unit 10: solutions
DESCRIPTION
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 . - PowerPoint PPT PresentationTRANSCRIPT
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Unit 10:Solutions
Chemistry
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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 + Sn;brass = Cu + Zn
e.g., water e.g., salt
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soluble: “will dissolve in”
miscible: refers to two liquids that mix evenly in all proportions
-- e.g., food coloring and water
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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.
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Classes of Solutions
aqueous solution: solvent = water
amalgam:e.g.,
tincture:
e.g.,
organic solution:
solvent = alcohol
dental amalgam
tincture of iodine (for cuts)
solvent = mercury
water = “the universal solvent”
solventcontains ________
Organic solvents includebenzene, toluene, hexane, etc.
carbon
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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
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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)
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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
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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
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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
Na1+
detergent
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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 Na1+ at polar end of soapmolecule. Soap is changed into an insolubleprecipitate (i.e., soap scum).
NONPOLARHYDROCARBON
TAILPOLAR HEAD
Na1+
micelle: a liquid droplet covered w/soap or detergent molecules
oil
H2OH2O
H2O H2O
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Solubility
Temp. (oC)
Solubility(g/100 g
H2O)
KNO3 (s)
KCl (s)
HCl (g)
SOLUBILITYCURVE
unsaturated: sol’n could hold more solute;
saturated: sol’n has “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
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Solids dissolved Gases dissolved in liquids in liquids
To
Sol.
To
Sol.
As To ,solubility ___
As To ,solubility ___
[O2]
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Using an available solubilitycurve, classify asunsaturated, saturated,or supersaturated.
80 g NaNO3 @ 30oC
45 g KCl @ 60oC
30 g KClO3 @ 30oC
70 g Pb(NO3)2 @ 60oC
per 1
00 g
H2O
unsaturated
saturated
supersaturated
unsaturated
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Per 500 g H2O,100 g KNO3 @ 40oC
saturation point@ 40oC for 100 g H2O
= 63 g KNO3
So saturation pt.@ 40oC for 500 g H2O
= 5 x 63 g= 315 g
100 g < 315 g
unsaturated
(Unsaturated, saturated, or supersaturated?)
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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
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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
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** Measure to part of meniscus w/zero slope.
water ingrad. cyl.
mercury ingrad. cyl.
measure tobottom
measure totop
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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”
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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)
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Na1+
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
OH1– NaOH
Mg2+
mol 1g 262.93.38 mol = 889 g Mg3(PO4)2
PO43– Mg3(PO4)2
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Find molarity if 58.6 g barium hydroxide arein 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+ OH1–
Ba(OH)2
= 0.342 mol
L 5.65mol 0.342 = 0.061 M Ba(OH)2
K1+ NO31–
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
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m m
V V
P P
mol mol
M L M L
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.
1 1 22
PbI2 KI
V of gasesat STP
V of sol’ns
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__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 neededto yield 89 g PbI2. Based on (1), find volumeof 4.0 M KI sol’n.
1 1 22
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
mol
L M
M LM L
PbI2 KI
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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)
M LM L
Cu CuSO4
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
mol
L M
+ __Al2(SO4)3(aq)
SO42– Al3+
3 3 2 1
= 0.346 L
L 1mL 1000
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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
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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
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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) = 0.500 MVD = 200 L
400 samples @ $6.35 ea. = $2,540
Moral:Buy the concentrateand mix it yourself to
any desired concentration.
(Expensive water!)
(VD)
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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
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Dissociation occurs when neutral combinations ofparticles separate into ions while in aqueous solution.
In general, _____ yield hydrogen (H1+) ions in aque-ous solution; _____ yield hydroxide (OH1–) ions.
acidsbases
sodium hydroxide NaOH hydrochloric acid HCl
sodium chloride NaCl Na1+ + Cl1–
Na1+ + OH1–
H1+ + Cl1–
sulfuric acid H2SO4 2 H1+ + SO42–
acetic acid CH3COOH CH3COO1– + H1+
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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 Na1+ + Cl1–
1000 0 0
980 20 20
CH3COOH CH3COO1– + H1+ Weak electrolytes exhibit little dissociation.
Strong electrolytes exhibit nearly 100% dissociation.
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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.
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nonelectrolytes: solutes that DO NOT dissociate
--
-- e.g., any type of sugar
DO NOT conduct elec. current (not enough ions)
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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)
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water + more salt
water + a little salt
water
BPFP
0oC (NFP)
Applications of Colligative Properties (NOTE: Data are fictitious.)
1. salting roads in winter
100oC (NBP)
–11oC 103oC
–18oC 105oC
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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
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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.
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h
h