Chemistry 20 Chapter 5 - Solutions PowerPoint Presentation by R. Schultz [email protected]

Chemistry 20

Chapter 5 - Solutions PowerPoint Presentation by R. Schultz

[email protected]

5.1 Classifying Solutions

• Solutions are homogeneous mixtures• solvent: part of solution present in

greatest quantity – e.g. water in an aqueous solution

• solute: the substance dissolved in the solvent – e.g. salt in a saltwater solution

• types of solutions: not just solid in liquid, but gas in gas, gas in liquid, liquid in liquid, ……. (chart page 167)

• Do questions 1, 2 page 168

5.1 Classifying Solutions• aqueous solutions will be our main focus• Dissolving processes:

ionicmolecular

a solute bonds break (endothermic)

c new solute/solvent bonds form (exothermic)b solvent intermolecular bonds break (endothermic)

Ec > Ea + Eb (exothermic); Ec< Ea + Eb (endothermic)


• discuss question 4, page 170• discuss cold packs:


• Electrolytes: substances that conduct electricity when dissolved in water(ionics and strong acids)

• Non-Electrolytes: don’t (moleculars)

• Theory: dissolved ionic compounds dissociate; the compound breaks apart to release free cations and anions

Dissociation equations:NaOH(s) Na+(aq) + OH¯(aq)

K2SO4(s) 2 K+(aq) +

SO42ˉ(aq)

5.1 Classifying Solutions• The dissociation equation shows what

happens when ionics dissolve in water • starts with the pure ionic compound in (s)

state and produces (aq) ions• Ionics with low solubility remain in (s) state

• Acids – molecular compounds that ionize to form H+(aq) and an anion in water. Ionization equation:

• HCl(g) H+(aq) + Clˉ(aq)


• Difference between ionization and dissociation (and equations)?:

• Do worksheet BLM 5.1.3B

ionization: acids – ions not initially present; they form in the presence of waterstarting compound in pure state or (aq)

dissociation: ionics – ions already present in crystal lattice; they just separatestarting compound in (s) state

Discuss questions 3, 4, 6, page 175

5.2 Solubility

• solubility:

• saturated solution

• supersaturated solution:

the ability to be dissolved often expressed in g (solute) per 100 mL solution

solution of maximum concentration under agiven set of conditions

solution of > maximum concentration under agiven set of conditions

5.2 Solubility

• Gas solubilities low in general, but very important in some cases e.g. O2 in H2O

• In a saturated solution dissolving never really stops; dynamic equilibrium exists between dissolving and crystallizing processes

undissolved solute dissolved solute

dissolving

crystallizing

5.2 Solubility

• dynamic equilibrium: 2 opposing processes, occurring at same rate so that no net change is observed

• experimental evidence: in a mixture of a saturated solution with some undissolved solute, the size of the crystals will increase over time, even though the mass of crystals remains unchanged

5.2 Solubility

• Example 11a, page 179

NaCl(s) Na+(aq) + Clˉ(aq)• Example 11c, page 179

(NH4)2CO3(s) 2 NH4+(aq) + CO3

2-

(aq)

• Try 11b, 11d, page 179

Note the difference in equations for saturated & unsaturated

5.2 Solubility

• Solubility Generalizations:

Solubility of Effect of Temperature

Effect of Pressure

Solids in liquids As temp solubility

No appreciable effect

Liquids in liquids No appreciable effect

No appreciable effect

Gases in liquids As temp solubility

As pressure of a gas solubility of

that gas

5.2 Solubility

• Question 7, page 183 – discuss

• Do questions 2a, 3c, 4, 5, 6

5.2 Solubility

• Do Lab 5.B.1 – handout to help with report

5.2 SolubilitySolubility Chart 2009-10 Data Booklet

only

H+, NH4+ H+, Na+H+, Na+

K+, NH4+ K+, NH4

+K+, NH4

+

5.3 Concentration of Solutions

• Concentration of solution: quantity of solute per quantity of solution (it’s always solution, not solvent)

• Can use terms “concentrated” or “dilute”, but most often using numbers

• Expressions of Concentration:• % by mass:

commercial hydrogen peroxide is 3

( )

% 100%( )

mm

mass of solute g

mass of gsolution

% mm


• For very small concentrations: parts per million, ppm, and parts per billion, ppb

• Used for toxic environmental pollutants

6( )# 10

( )

mass of solute gppm ppm

mas solutio gns of

9( )# 10

( )

mass of solute gppb ppb

mas solutio gns of


• Examples:• Practice Problem 2, page 186

• Practice Problem 6a, page 188

• Do worksheet BLM 5.3.2, #1 – 4, and 6

55.0

% 100% 35.5%55.0 100.0

m mm m

g

g g

6

56 6

( )# 10 ; Note: units don't have to be g, just be same

( )

20 2.0 10# 10 # 10 0.33

60 60

mass of

sol

solute gppm ppm

mass of g

mg kgppm ppm ppm ppm ppm

kg kg

ution


• moles – because we are doing the course in different order from the text we need to deal with moles now

• recall from Science 10: a mole (mol) is a gigantic # of particles (6.02 x 1023)

• atoms and molecules are so small that gigantic #’s are needed to make them have a measurable mass

• Green Pea Analogy, Moles Song

you don’t need to know this # by memory

5.4 Preparing and Diluting Solutions

• October 23 each year from 6:02 am to 6:02 pm is International Mole Day since date and time is 6:02 10 23

• National Chemistry Week in USA is week in which Mole Day falls

• Could you write a moles song, a moles poem, dress up as a mole – I challenge you

/


1 mol of moles placed head to tail would cover a distance of 11 million light years!They would have a mass of approximately 90% of the Moon


• molar mass: mass of 1 mol of a substance

• Examples:• Molar Mass of sodium 22.99• Molar Mass of nitrogen

• Molar Mass of (NH4)2CO3

gmol

nitrogen is N2 2 x 14.01 = 28.02 gmol

gmol

2 N 2 14.01

8 H 8 1.01

1 C 1 12.01

3 O 3 16.00

96.11

gmol

gmol

gmol

gmol

gmol

you will need to calculate molar masses often as part of other calculations in every further chapterin this chapter you need to show this workingin further chapters no


• Calculating # of moles of a substance

• example: calculate # moles in 3.50 g Na3PO4 • formula method unit cancelling method

m

nM

number of moles (mol)

mass of sample (g)

molar mass gmol

3 4Na PO 3 22.99

1 30.97

4 16.00

163.94

gmol

gmol

gmol

gmol

M

3.50

0.0213163.94 g

mol

m gn mol

M

3 4Na PO 3 22.99

1 30.97

4 16.00

163.94

gmol

gmol

gmol

gmol

M

need mol; have g and

to get mol, do g since

g = g x = molgmol

molg

gmol

3.500.0213

163.94 gmol

gmol

gmol


• calculating mass of a substance from # of moles

example: calculate mass of 1.75 mol of Ca(NO3)2 formula method unit cancelling method

m

nM

m n M

3 21 40.08

2 14.01

6 16.00

164.10

gmolCa NO

gmol

gmol

gmol

M

1.75 164.10 287gmolm n M mol g

m

n M

3 21 40.08

2 14.01

6 16.00

164.10

gmolCa NO

gmol

gmol

gmol

M

need g; have mol and

to get g do mol x gmol

gmol

1.75 164.10 287gmolmol g

Do worksheet BLM 3.0.2


• Back to concentrations – molar concentration

• don’t memorize this formula – get it upside down and you’re sunk

• molar concentration units (mol/L) – that tells you everything you need

n

cv

molar concentration (mol/L)

# moles solute (mol)

volume of solution (L)


• Example:• Practice Problem 12a, page 191

12 22 1112 12.01

22 1.01

11 16.00

342.34

gmolC H O

gmol

gmol

gmol

M

4.63

0.0135342.34 g

mol

m gn mol

M

0.0135

0.8050.0168

molL

n molc

v L

Do worksheet, BLM 5.3.3 a, b, e

5.3 Concentration of Solutions• Molar concentration of ions• Solving these always requires a

dissociation or ionization equation• Example: calculate the molar

concentration of each ion in 0.125 mol/L Fe2(SO4)3

• 1 Fe2(SO4)3(s) 2 Fe3+(aq) + 3 SO42(aq)

looking for

givenyou will use this ratio often in Chapters 5, 7, and 8 (get used to it here)

3 2

0.125 0.2501

mol molL LFe

2

4

30.125 0.375

1mol mol

L LSO

“[ a ]” stands for molar concentration of a


• Example: Calculate the molar concentration of Ca3(PO4)2 required to produce a solution with [Ca2+]=0.425 mol/L

• 1 Ca3(PO4)2(aq) 3 Ca2+(aq) + 2 PO3-(aq)

3 4 2

10.425 0.142

3mol mol

L LCa PO

Do worksheet BLM 5.3.5A – answers on my website


• Molar concentration of ions in solution (continued)

• Example: Practice Problem 17, page 193

2 32 22.99

1 12.01

3 16.00

105.99

gmolNa CO

gmol

gmol

gmol

M

2.7

0.0255105.99 g

mol

m gn mol

M

0.0255

0.1460.175

molL

n molc

v L

this looks like final answer, but need concentration of Na+, not solute

1 Na2CO3(s) 2 Na+(aq) + 1 CO32-(aq)

2

0.146 0.291

mol molL LNa

Do worksheet BLM 5.3.5 2, 5, 6


• Calculating mass from concentration and volume

• Example: Practice Problem 23a, page 194

1 22.99

1 35.45

58.44

gmolNaCl

gmol

gmol

M

0.200 0.125 0.0250molL

nc n c v

vn L mol

0.0250 58.44 1.46gmol

mn m n M mol g

M

unit cancellation is better


• Calculating volume from mass and concentration

• Example: Practice Problem 27b, page 195

31 107.87

1 14.01

3 16.00

169.88

gmolAgNO

gmol

gmol

gmol

M

2.0

0.0118169.88 g

mol

m gn mol

M

0.0118

0.052 or 52 0.225 mol

L

n n molc v L mL

v c

Do worksheet BLM 5.3.6 first side only


• standard solution: carefully made solution with known concentration

• you’ll learn, in the lab, how to prepare a standard solution by 2 methods

• Method 1: Preparation from solid – calculation already done (slide 27)


• Method 2: Preparation by Dilution

• This is done for 2 possible reasons:– calculated mass of solute is too small to be

weighed out– you may already have a more concentrated

solution available and can prepare your by diluting the other


• Dilution calculation:

• since in dilution, moles of solute can’t change, ni = nf

• the dilution equation!

initial # moles of solute, i i i in n c v

final # moles of solute, f f f fn n c v

i i f fc v c v


• Example: Practice Problem 31a, page 198

• biggest problem students have with these is misidentifying the variables

• I encourage you to list them:ci = 1.25 mol/L

vi = ?

cf = 1.00 mol/L

vf = 50 mL

i i f fc v c v

f fi

i

c vv

c

1.00 5040

1.25

molLf f

i molLi

c v mLv mL

c

units don’t have to “match”; as long as both v’s and both c’s have same units

formulas with all multiplication are easy to rearrange do this 1st

Do worksheet BLM 5.4.1 all


• Investigation 5.D – Solution Preparation and Dilution, page 200


• Chapter Review

• Calculation Review:worksheet BLM 5.3.7


Documents

Chemistry 20 Chapter 5 - Solutions PowerPoint Presentation by R. Schultz [email protected]