Download Ch. 14 Study Guide

Chemistry 201: General Chemistry II - Lecture
Dr. Namphol Sinkaset
Chapter 14 Study Guide
Concepts
1. Solutions are homogeneous mixtures of two or more substances.
2. solute: substance present in smaller amount.
3. solvent: substance present in greater amount.
4. Liquid solutions are easiest to study because: (1) components are localized, (2) components mix well, and (3) they are easy to handle/transfer.
5. Solutions do not have to be liquid based.
6. solubility: amount of substance that will dissolve in a given amount of solvent.
7. Formation of a gas phase solution comprised of two noble gases is driven by entropy,
not energy.
8. entropy: measure of energy of randomization or energy dispersal in a system.
9. Formation of liquid solutions are driven by entropy and intermolecular forces.
10. In solution, there are three types of particle interactions.
11. “Like dissolves like” rule: when solute and solvent have similar intermolecular
forces, they tend to form a solution.
12. miscible: term used to describe two liquids that are soluble in all proportions.
13. The relationship between solvent-solute, solvent-solvent, and solute-solute interactions
determines when a solution will form. If solvent-solute interactions are about equal to
solvent-solvent and solute-solute interactions, entropy will decide whether a solution
forms.
14. Energy changes are observed when solutions form. The enthalpy of solution, ∆Hsoln ,
is the enthalpy associated with solution creation, and it can be estimated with a Hess’s
Law type calculation.
15. Formation of a solution can be broken up into three steps: (1) separation of solid into
its constituent particles, ∆Hsolute ; (2) separation of solvent particles to make room for
solute, ∆Hsolvent ; (3) mixing solute particles with solvent particles, ∆Hmix .
16. For ionic, water-based solutions, ∆Hsolvent and ∆Hmix can be combined into ∆Hhydration .
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17. heat of hydration: enthalpy change when 1 mole of gaseous solute ions are dissolved
in water.
18. For ionic compounds, ∆Hsolute = −∆Hlattice .
19. When excess solid is added to a solution, an equilibrium is set up between the dissolution and deposition processes. If an equilibrium is established, the solution is said to
be saturated.
20. If an equilibrium cannot be established, then the solution is unsaturated.
21. Supersaturated solutions hold more solute than is theoretically possible.
22. Generally, higher temperatures result in higher solubility of solids.
23. Henry’s Law states that if we push a gas hard enough (with external pressure), more
of it should dissolve in solution. Note however that higher temperatures generally
decrease gas solubility.
24. Since molarity varies with temperature, sometimes it’s necessary to express concentrations in temperature-independent units.
25. mass percent: ratio of mass of component to mass of solution times 100%.
26. molality (m): moles of solute per 1 kg solvent.
27. mole fraction (X): ratio of moles of component to total moles of solution.
28. ppm and ppb are parts by mass concentrations.
29. Raoult’s Law expresses how the vapor pressure of a solvent is affected when a nonvolatile solute is added.
30. For solutions of two volatile liquids, Raoult’s Law can be combined with Dalton’s Law
of Partial Pressures to get the total vapor pressure of the solution.
31. Ideal solutions obey Raoult’s law at all concentrations for both solute and solvent.
Typically ideal solutions have ∆Hsoln = 0. Physically, this means that the IM forces
between solute-solute and solvent-solvent are equal to the IM forces between solventsolute, i.e. a solute particle doesn’t feel more or less attraction to a solvent particle
and vice versa.
32. Negative deviations from Raoult’s Law are due to stronger forces of attraction in
solution than in the pure liquid.
33. Positive deviations from Raoult’s Law are due to weaker forces of attraction in
solution than in the pure liquid.
34. colligative properties: properties that depend on the amount of solute and not the
type of solute.
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35. Freezing point depression and boiling point elevation are two colligative properties that deal with changes in freezing and boiling temperatures as a function of
added solute.
36. osmosis: the flow of solvent from a solution of lower solute concentration to one of
higher solute concentration.
37. The pressure needed to stop the flow of solvent is called the osmotic pressure.
38. If the solute is an electrolyte (salt that can dissociate), then its effective molality is
higher.
39. The formation of ion pairs can lead to deviations from the expected freezing/boiling
point depression/elevation. Ion pairs are closely associated ions of opposite charge
that act as one species instead of completely dissociating.
40. van’t Hoff factor: a number that gives an idea of the amount of dissociation in an
electrolyte.
Equations
1. ∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix (enthalpy of solution calculation)
2. Sgas = kH Pgas at constant T (Henry’s Law)
◦
3. Psolution = Xsolvent Psolvent
(Raoult’s Law)
4. ∆Tf = mKf (freezing point depression)
5. ∆Tb = mKb (boiling point elevation)
6. Π = M RT (osmotic pressure)
moles particles in soln
7. i = moles
formula units dissolved (van’t Hoff factor)
8. ∆Tf = imKf (freezing point depression, for ionics)
9. ∆Tb = imKb (boiling point elevation, for ionics)
10. Π = iM RT (osmotic pressure, for ionics)
Representative Problems
68. In an aqueous solution of sulfuric acid, the concentration is 1.89 mole % of
acid. The density of the solution is 1.0645 g/mL. Calculate the following: (a)
the molal concentration of H2 SO4 , (b) the weight percent of the acid, and (c)
the molarity of the solution.
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Whenever you’re given mole percent or weight percent, the best starting point is the
assume that you have 100 units of it. If we do this for 1.89 mole % sulfuric acid, then it
means we have:
1.89 mole H2 SO4
1.89 mole H2 SO4
=
100 total moles
1.89 mole H2 SO4 + 98.11 mole H2 O
From here, it’s probably easiest to go the weight percent, which is mass H2 SO4 divided by
the total mass and then multiplied by 100%. To get there, we need to calculate the masses
from the moles that we got above.
1.89 mole H2 SO4 ×
98.0716 g
= 185.3 g
1 mole H2 SO4
98.11 mole H2 O ×
18.015 g
= 1767.4 g
1 mole H2 O
For (b), we have:
mass H2 SO4
× 100%
total mass
185.3 g H2 SO4
=
× 100%
185.3 + 1767.4 g
= 0.09489 × 100%
= 9.49%
weight % H2 SO4 =
For (a), we want to calculate the molality, which is mole per kg solvent. We already have
all the necessary numbers, so we just plug it in.
mole H2 SO4
kg H2 O
1.89 mole
=
1.7674 kg
= 1.07
m H2 SO4 =
For (c), we need to use the density. However, note that it’s the density of the sulfuric acid
solution, not just plain water. Therefore, we need to have total mass in the denominator.
Recall that molarity is in mole per L.
1.89 mole H2 SO4 1.0645 g 1000 mL
×
×
= 1.030 mole/L = 1.03 M
1952.7 g
1 mL
1L
72. Pentane (C5 H12 ) and heptane (C7 H16 ) are two hydrocarbon liquids present
in gasoline. At 20 ◦ C, the vapor pressure of pentane is 420 torr and the vapor
pressure of heptane is 36.0 torr. What will be the total vapor pressure (in torr)
of a solution prepared by mixing equal masses of the two liquids?
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This a Raoult’s Law and Dalton’s Law of Partial Pressures problem. First it says that
we’re mixing equal masses of the two liquids, so we need to pick something. I will choose a
total mass of 100.0 g which means that each component will have a mass of 50.0 g. Raoult’s
Law deals with mole fractions, so we need to calculate the mole fractions of each component.
We start by finding the moles of each component.
50.0 g pentane ×
1 mole pentane
= 0.693 mole pentane
72.15 g
50.0 g heptane ×
1 mole heptane
= 0.499 mole heptane
100.2 g
Now we calculate mole fraction of each.
mole pentane
total moles
0.693 mole
=
1.192 mole
= 0.581
Xpentane =
Since the mole fractions of all components must add up to one, we know that the mole
fraction of heptane must be 0.419 (1.000 - 0.581). Now we calculate the new vapor pressures
of the components using Raoult’s Law.
◦
Ppentane = Xpentane Ppentane
= (0.581)(420 torr)
= 244 torr
◦
Pheptane = Xheptane Pheptane
= (0.419)(36.0 torr)
= 15.1 torr
Now we use Dalton’s Law of Partial Pressures to get the total vapor pressure of the
solution.
Ptotal =
=
=
=
Ppentane + Pheptane
244 torr + 15.1 torr
259.1 torr
259 torr
81. A solution of 12.00 g of an unknown nondissociating compound dissolved
in 200.0 g of benzene freezes at 3.45 ◦ C. Calculate the molecular mass of the
unknown.
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We use the freezing point depression equation for this problem. Benzene freezes at
5.45 C and has a Kf of 5.07 ◦ C/m.
◦
∆Tf
5.45 C - 3.45 ◦ C
2.00 ◦ C
m
◦
=
=
=
=
Kf m
(5.07 ◦ C/m)m
(5.07 ◦ /m)m
0.394
Now, we need to find out how many moles of the compound exist in this solution. The
problem states that we have only 200.0 g of benzene, or 0.2000 kg.
0.2000 kg ×
0.394 mole compound
= 0.0788 mole compound
1 kg
Molecular mass is expressed in g/mole. We’re given that 12.00 g of the unknown was
added, and from out calculation above, this corresponds to 0.0788 mole.
12.00 g
0.0788 mole
= 152 g/mole
MW unknown =
89. What is the osmotic pressure, in torr, of a 2.0% solution of NaCl in water
when the temperature of the solution is 25 ◦ C? Assume the density of the solution
is 1.00 g/mL.
This is an osmotic pressure problem that involves an ionic compound. Colligative
property problems with an ionic compound involve using the van’t Hoff factor. Looking up
the van’t Hoff factor for NaCl, we see that i = 1.9. We plug in what we know into the
osmotic pressure equation.
Π = iM RT
= (1.9) M (0.082056 L·atm/K·mole) (298.15 K)
We are given the concentration in % mass, so we need to convert to molarity.
1 mole NaCl
1.00 g solution
1000 mL
2.0 g NaCl
×
×
×
= 0.34223 M NaCl
100.0 g solution 58.44 g NaCl 1.00 mL solution
1L
We substitute into the osmotic pressure equation we’ve started:
Π = (1.9) M (0.082056 L·atm/K·mole) (298.15 K)
= (1.9) (0.34223 M) (0.082056L·atm/K·mole) (298.15 K)
= 15.908 atm
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We finish the problem by converting to torr:
15.908 atm ×
760 torr
= 1.2 × 104 torr
1 atm
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