Solution Chemistry
1.
What components make up a solution?
a. Solute – substance that is dissolved (present in a smaller quantity).
b. Solvent – substance that does the dissolving (present in greater
quantity).
2.
Ways to describe a solution
a.
b.
c.
d.
3.
An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by
mass. The density of the solution is 1.05 g/cm3. Calculate the
molality, molarity, and mole fraction of ethylene glycol.
In this case the solute is the ethylene glycol and the solvent is water.
We can use the percent by mass of the ethylene glycol o determine the
moles of ethylene glycol and kg of solvent.
Remember that mass percent is:
For ease, we will assume that we have 100 g of solution. This makes
our lives easier because this means that we have 40.0 g of ethylene
glycol.
The important part to this question is realizing that in molality the
denominator is terms of solvent only and in mass% the denominator is
interms of the whole solution.
This means that if we assume a 100g solution and 40g of it were
solute. The remaining 60g are solvent. Leaving us with the following
fraction:
Now we turn to molarity:
I will start with the same initial point as for molality. The key difference
here is that in molarity, the denominator is terms of the entire solution.
Lastly we will solve for the mole fraction:
In this case we will use the percent by mass information to solve for
the moles of ethylene glycol and the moles of water. Our assumption
will still be that we are dealing with 100g of solution.
Now take this information and plug into the mole fraction equation:
4.
Adage used to determine if a solvent will dissolve a solute
“Like Dissolves Like”
This means that polar substances will dissolve polar substances and
non-polar substances will dissolve non-polar substances. However,
polar and non-polar cannot dissolve each other (think oil and water).
5.
Which solvent, water or carbon tetrachloride, would you choose to
dissolve each of the following.
To do these problems you need to remember that like dissolves like.
SO we will want to stick non polar substances in the non-polar solvent
(CCl4) and the polar/ionic compounds in the polar solvent (H2O).
a. KrF2
Linear Non-polar CCl4
b. SF2
Bent polar H2O
c. SO2
Bent Polar H2O
d. CO2
Linear Non-polar CCl4
e. MgF2
Ionic H2O
f. CH2O
Polar C=O bond, symmetry doesn’t cancel Polar H2O
g. CH2CH2
All non polar bonds Non-polar CCl4
6.
What are the three steps to solution formation
When we create a solution there are 3 specific steps that must occur in
order for the solute to dissolve into the solvent.
a. Expanding the Solute (i.e. breaking up the solute)
This means that if you have NaCl going into water. The first step
would be separating the Na+ from the Cl-
b. Expanding the Solvent (i.e. overcoming intermolecular forces in the
solvent)
This means that the water (that the NaCl was dumped into) has to
break up some of its H-bonds (etc.) to make room for the Na+ and
Cl- ions.
c. Solution Formation
The solute and the solvent come together.
7.
Which steps typically require energy?
Expanding the solute and expanding the solvent typically requires
energy. The solution’s formation typically gives off energy.
8.
What is and how do you determine
a. Enthalpy of Hydration
This the sum of the enthalpies of expanding the solvent and
forming the of the solution.
b. Enthalpy of Solution
This is the sum of enthalpies for each of the three steps in the
process.
9.
Generally, ∆Hsol’n values are positive for ionic substances dissolved in
water.
10.
Why does a solution that has a positive ∆Hsoln, like NaCl, proceed
spontaneously at room temperature?
Remember that the reaction would look like:
NaCl(s) Na+*(aq) + Cl-(aq)
For this reaction we are going from a solid component to two aqueous
componenets. This means that there would be a positive change in
entropy.
+∆S is balancing out the +∆H. Remember the formula, ∆G = ∆H - T∆S
As long as ∆G comes out negative the reaction will be spontaneous.
11.
Explain the following information
The reason for these flipped results is that when, in the case of LiF,
salvation occurs the water molecules become more structured around
the Li+ creating a new “molecule” : [Li(H2O)4 ]+
So instead of taking a solid and breaking it up into 2 aqueous pieces.
We take take 5 particles (1 Li+ and 4 H2O molecules and combine them
into one rigid structure). Hence an overall loss in entropy.
From this we can glean that:
When the charge density (area over which the anion/cation charge is
spread) is increased the hydration effect is increased as well…
meaning we will have a greater potential of creating a rigid structure as
the charge density increases and therefore having an overall decrease
in the entropy.
We can reconcile the data given as follows:
Li+ and F- are smaller than K+ and Cl-, respectively. This they would
have a greater charge density.
Ca2+ and S2- have higher charges than K+ and Cl- spread over
relatively similar areas. This would, once again, lead to an overall
higher charge density.
12.
Which ion would more strongly hydrated in the following pairs.
a. Na + or Mg2+
More positive charge over similar area = greater charge density
b. Mg2+ or Be2+
Smaller area for given positive charge = greater charge density
c. Fe2+ or Fe3+
Greater charge over smaller area = greater charge density
d. F- or BrSmaller area for given negative charge = greater charge density
e. Cl- or ClO4Smaller area for given negative charge = greater charge density
f. ClO4- or SO42Smaller area for given negative charge = greater charge density
13.
What sign does entropy have when benzene is added to water? Why?
-∆S due to the “cage formation”. Because benzene is nonpolar and
water is polar (meaning there would be a repulsion of sorts between
the two substances), water creates these “pockets” in which the
benzene is sectioned off in – thereby minimizing the interactions. This
caging off of the benzene does require water to become more ordered
however which would cause a decrease in entropy.
14.
Define
a. Hydrophilic – Literally this means, water loving. These are soluble
polar/ionic substances that dissolve in water.
b. Hydrophobic – This means water fearing. Nonpolar substances
would fit into this category.
15.
How does an increase in pressure/temperature affect the solubility of
a. Solid –
An increased pressure has no effect. (The amount of water vapor
above the solution that would condense is neglible).
Picture a water solution that is oversaturated with NaCl –
oversaturated meaning that a clump of NaCl would exist at the
bottom of the container. If I were to press down on the solution with
a piston (increasing the pressure) the excess salt would not be
more inclined to dissolve.
Generally speaking, an increase in temperature will increase the
solubility of a solid. Remember that this is general, and not true for
every ionic compound.
b. Gas –
An increase in pressure would Increase the solubility of a gas.
If the area above a is decreased, then there would be less area for
the gas to exist and so more particles would have to dissolve.
An increase in temperature would gas the gas to be less soluble.
Consider a glass if soda left out on a counter to get warm… it would
go flat. The increase in temperature (due to heat flow from the
environment) would impart more kinetic energy to the gas particles
in the solution… more kinetic energy permits greater evaporation
(i.e. solubility decreases).
16.
Henry’s Law
a. When is it valid?
If the solution is dilute and there is no reaction between the solute
and solvent.
b. What relationship does it establish?
The amount of gas dissolved in the solution is directly proportional
to the pressure of the gas above the solution.
17.
Calculate the solubility of N2 in water when the partial pressure of of
nitrogen above water is 1.10 atm at 0°C.
(kH = 962 Latm/mol)
This is a Henry’s Law question. Because we were given
concentration, we will use the formula P = kHC. This question literally
boils down to plug and chug.
18.
True or False.
a. Increasing the temperature will increase solubility of a substance.
False. This is not always true. It depends on the signs of ∆H and
∆S.
b. Increasing the temperature increases the speed at which a
substance dissolves.
True. There may be more or less that dissolves at the raised
temperature; but the amount that does dissolve will do so more
quickly.
c. The solubility of a gas will increase with an increase in temperature.
False.
19.
What is Vapor Pressure?
This is the pressure above a solid or liquid due to evaporation.
20.
Can a solute affect the vapor pressure of a solvent?
Yes.
21.
What do solutes do to the vapor pressure of the solvent?
Lower it. Please note that it is the vapor pressure of the solvent is
lowered, not necessarily the vapor pressure for the overall solution
22.
How does it lower vapor pressure of solvent?
Consider:
Less of the solvent is able to escape due to it’s path for evaporation
being blocked by the new solute particles. Less particles escaping
means lowered vapor pressure.
23.
What are two categories of solute?
a. Non-volatile Solute
This type of solute has no tendency to escape from the solution into
the vapor phase – meaning it makes no contribution to the vapor
pressure above the solution.
i. What is Raoult’s Law for this type of solute?
b. Volatile Solute
This type of solute does have a tendency for evaporation so it will
contribute to the overall vapor pressure above the solution.
i. What is Raoult’s Law for this type of solute?
24.
What type of solution obeys Raoult’s Law?
Ideal Solutions. This is a type of solution where the solute and solvent
have similar attractions. Meaning they are not strongly attracted to or
repelled by one another.
a. How does an ideal solution differ from a non ideal solution?
In an ideal solution there is no interaction between molecules. It
would be like adding water to water.
b. How do these interactions affect the predictions of Raoult’s Law?
If there are strong interactions, the IMFs are higher than for the
pure substances. This means that the molecules, being highly
attracted to one another, are less likely to escape (vapor pressure
decreases). These kinds of interactions are exothermic and said to
have a, “negative deviation”, from Raoult’s Law predictions.
If there are weak interactions, the IMFs are lower than for the pure
substances – sometimes even repulsive. This makes it easier for
molecules to escape (vapor pressure increases). These kinds of
interactions are endothermic and said to have a “positive deviation”
from Raoult’s Law.
c. Label which of the following graphs obeys Raoult’s Law, which has
strong exothermic interactions and which has endothermic
interactions that cause a deviation from Raoult’s Law.
d. Give an example of each type of interaction
i. Benzene and Toluene = Ideal
ii. CH3CH2OH + C6H14 = endothermic
(polar and nonpolar combination)
iii. C3H6O + H2O = exothermic
(polar and polar combination)
25.
What is a colligative property?
This is a property that is affected by the addition of a solute. These
properties, in particular, depend on the number, not identity, of solute
particles in an ideal solution.
26.
What are three examples of colligative properties?
a. Boiling Point Elevation
b. Freezing Point Depression
c. Osmotic Pressure
27.
How does adding a solute affect the boiling point of a solvent?
It increases the boiling point. Remember that in order to boil the vapor
pressure of the liquid must equal the vapor pressure of the
atmosphere. Because adding in a solute drops the vapor pressure
down, it takes more heat energy to reach the same final vapor
pressure – and a higher temperature is required to reach the boiling
point.
28.
How does adding a solute affect the freezing point of a substance?
It lowers the freezing point. Remember that freezing depends on the
IMFs between particles. Once a solute has been put into the solvent, it
disrupts the typical IMFs experienced within the pure solvent and thus,
makes it harder for the substance to solidify – necessitating lower
temperatures to freeze.
29.
What is osmotic pressure?
The pressure that just stops osmosis. As a reminder, osmosis is the
spontaneous migration of a solvent from a less concentrated solution
to a more concentrated solution.
a. What does isotonic mean?
This occurs when two substances have identical osmotic
pressures.
30.
What is the Van’t Hoff Factor?
Our colligative properties are affected by the total number of particles
present in the solution. The Van’t Hoff factor takes into account the
number of particles present in solution if the compound dissolved
breaks into ions.
Technically
Generally, i = number of ions in a formula
If the compound does not break into ions (like ethanol CH3CH2OH) it
has an i value equal to one.
31.
What is the equation for determining the
a. Boiling Point Elevation
b. Freezing Point Depression
c. Osmotic Pressure
32.
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of
a solution made by adding 164 g of glycerin to 338 mL of H2O at
39.8°C? The vapor pressure of pure water at 39.8°C is 54.74 torr and
its density is 0.992 g/cm3.
This is a Raoult’s Law problem – we can identify this as it is looking at
the affects of a solute on the vapor pressure of the solute.
We are told that it is a nonvolatile solute, this means that we will use:
First we will determine the mole fraction of water in the solution.
Plugging this value into the formula we get:
33.
At a certain temperature the vapor pressure of pure benzene (C6H6) is
0.930 atm. A solution was prepared by dissolving 10.0 g of a nondissociating, nonvolatile solute in 78.11 g of benzene at that
temperature. The vapor pressure of the solution was found to be
0.900 atm. Assuming that the solution behaves ideally, determine the
molar mass of the solute.
Once again, based on the information present we know we are dealing
with a Raoult’s Law problem.
As this is a nonvolatile solute we will be dealing with
From the information provided we are able to provide χsolvent..
Remember that the mole fraction is equal to:
We can determine the moles of solvent present in the solution using
the initial mass given:
Plugging what we know into the mole fraction equation:
Given that we know that there were 10.0g of solute added to the
solution we can solve for the molar mass of the solute.
34.
When pure methanol is mixed with water, the solution gets warmer to
the touch. Would you expect this solution to be ideal? Why or why
not?
No. We would expect this solution to have a negative deviation from
an ideal sol’n as it was an exothermic reaction. In an exothermic
reaction the solute-solvent interactions are strong.
Ideal solutions have a ∆Hsol’n = 0 (i.e. no net heat flow)
35.
Given
Which of the following statements is false concerning solutions of A
and B?
a. The solutions exhibit negative deviations from Raoult’s Law.
True.
b. ∆Hmix for the solutions should be exothermic.
True – negative deviation = exothermic.
c. The intermolecular forces are stronger in solution than in either
pure A or B.
True – negative deviation indicates stronger interactions between
solute and solvent.
d. Pure liquid B is more volatile than pure liquid A.
True. Higher PBo vapor pressure compared to PAo.
e. The solution with χB = 0.6 will have a lower boiling point than either
pure A or pure B.
False. Lower vapor pressure means higher boiling point.
36.
A solution is prepared by dissolving 27.0 g of urea [(NH2)2CO], in 150.0
g of water. Calculate the boiling point of the solution. Urea is a
non-electrolyte. kb=0.51 (°C kg)/mol.
As this is a situation where a solute has been added to a solvent, we
are going to see a boiling point elevation and will need to calculate the
change in the temperature using
Because urea is a non electrolyte we know that i = 1. The only thing
we need to calculate, before plugging in, is the molality.
Plugging in all the data:
Thus the new boiling point would be:
37.
What mass of glycerin (C3H8O3), a nonelectrolyte, must be dissolved in
200.0 g of water to give a solution with a freezing point of -1.50°C?
Once again we are looking at how a colligative property, freezing point
depression. This means that we will be using:
In this case we have the change in temperature, i, and we can use the
molality to solve for the mass of glycerin.
Now converting to mass:
38.
Which of the following will have the lowest total vapor pressure at
25oC? Which has the highest vapor pressure at 25oC? At 25oC, the
vapor pressure of pure water is 23.8 torr.
a. Pure water.
b. A solution of glucose in water with χglucose = 0.01
If the mole fraction of glucose is 0.01. The mole fraction of
H2O = 0.99 (remember that the mole fraction has to add up to one)
c. A solution of sodium chloride in water with χNaCl = 0.01
This one is a little bit trickier. The mole fraction of water would be
0.99, as it was in the case above – thus calculating you would
determine the same vapor pressure using Raoult’s Law. The key
difference here is that the NaCl breaks up into 2 ions and glucose is
a non-electrolyte. This means that there are more particles
blocking the path for evaporation with the NaCl, thus it would have
lower overall vapor pressure.
d. A solution of methanol in water with χmethanol = 0.2
(at 25oC, the vapor pressure of pure methanol is 143 torr)
In this case, we are given a vapor pressure for methanol, this
means that it is a volatile solute. So we would have to use
Plugging in, we get:
Because methanol adds to the total vapor pressure – this solution
would have the highest total.
The final order would be:
39.
Consider the following
0.010 m Na3PO4 in water
i = 4 ion molality = 4 (0.010m) = 0.040 m
0.020 m CaBr2 in water
i = 3 ion molality = 3 (0.020m) = 0.060 m
0.020 m KCl in water
i = 2 ion molality = 2 (0.020m) = 0.040 m
0.020 m HF in water (HF is a weak acid)
i = 1 ion molality = 0.020 m
a. Assuming complete dissociation of soluble salts, which solution(s)
would have the same boiling point as 0.040 m C6H12O6 in water?
(non-polar electrolyte).
We are basically looking for which substances have the same
molality as 0.040 m C6H12O6.
Na3PO4 and KCl both have a total molality (in terms of ions) equal
to 0.040 m
Thus these two compounds would have the same boiling points.
b. Which solution would have the highest vapor pressure at 25°C?
Smallest amount of solute has the highest vapor pressure.
Thus the solution containing HF would have the highest overall
vapor pressure.
c. Which solution would have the largest freezing point depression?
The greater the concentration of solute particles the greater the
freezing point depression. This means that the CaBr2 solution
would have the lowest overall freezing point as it has the greatest
concentration of ions in solution.
40.
From the following :
pure water
solution of C6H12O6 (χ = 0.01) in water
i=1
solution of NaCl (χ = 0.01) in water
i=2
solution of CaCl2 χ = 0.01) in water
i=3
Choose the one with the following:
a. Highest freezing point
Pure water. Any added solute would drop the freezing point.
b. Lowest freezing point
The CaCl2 solution would have the lowest freezing point as this
solution would contain the greatest number of particles. The
greater the concentration of particles the lower the freezing point.
c. Highest Boiling Point
Once again, the CaCl2 solution would have the highest boiling
point as this solution would contain the greatest number of
particles. The greater the concentration of particles the higher the
boiling point elevation.
d. Lowest Boiling Point
Pure water. Any added solute would raise the boiling point.
e. Highest Osmotic Pressure
Again, the CaCl2 solution would have the highest osmotic pressure
as it has the greatest number of particles in solution. The osmotic
pressure would be higher because more solvent would want to shift
in order to compensate for the difference in concentration.
41.
Determine the boiling point for a solution of 20.0 g of NaCl and 40.0 g
CaF2 in 1.00 L of H2O. The Kb for water is 0.51 oC kg/mol and the
density equals 1.00 g/cm3.
For this problem we are dealing with a boiling point elevation. Thus we
will be using the formula:
The trick here is that there is not a clear i value because we have two
solutes. Remember that the whole point of the Van’t Hoff factor is to
change the molality of the overall solute into the molality of the ions.
Plugging into out formula: