Unit 10:Solutions
Chemistry
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
soluble: “will dissolve in”
miscible: refers to two liquids that mix evenly in all proportions
-- e.g., food coloring and water
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.
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
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
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)
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
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
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
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
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
Solids dissolved Gases dissolved in liquids in liquids
To
Sol.
To
Sol.
As To ,solubility ___
As To ,solubility ___
[O2]
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
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?)
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
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
** Measure to part of meniscus w/zero slope.
water ingrad. cyl.
mercury ingrad. cyl.
measure tobottom
measure totop
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”
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)
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
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
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
__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
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
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
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
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)
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
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+
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.
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.
nonelectrolytes: solutes that DO NOT dissociate
--
-- e.g., any type of sugar
DO NOT conduct elec. current (not enough ions)
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)
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
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
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.
h
h