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Characteristics of solutions
• Solution – homogeneous mixture
a) parts of a solutioni) solute – substance being
dissolvedii) solvent – substance doing
dissolving
both can be either solid, liquid, or gas
Solubility
• Soluble – substance can dissolve in a solvent
ex: salt in water
• Insoluble – substance cannot dissolve in a solvent
ex: sand in water
Solvation In Aqueous Solutions• Solvation – process of surrounding
solute particles with solvent particles
Why are some substances soluble in a solvent and some others are not?
must be compatibility between solute and solvent
Dissolution of sodium Chloride
“like dissolves like”
• Defn – rule used to determine if substance will dissolve in another
- based on attractive forces between solute and solvent
“like dissolves like”
• polar solvents – dissolve polar molecular compounds and ionic compoundsex: salt and water, alcohol and vinegar
• nonpolar solvents – dissolve nonpolar compounds onlyex: oil and gasoline
Factors Affecting Rate of Solvation• How can you dissolve something
faster???
a) increase temp of solvent
this accelerates particles creating more particle collisions
Factors Affecting Rate of Solvation
b) agitate the solution more particle collisions between
solute and solvent
c) Increase surface area of solutebreaking into smaller pieces allows more solute to be in contact w/ solvent
Solubility
• Defn – max amt of solute that can dissolve in a solvent at a specific temp
how much solute can be put into solvent?
Unsaturated Solution
• Defn – less than max amt of solute dissolved
if I put sugar into water and all sugar is dissolved, solution is unsaturated
Saturated Solution
• Defn – contains max amt of solute dissolved
if I put sugar into water and not dissolves (you can see the sugar), the solution is saturated
Supersaturated Solution
• Defn – contains more solute than saturated solution at the same conditions
a saturated solution made at high temp cools slowly. Slow cooling allows excess solute to remain dissolved in solution at lower temperature
very unstable
Solubility Curve (generic)Saturated-Line represents max amount solute that will dissolve at a given temperature
Temperature
Solu
bili
ty(g
solu
te/
10
0 g
H2O
)
Unsaturated(below line)
Supersaturated(above line)
How does temp affect solubility?• The higher the temp, higher the
solubility
(for most cases)
Solution Concentration
• Concentration – how much solute dissolved in amount of solvent
what is difference between concentrated and diluted?
Concentrated vs. Dilute
Concentration
• 3 different units of concentration
a) percent by mass
b) molarity (M)
c) molality (m)
Percent by mass
• Formula
OR
100 xsolution of mass
solute of mass
100 x solvent mass solute mass
solute mass
Percent by mass
• Ex prob: If 3.6 g NaCl is dissolved in 100 g H2O, what is the percent by mass?
What is the solute? NaCl
What is the solvent? H2O
Percent by mass
• Mass of solute (NaCl) =3.6 g
• Mass solvent (H2O) =
100 g• Mass solution =
3.6 + 100 = 103.6 g
Percent by mass
Percent by mass = 3.6 g x 100
103. 6 g
= 3.5 % NaCl
Molarity
• Defn - # of moles per liter of solution
• Formula mol solute L solution
• unit mol = M (capital M) L
Molarity Ex prob #1
• A solution has a volume of 250 mL and has 0.70 mol NaCl. What is the molarity?
2.8 mol/Lor
2.8 M0.250 L
=0.70 mol
Molarity ex prob #2
• What is the molarity of a solution made of 47.3 g NaOH in 500 mL water?
step 1: convert grams to moles47.3 g NaOH
40 g NaOH
1 mol NaOH=1.1825 mol NaOH
Molarity ex prob #2
Step 2: divide moles by volume (L)
2.37 mol/L NaOHor
2.37 M NaOH0.500 L
=1.1825 mol
Molarity ex prob #3
• How many moles of solute are present in 1.5 L of 2.4 M NaCl? How many grams?
# moles = volume x molarity
1.5 LL
2.4 mol NaClx = 3.6 mol NaCl
Molarity ex prob #3
• moles to grams
3.6 mol NaCl
1 mol NaCl
58.5 g NaCl= 210.6 g NaCl
Diluting Solutions
• Defn – add more solvent to original solution
• FormulaM1V1 = M2V2
M1 is more concentrated than M2
Diluting Solutions
• What volume of a 2.0 M stock solution is needed to make 0.50 L of a 0.300 M solution?
M1=
V1=
M2=
V2=
2.0 M 0.300 M
? 0.50 L
(2.0 M) V1 = (0.300 M)(0.50 L)
V1 = 0.075 L
Diluting Solutions
• If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution?M1=
V1=
M2=
V2=
3.5 M ?
20.0 mL 100 mL
(3.5 M)(20.0 mL) = M2 (100.0 mL)
M2 = 0.7 M
Molality
• Defn - # moles of solute in one kg solvent
• Formula mol solutekg solvent
• Units mol = m (lower case m)
kg
Molality ex problem
• What is the molality of a solution with 8.4 g NaCl in 255 g of water?
Step 1: convert grams to moles
8.4 g NaCl
58.5 g NaCl
1 mol NaCl= 0.14 mol NaCl
Molality ex problem
Step 2: divide by mass (kg)
0.55 mol/kg NaClor
0.55 m NaCl0.255 kg
=0.14 mol NaCl
Colligative Properties of Solutions
• Solutes affect the physical properties of their solvents
• Colligative properties (defn) – properties that depend only on the number of solute particles present, not their identity
• Ex: boiling point, freezing point
Electrolytes
• Defn – substances that break up (ionize) in water to produce ions; can conduct electricity- consist of acids, bases, ionic compounds
Ex: NaCl Na1+ + Cl1-
H2SO4 2 H+ + SO42-
Nonelectrolytes
• Defn – do not break up (ionize) in water, they stay the same; doesn’t conduct electricity- usually molecular/covalent compounds
Ex: sugar C6H12O6 C6H12O6
ethanol C2H5OH C2H5OH
Determining # of solute particles
• For ionic cmpds/acids sum # moles of ions
ex: NaCl 1 Na+ + 1 Cl1- = 2 particles
CaBr2 1 Ca2+ + 2 Br1- = 3 particles
Determining # of solute particles
• For covalent compounds # is always 1
ex: sugar C6H12O6 1 particle
ethanol C2H5OH 1 particle
Freezing Point Depression
• Defn (ΔTf) – difference in temp between solution’s fp and pure solvent’s fp
• Formula ΔTf = Kf x m x i
# particlesmolalitymolal fp constant
Ex problem
• What is the freezing pt of water if 12.3 g NaCl is added to 200 g water?(Kf = 1.86 °C/m, fp = 0°C)
1 mol NaCl
58.5 g NaCl
12.3 g NaCl = 0.21 mol NaCl
m = 0.21 mol NaCl
0.200 kg water= 1.05 m
Ex problem
• ΔTf = Kf x m x i
ΔTf = (1.86 °C/m)(1.05 m) (2)
= 3.91 °C
New f.p. = 0°- 3.91° = -3.91°C
Boiling Point Elevation
• Defn (ΔTb) – difference in temp between solution’s bp and pure solvent’s bp
• Formula ΔTb = Kb x m x i
# particlesmolalitymolal bp constant
Ex problem
• What is the new boiling point of acetone if 13.7 g C6H12O6 is dissolved in 200 g acetone? (Kb = 1.71°C/m, bp = 56°C)
1 mol C6H12O6
180 g C6H12O6
13.7 g C6H12O6 = 0.076 mol
m = 0.076 mol C6H12O6
0.200 kg acetone= 0.381 m
Ex problem
• ΔTb = Kb x m x i
ΔTb = (1.71 °C/m)(0.381 m)(1)
= 0.65 °C
New bp = 56°+ 0.65° = 56.65°C