Solutions

Solution Definitions

• Solution: Solution: homogeneous mixture, evenly mixed at the particle level (like salt water).

Solvent:Solvent: present in greater amount that can dissolve the solute

Solute: Solute: substance being dissolved

Classes of Solutionsaqueous solution:

water = “the universal solvent”

solvent = water

amalgam: solvent = Hg

e.g., dental amalgam

tincture: solvent = alcohol

e.g., tincture of iodine (for cuts)

organic solution: solvent contains carbon

e.g., gasoline, benzene, toluene, hexane

Solution Definitions

alloy:

solvent: the substance that dissolves the solute

water salt

a solid solution of metals

-- e.g., bronze = Cu + Sn; brass = Cu + Zn

“will dissolve in”

refers to two gases or two liquids that forma solution; more specific than the term “soluble”

miscible:

soluble:

Solubility

Solubility - Solubility - maximum grams of solute that will dissolve in 100 g of solvent at a given temperature

varies with temperature

SolubilitySATURATED SOLUTION

no more solute dissolves

UNSATURATED SOLUTIONmore solute

dissolves

SUPERSATURATED SOLUTION

becomes unstable, crystals form

increasing concentration

Temp. (oC)

Solubility(g/100 g H2O)

KNO3 (s)

KCl (s)

HCl (g)

SOLUBILITYCURVE

Solubility how much solute dissolves in a given amt.

of solvent at a given temp.

unsaturated: solution could hold more solute; belowbelow line

saturated: solution has “just right” amt. of solute; onon line

supersaturated: solution has “too much” solute dissolved in it;

above the line

Solubility Table

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517

0 10 20 30 40 50 60 70 80 90 100Temperature (°C)

Solubility vs. Temperature for Solids

Sol

ubili

ty (

gram

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

NaCl KClO3

SO2

shows the dependence

of solubility on temperature

gases

solids

per100 gH2O

Classify as unsaturated, saturated, or supersaturated.

80 g NaNO3 @ 30oC unsaturated

60 g KCl @ 60oC saturated

50 g NH3 @ 10oC unsaturated

70 g NH4Cl @ 70oC supersaturated

So sat. pt. @ 40oC for 500 g H2O = 5 x 66 g = 330 g

120 g < 330 g unsaturated

saturation point @ 40oC for 100 g H2O = 66 g KNO3

Per 500 g H2O, 120 g KNO3 @ 40oC

SolubilitySolubility

Solids are more soluble at...Solids are more soluble at...• high temperatures.

Gases are more soluble at...Gases are more soluble at...

low temperatures

(A) Per 100 g H2O, 100 g Unsaturated; all

soluteNaNO3 @ 50oC. dissolves; clear

solution.

(B) Cool solution (A) very Supersaturated; extraslowly to 10oC. solute remains in solution;

still clear.

Describe each situation below.

(C) Quench solution (A) in Saturated; extra solute an ice bath to 10oC. (20 g) can’t remain in

solution, becomes visible.

Non-Solution Definitions

insoluble: “will NOT dissolve in”

e.g., sand and water

immiscible: refers to two gases or two liquids that will NOT form a solution

e.g., water and oil

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

Water

HOT

Solubility

A B

Before

Water

COLD

Add 1 drop of red food coloring

Miscible – “mixable”

two gases or two liquids that mix evenly

Experiment 1:

Water

HOT

AFTER

Water

COLD

A B

Solubility

Water Water

Oil

T30 sec

AFTER

Before

Add oil to water and shake

Immiscible – “does not mix”

two liquids or two gases that DO NOT MIX

Experiment 2:

T0 sec

Solutions• What the solute and the solvent are

determines:–whether a substance will dissolve. –how much will dissolve.

As size , rate

As T , rate

3. mixing

Factors Affecting the Rate of Dissolution

1. temperature

2. particle size

4. nature of solvent or solute

More mixing, rate

Solid Solubility

Timberlake, Chemistry 7th Edition, page 297

KI

NaNO 3

KN

O 3

Na 3PO 4

NaCl

Temperature (oC)

Sol

ubili

ty (

g so

lute

/ 10

0 g

H2O

)200

180

160

140

120

100

80

60

40

20

020 40 60 80 100

Gas SolubilityCH4

O2

CO

He

Temperature (oC)

Sol

ubili

ty (

mM

)

2.0

1.0

0 10 20 30 40 50

Higher Temperature

…Gas is LESS Soluble

Particle Size and Mixing

• In order to dissolve - the solvent molecules must come in contact with the solute.

• Stirring moves fresh solvent next to the solute.

• The solvent touches the surface of the solute.

• Smaller pieces increase the amount of surface of the solute.

Molecular Polarity

• Nonpolar molecules: electrons are shared equally– Molecule has no charge

• e.g. Ethane

O2-

H+

H+

+

Polar Molecules

• Electrons are not equally shared• Molecule has slight charge

“Like dissolves like”(Polarity)

Solubility Rules

1. Most nitrate (NO31-) salts are soluble.

2. Most salts of Na+, K+, and NH4+ are soluble.

3. Most chloride salts are soluble. Notable exceptionsare AgCl, PbCl2, and Hg2Cl2.

1. Most sulfate salts are soluble. Notable exceptionsare BaSO4, PbSO4, and CaSO4.

1. Most hydroxide compounds are only slightlysoluble.* The important exceptions are NaOH andKOH, Ba(OH)2 and Ca(OH)2 are only moderatelysoluble.

1. Most sulfide (S2-), carbonate (CO32-), and phos-

phate (PO43-) salts are only slightly soluble.

*The terms insoluble and slightly soluble really mean the same thing. Such a tiny amount dissolves that it is not possible to detect it with the naked eye.

General Rules for Solubility ofIonic Compounds (Salts) in Water at 25 oC

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 218

Oil and Water Don’t Mix• Oil is nonpolar• Water is polar• “Like dissolves like” • polar + polar = solution• nonpolar + nonpolar =

solution• polar + nonpolar =

suspension (won’t mix evenly)

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 470

SolvationSolvation

Soap / DetergentSoap / Detergent• polar “head” with long nonpolar “tail”• dissolves nonpolar grease in polar water

micelle

Solubility of Common Compounds

NO3- salts

Na+, K+, NH4+ salts

Soluble compounds

Cl-, Br-, I- salts

S2-, CO32-, PO4

3- salts

OH1- salts

SO42- salts

Except forthose containing

Ag+, Hg22+, Pb2+

Except forthose containing Ba2+, Pb2+, Ca2+

Insoluble compounds

Except forthose containing

Na+, K+, Ca2+

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 218

Making Molar

Solutions

…from liquids(More accurately, from stock solutions)

Concentration…a measure of solute-to-solvent ratio

concentrated vs. dilute “lots of solute” “not much solute”

“watery”

Add water to dilute a solution; boil water off to concentrate it.

Concentration“The amount of solute in a solution”

mol

L M

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

B. parts per million (ppm) also, ppb and ppt – commonly used for minerals or contaminants in water supplies

C. molarity (M) = moles of solute L of sol’n

– used most often in this class

D. molality (m) = moles of solutekg of solvent

M = mol L

% by mass – medicated creams% by volume – rubbing alcohol

V = 1000 mL

n = 2 moles

Concentration = 2 molar

V = 1000 mL

n = 4 moles

[ ] = 4 molar

V = 5000 mL

n = 20 moles

[ ] = 4 molar

Molarity (M) = # of moles

volume (L)V = 250 mL

n = 8 moles[ ] = 32 molar

Glassware

precise; expensive; good for making solutions

Range:

Glassware – Precision and Costbeaker vs. volumetric flask

When filled to 1000 mL line, how much liquid is present?

beaker 5% of 1000 mL = 50 mL

volumetric flask1000 mL + 0.30 mL

950 mL – 1050 mL 999.70 mL– 1000.30 mL

imprecise; cheap

Range:

How to mix a Standard Solution

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

Wash bottle

Volume marker(calibration mark)

Weighedamount of solute

How to mix solid chemicals

How many grams of sodium hydroxide would we need to make 100. mL of 3.0 M NaOH?

M = mol L 3 M =

? mol 1 L

How much will this weigh?

1 Na @ 23g/mol + 1O @ 16g/mol + 1 H @ 1 g/mol

MMNaOH = 40g/mol

40.0 g NaOH1 mol NaOH

X g NaOH = 3.0 mol NaOH =

To mix this:add 120 g NaOH into 1L volumetric flask with~750 mL cold H2O. Mix, allow to return to room temperature – bring volume to 1 L.

120 g NaOH

40.0 g NaOH1 mol NaOH

How many moles of solute are required to make 1.35 L of 2.50 M solution?

mol = M L

B. What mass of magnesium phosphate is this?

A. What mass of sodium hydroxide is this?

mol

L M

3.38 mol= 2.50 M (1.35 L) =

X g NaOH = 3.38 mol NaOH = 135 g NaOH

262.9 g Mg3(PO4)2

1 mol Mg3(PO4)2

X g Mg3(PO4)2 = 3.38 mol Mg3(PO4)2 = 889 g Mg3(PO4)2

0.342 mol

5.65 L

Find molarity if 58.6 g of barium hydroxide are in 5.65 L solution.

171.3 g Ba(OH)2

1 mol Ba(OH)2

Step 2). What is the molarity of a 5.65 L solution containing 0.342 mol solute?

Step 1). How many moles barium hydroxide is this?

0.0605 M Ba(OH)2=

X mol Ba(OH)2 = 58.6 g Ba(OH)2 = 0.342 mol Ba(OH)2

M =

mol

LM

=

You have 10.8 g potassium nitrate. How many mL of solution will make this a 0.14 M solution?

convert to mL

MolalityMolality

solvent ofkg

solute of moles(m)molality

mass of solvent only

1 kg water = 1 L waterkg 1

mol0.25 0.25m

MolalityMolality

Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.

75 g MgCl2 1 mol MgCl2

95.21 g MgCl2

= 3.2m MgCl2

0.25 kg water

kg

molm

MolalityMolality

How many grams of NaCl are required to make a 1.54 m solution using 0.500 kg of water?

0.500 kg water 1.54 mol NaCl

1 kg water

= 45.0 g NaCl

58.44 g NaCl

1 mol NaCl

kg 1

mol1.5 1.5m

500 mLvolumetric

flask

Preparing Solutions

500 mL of 1.54M NaCl

500 mLwater

45.0 gNaCl

– mass 45.0 g of NaCl– add water until total volume is

500 mL

– mass 45.0 g of NaCl– add 0.500 kg of water (volume

will be just over 500 total mL)

500 mLmark

1.54m NaCl in 0.500 kg of water

molality molarity

2211 VMVM

Dilution

• Preparation of a desired solution by adding water to a concentrate.

• Moles of solute remain the same.

Dilution

• What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:

M1 = 15.8M

V1 = ?

M2 = 6.0M

V2 = 250 mL

WORK:

M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

occurs when neutral combinations of particles separate into ions while in aqueous solution.

sodium chloride

sodium hydroxide

hydrochloric acid

sulfuric acid

acetic acid

Dissociation:

NaCl Na1+ + Cl1–

NaOH Na1+ + OH1–

HCl H1+ + Cl1–

H2SO4 2 H1+ + SO42–

CH3COOH CH3COO1– + H1+

NaCl Na1+ + Cl1–

CH3COOH CH3COO1– + H1+

Weak electrolytes exhibit little dissociation.

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

Strong electrolytes exhibit nearly 100% dissociation.

NOT in water: 1000 0 0in aq. solution: 1 999 999

NOT in water: 1000 0 0

in aq. solution: 980 20 20

electrolytes: solutes that dissociate in solution-- conduct electric 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.

nonelectrolytes: solutes that DO NOT dissociate-- DO NOT conduct electric current (not enough ions) e.g., any type of sugar

…normal boiling point (NBP) …higher BP

FREEZING PT.DEPRESSION

BOILING PT.ELEVATION

Colligative Properties depend on concentration of a solution

Compared to solvent‘s: a solution w/that solvent has a:…normal freezing point (NFP) …lower FP

1. salting roads in winter

FPBP

water 0oC (NFP) 100oC (NBP)

2. antifreeze (AF) /coolant

FPBP

water 0oC (NFP) 100oC (NBP)

water + a little AF –10oC 110oC

50% water + 50% AF –35oC 130oC

water + a little salt

water + more salt

–11oC 103oC

–18oC 105oC

Applications (NOTE: Data are fictitious.)

3. law enforcement

white powder

startsmelting

at…

finishesmelting

at…

penalty, ifconvicted

A 109oC 175oC comm. service

B 150oC 180oC 2 years

C 194oC 196oC 20 years

Calculations for Colligative PropertiesCalculations for Colligative Properties

The change in FP or BP is found using… Tx = Kx m i

Tx = change in To (below NFP or above NBP)

Kx = constant depending on… (A) solvent

(B) freezing or boilingm = molality of solute = mol solute / kg solvent

i = integer that accounts for any solute dissociationany sugar (all nonelectrolytes)……………...i = 1

table salt, NaCl Na1+ + Cl1–………………i = 2

barium bromide, BaBr2 Ba2+ + 2 Br1–……i = 3

Freezing Point Depression Boiling Point Elevation

Tf = Kf m i Tb = Kb m i

Then use these in conjunction with the NFP and NBP tofind the FP and BP of the mixture.

(Kb = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water)

(Kf = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water)

m 0.373 kg 50.2

g 180g 168

OH kg

OHC mol m

2

6126 (NONELECTROLYTE)

168 g glucose, C6H12O6, (a sugar, nonelectrolyte) are mixed w/2.50 kg H2O.

Find BP and FP of mixture. For water, Kb = 0.512, Kf = –1.86.

i = 1

Tb = Kb m i = 0.512 (0.373) (1) =

0.19oC

BP = (100 + 0.19)oC = 100.19oC

Tf = Kf m i = –1.86 (0.373) (1) = –

0.694oC

FP = (0 + –0.69)oC = –0.694oC

m 0.316 kg 50.2

g 212.8g 168

OH kg

CsBr mol m

2

168 g cesium bromide are mixed w/2.50 kg H2O. Find BP and FP of mixture.

For H2O, Kb = 0.512, Kf = –1.86.

Tb = Kb m i = 0.512 (0.316) (2) =

0.32oC

BP = (100 + 0.32)oC = 100.32oCTf = Kf m i = –1.86 (0.316) (2) = –

1.18oC

FP = (0 + –1.18)oC = –1.18oC

Cs1+ Br1– i = 2

CsBr Cs1+ + Br1–

1 Pb(NO3)2(aq) + KI (aq) PbI2(s) + KNO3(aq)__ __ __ __2 1 2

Molarity and Stoichiometry

M M

V V

P P

mol mol

M L M L

M =mol

L

mol = M L

What volume of 4.0 M KI solution is required to yield 89 g PbI2?

461 g PbI2

Strategy:

1 Pb(NO3)2(aq) + 2 KI (aq) 1 PbI2(s) + 2 KNO3(aq)

X mol KI = 89 g PbI2

1 mol PbI2

1 mol PbI2

2 mol KI= 0.39 mol KI

(1) Find mol KI needed to yield 89 g PbI2.(2) Based on (1), find volume of 4.0 M KI solution.

What volume of 4.0 M KI solution is required to yield 89 g PbI2?

89 g? L 4.0 M

M = molL

L = molM

= 0.39 mol KI

4.0 M KI= 0.098 L of 4.0 M KI

= 0.173 mol CuSO4

How many mL of a 0.500 M CuSO4 solution will react w/excess Al to produce 11.0 g Cu?

__CuSO4(aq) + __Al (s)

Al3+ SO42–

CuSO4(aq) + Al (s) Cu(s) + Al2(SO4)3(aq)3 2 3 1x mol 11 g

__Cu(s) + __Al2(SO4)3(aq)

X mol CuSO4 = 11 g Cu1 mol Cu

63.5 g Cu

3 mol CuSO4

3 mol Cu

M = molL

L = molM

0.173 mol CuSO4

0.500 M CuSO4

= 0.346 L

0.346 L 1000 mL

1 L= 346 mL

Limiting Reactants• 79.1 g of zinc react with 0.90 L of 2.5M HCl.

Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

79.1g Zn

1 molZn

65.39g Zn

= 27 L H2

1 molH2

1 molZn

22.4 LH2

1 molH2

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

22.4L H2

1 molH2

0.90L

2.5 molHCl

1 L= 25 L

H2

1 molH2

2 molHCl

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

Zn: 27 L H2 HCl: 25 L H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 25 L H2

left over zincCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem