Chemistry 20 Chapter 5 - Solutions PowerPoint Presentation by R. Schultz...
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Transcript of Chemistry 20 Chapter 5 - Solutions PowerPoint Presentation by R. Schultz...
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)
5.1 Classifying Solutions
• discuss question 4, page 170• discuss cold packs:
5.1 Classifying Solutions
• 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)
5.1 Classifying Solutions
• 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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
/
5.4 Preparing and Diluting Solutions
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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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)
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
5.3 Concentration of Solutions
• 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
5.4 Preparing and Diluting Solutions
• 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)
5.4 Preparing and Diluting Solutions
• 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
5.4 Preparing and Diluting Solutions
• 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
5.4 Preparing and Diluting Solutions
• 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
5.4 Preparing and Diluting Solutions
• Investigation 5.D – Solution Preparation and Dilution, page 200
5.4 Preparing and Diluting Solutions
• Chapter Review
• Calculation Review:worksheet BLM 5.3.7
5.4 Preparing and Diluting Solutions