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 Br-

Smaller area for given negative charge = greater charge density

e. Cl- or ClO4

-

Smaller area for given negative charge = greater charge density

f. ClO4- or SO4

2-

Smaller 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 non-dissociating, 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 PB

o 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: