Upload
tranliem
View
213
Download
0
Embed Size (px)
Citation preview
1
SolutionsChapter 14
Hein and Arena
Version 1.1
Eugene Passer
Chemistry Department
Bronx Community College
© John Wiley and Sons, Inc
2
General Properties of
Solutions
3
• A solute is a compound that is dissolved
in a solvent.
• The solvent is the dissolving agent or the
most abundant component in the solution.
• A solution is a system in which one or
more solutes are homogeneously
dissolved in a solvent.
4
Solubility
(…and the dissolution process)
5
Solubility describes the amount of a substance
that will dissolve in a specified amount of
solvent.
6Na+ and Cl- ions hydrated by H2O molecules.15.2
Dissolution of sodium chloride in water.
7
– very soluble ( greatest degree of dissolution)
– soluble
– moderately soluble
– slightly soluble
– Insoluble (least degree of dissolution)
• Terms that describe the extent of
solubility of a solute in a solvent:
8
• Terms that describe the solubility of
liquids:
– immiscible: liquids that are insoluble in
each other.
oil and water
methyl alcohol and water
– miscible: liquids that are soluble in each
other.
9
Factors Related to
Solubility
10
The Nature of theSolute and Solvent
11
• The general rule for predicting solubility
is “like dissolves like”.
12
• Polar compounds tend to be more
soluble in polar solvents than nonpolar
solvents.
• Nonpolar compounds tend to be more
soluble in nonpolar solvents than in
polar solvents.
13
Rate of
Dissolving Solids
14
– Particle size; smaller particles dissolve
faster due to greater surface area.
– Temperature; higher temperatures
generally increase the dissolution rate.
– Mixing the solution increases contact
between solvent molecules and the
solute surface.
• Factors that affect solubility are:
15
Solutions:
A Reaction Medium
16
sodium chloride reacts
with silver nitrate when dissolved
in water
NaCl(aq) + AgNO3(aq) → AgCl(s) +NaNO3(aq)
17
Concentration
of Solutions
18
• A dilute solution contains a relatively
small amount of dissolved solute.
• A concentrated solution contains a
relatively large amount of dissolved
solute.
19
Mass Percent Solution (wt/wt)
20
Mass percent expresses the concentration
of a solution as the percent of solute in a
given mass of solution.
g solute g solutemass percent = x 100 = x 100
g solute + g solvent g solution
21
What is the mass percent of sodium hydroxide in a
solution that is made by dissolving 8.00 g NaOH in
50.0 g H2O?
grams of solute (NaOH) = 8.0 g
grams of solvent (H2O) = 50.0 g
2
8.00 g NaOH x 100 = 13.8% NaOH solution
8.00 g NaOH + 50.0 g H O
g solutemass percent = x 100
g solute + g solvent
22
What masses of potassium chloride and water are
needed to make 250. g of 5.00% (by wt/wt) solution?
The percent expresses the mass of the solute.
250.g = total mass of solution
5.00 x 250. g = 12.5 KCl solute
100
g solutemass percent = x 100
g solution
mass percent x g solution = g solute
100
250.g – 12.5 g = 238 g H2O
Dissolving 12.5 g
KCl in 238 g H2O
gives a 5.00% KCl
solution.
23
A 34.0% (by wt/wt) sulfuric-acid solution has a
density of 1.25 g/mL. How many grams of H2SO4 are
contained in 1.00 L of this solution?
Grams of solution are
determined from the
solution density.
Step 1. Determine
grams of solution.
1.00 L = 1.00 x 103 mL
MassDensity =
VolumeDensity x Volume = Mass
1.25 g
1 mL
3(1.00 x 10 mL) = 1250 g (mass of solution)
24
Solve the mass percent equation for grams of solute.
[(g H2SO4)/(g solution)] x 100 = mass %
g H2SO4 = [34.0 % (g solution)]/100
g H2SO4 = [34.0%(1250g solution)]/100 = 425gH2SO4
A 34.0% (by wt/wt) sulfuric-acid solution has a
density of 1.25 g/mL. How many grams of H2SO4 are
contained in 1.00 L of this solution?
25
Mass/Volume Percent(wt/v)
26
Mass /volume percent expresses the
concentration of solute as g solute
per ml solution.
g solutemass/volume percent = x 100
mL solution
27
Solve the mass/volume equation for mL of solution.
g solutemL solution = x 100
m/v percent
A 3.0% (by wt/vol) H2O2 solution is commonly used
as a topical antiseptic to prevent infection. What
volume of this solution will contain 10. g of H2O2?
2 210 g H O
x 100 =3.0 m/v percent
mL solution = 330 mL
g solutemass/volume percent = x 100
mL solution
28
Volume Percent (vol./vol.)
29
volume of liquid in questionvolume percent = x 100
total volume of solution
The volume percent is the volume of a
liquid per total volume of solution.
30
Molarity
31
Molarity of a solution is the number of
moles of solute per liter of solution.
number of moles of solute molesmolarity = M = =
liter of solution liter
32
What is the molarity of a solution containing 1.4 mol
of acetic acid (HC2H3O2) in 250. ml of solution?
It is necessary to convert 250. mL to L since
molarity = mol/L.
mol molThe conversion is: = M
mL L
5.6 mol = 5.6 M
L
1.4 mol
250. mL
1000 mL =
L
33
volume = 600 mL
The data are:
How many grams of potassium hydroxide are required
to prepare 600. mL of 0.450 M KOH solution?
Convert: mL L mol g
0.450 molM =
L
The calculation is:0.450 mol
L
600 mL1 L
1000 mL
56.11 g KOH =
mol
15.1 g KOH
56.11 gmolar mass KOH =
mol
34
0.325 L
Calculate the number of moles of nitric acid in 325 mL
of 16 M HNO3?
Substitute the data given in the problem and solve:
moles = 316 mol HNO =
1 L
3 5.2 mol HNO
moles = liters x MUse the equation:
Convert: mL L
1 L(325 mL) = 0.325 L
1000 mL
35
2 3
1 L =
0.250 mol K CO
2 3mass K CO = 16.0 g
The conversion is:
What volume of 0.250 M solution can be prepared
from 16.0 g of potassium carbonate?
2 3 2 3Convert: g K CO mol K CO L solution
0.250 molM =
L2 3
138.2 gmolar mass K CO =
mol
The data are:
2 316.0 g K CO 2 3
2 3
1 mol K CO
138.2 g K CO
0.463 L
36
How many mL of 2.00 M HCL will react with 28.0 g
NaOH?
Convert: g NaOH mol NaOH
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(aq)
Step 1 Write and balance the equation for the
reaction.
Step 2 Find the number of moles of NaOH in 28.0 g
NaOH.
0.700 mol NaOH 28.0 g NaOH1 mol
=40.0 g
37
How many mL of 2.00 M HCL will react with 28.0 g
NaOH?
Step 3 Solve for moles and volume of HCl.
0.700 mol NaOH1 mol HCl
1 mol NaOH
1 L HCl =
2.00 mol HCl
0.350 L HCl
Convert: mol NaOH mol HCl L HCl mL HCl
350 mL HCl0.350 L HCl1000 mL
=1 L
1
38
How many mL of 0.5000 M H2SO4 will be required to
completely react with 31.60 g Al(OH)3? The formula
weight of Al(OH)3 is 78.01 g/mol.
Answer
1215 mL
3H2SO4(aq) + 2Al(OH)3(aq) → Al2(SO4)3 (aq) + 6H2O(aq)
39Note: volumes are not always additive.
40
Introduction to pH and pOH(Please see pages 363 - 365 of Ch. 15)
41
pH is the negative logarithm of the hydrogen
ion concentration.
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
[ ] = molarity
pOH is the negative logarithm of the hydroxide
ion concentration.
42
Calculation of pH and pOH
43
What is the pH of a solution with a [H+] of 1.0 x 10-11?
pH = - log(1.0 x 10-11)
pH = 11.00
pH = - log[H+]
44
What is the pOH of a solution with a [OH-] of
1.0 x 10-3 and what is the pH of this solution?
pOH = - log[1.0 x 10-3] = 3.00
pH + pOH = 14.00
pH = 14.00 – pOH = 14.00 - 3.00 = 11.00
pOH = -log[OH-]
45
What is the pH of a 0.250 M HClO4 solution?
pH = - log(0.250) = 0.60
pH = - log[H+]
0.250 mol HClO4 x
L
1 mol. H+
1 mol HClO4
= 0.250 M H+
HClO4 H+ + ClO4-2
46
What is the pH of a 0.250 M Ca(OH)2 solution?
pOH = - log(0.500) = 0.30
pH + pOH = 14.00
pOH = - log[OH-]
0.250 mol Ca(OH)2 x
L
2 mol. OH+
1 mol Ca(OH)2
= 0.500 M. OH-
Ca(OH)2 Ca+2 + 2OH-
pH = 14.00 - 0.30 = 13.70
pH + 0.30 = 14.00
47
Dilution and Concentration Problems
48
• If a solution is diluted by adding pure
solvent:
– the volume of the solution increases.
– the number of moles of solute remain the
same.
– the ratio of mol./L gets smaller.
– the molarity decreases.
49
• If a solution is concentrated by
removing the solvent via evaporation:
– the volume of the solution decreases.
– the number of moles of solute remain the
same.
– the ratio of mol./L gets larger.
– the molarity increases.
50
Calculate the molarity of a sodium hydroxide solution
that is prepared by mixing 100. mL of 0.20 M NaOH
with 150. mL of water. Assume volumes are additive.
Step 1 Calculate the moles of NaOH in the original
solution.
0.020 mol NaOH 0.100 L0.20 mol NaOH
=1 L
molM =
Lmol = L M
0.020 mol is the number of moles of NaOH in the
original 100. mL of the 0.20 M NaOH solution.
51
Calculate the molarity of a sodium hydroxide solution
that is prepared by mixing 100. mL of 0.20 M NaOH
with 150. mL of water. Assume volumes are additive.
Step 2 Solve for the new molarity.
New solution volume = 100. mL + 150. mL = 250. mL
0.020 mol NaOH
0.250 L= 0.080 M NaOH
52
53
Quiz 10
1. How many moles of CO2 are present in 9.55L at 45°C and 752
torr? [5 pts.]
2. How many grams of CO2 are present in 9.55L at 45°C and 752
torr? See question 1 above. [5 pts.]
3. How many grams of BaCl2 are present in 75.0 g of a 12.0 % by
mass solution of BaCl2? [5 pts.]
4. How many milliliters of a 0.250M solution of H2SO4 would be
needed to completely react with 25.5 mL of 0.750 M NaOH?
Given:
H2SO4 + 2 NaOH Na2SO4 + 2 H2O
[5 pts.]
54
Midterm 3 Topic’s Review
1. Electron Configurations
2. Valence Shells and Valence electrons.
3. Full Octet
4. Periodic Trends as per Rule of Thumb
a. atomic radii
b. ionic radii
c. electronegativity
d. ionization energy
5. Bond Classifications: ionic or covalent
6. Lewis Structures and VSEPR
7. Molecular Shapes and Polarity
8. PV = nRT (i.e. STP and non-STP conditions)
9. % by Mass
10. Molarity