electrolytes.


Unit 10 solutions

nonelectrolytes: solutes that DO NOT dissociate

--

DO NOT conduct elec.

current (not enough ions)

-- e.g., any type of sugar


Unit 10 solutions

Colligative Properties

properties thatdepend on the conc. of a sol’n

Compared to

solvent’s…

a sol’n w/that

solvent has a…

…normal freezing

point (NFP)

…lower FP

(freezing point depression)

…normal boiling

point (NBP)

…higher BP

(boiling point elevation)


Unit 10 solutions

FP

BP

water

water + a little salt

water + more salt

Applications of Colligative Properties

(NOTE: Data is fictitious.)

1. salting roads

in winter

0oC (NFP)

100oC (NBP)

–11oC

103oC

–18oC

105oC


Unit 10 solutions

FP

BP

water

water + a little AF

50% water + 50% AF

Applications of Colligative Properties (cont.)

2. Antifreeze (AF)

(a.k.a., “coolant”)

0oC (NFP)

100oC (NBP)

–10oC

110oC

–35oC

130oC


Unit 10 solutions

Applications of Colligative Properties (cont.)

3. law enforcement

starts melting at…

finishes melting at…

penalty, if convicted

white powder

comm.

service

A

120oC

150oC

B

130oC

140oC

2 yrs.

C

20 yrs.

134oC

136oC


Unit 10 solutions

Calculations with

Colligative Properties


Unit 10 solutions

Adding a nonvolatile solute to a solvent decreases

the solution’s freezing point (FP) and increases its

boiling point (BP).

The freezing point depression and

boiling 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

m = molality of solute


Unit 10 solutions

i = van’t Hoff factor

(accounts for # of

particles in solution)

In aq. soln., assume that…

1

-- i = __ for nonelectrolytes

2

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

Jacobus Henricus van’t Hoff

(1852 – 1911)

3

-- i = __ for CaCl2, etc.

In reality, the van’t Hoff factor isn’t always

an integer. Use the guidelines unless

given information to the contrary.


Unit 10 solutions

Kf = 1.86oC/m

Kb = 0.52oC/m

0.690 m

Find the FP and BP of a soln. containing 360 g barium

chloride and 2.50 kg of water.

DTx = Kx m i

360 g BaCl2

= 1.725 mol BaCl2

i = 3

DTf = Kf m i

= 1.86(0.690)(3)

= 3.85oC

FP = –3.85oC

DTb = Kb m i

= 0.52(0.690)(3)

= 1.08oC

BP = 101.08oC


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