1. Solutions
L. Breen
Chemistry 1405

2. A solution is a system in which one or more substances (solute) are homogeneously mixed or dissolved in another substance (solvent).
The solute is the component that is dissolved – that is, the least abundant component of the solution.
The solvent is the dissolving agent or the most abundant component in the solution.

3. Examples of Solutions

4. Solubility
When one substance (solute) dissolves in another (solvent) it is said to be soluble
salt is soluble in water
bromine is soluble in methylene chloride
When one substance does not dissolve in another it is said to be insoluble
oil is insoluble in water
The solubility of one substance in another depends on two factors – nature’s tendency toward mixing, and the types of intermolecular attractive forces

5. Effect of Temperature on the Solubility of a Solid in a Liquid
For most solids dissolved in a liquid, an increase in temperature results in increased solubility.
Some solids increase in solubility only slightly with increasing temperature.
Some solids decrease in solubility with increasing in temperature.

6. As temperature increases, so does the solubility of a solid in a liquid.

7. Solubility Curves large increase in solubility with temperature
slight increase in solubility with temperature
decrease in solubility with increasing temperature

8. Temperature and Solubility of Gases in Liquids
As the temperature goes up, the solubility of gases decreases.
The extra energy from the heat provides enough energy for the gas molecules to break the intermolecular forces holding them in solution allowing them to escape.

9. large decrease in solubility with increasing temperature

10. Pressure Effects
Solubility of a gas in a liquid is also a function of the pressure of the gas.
The higher the pressure, the more molecules that are close to the liquid’s surface and the greater the solubility of the gas.
The lower the pressure, the fewer molecules of gas are close to the liquid’s surface and the lower the solubility.

11. Pressure and solubility
Solids and liquids are not influenced by changes in pressure the same way as gases.
Increase in pressure results in an increase in the solubility of gases.
That’s why soda is packed under pressure.

12. Terms that describe the extent of solubility of a solute in a solvent:
very soluble – greater than 10 g/100g solvent
Soluble – 1-10 g/100 g solvent
slightly soluble – g/100 g solvent
Insoluble – less than 0.1 g/100g solvent

13. methyl alcohol and water
Terms that describe the solubility of liquids:
miscible: liquids that are capable of mixing and forming solutions.
methyl alcohol and water
immiscible: liquids that are insoluble in each other.
oil and water

14. The general rule for predicting solubility is “like dissolves like”.
Polar compounds tend to be more soluble in polar solvents than nonpolar solvents.
NaCl (sodium chloride) is
soluble in water
slightly soluble in ethyl alcohol
insoluble in ether and hexane
Solvent Polarity

15. p. 572 of 10th edition, 9th edition page 533.
Nonpolar compounds tend to be more soluble in nonpolar solvents than in polar solvents.
benzene is:
insoluble in water
soluble in ether
Solvent Polarity
On your next test, you will be expected to know how to use the solubility rules in Table 13.4,
p. 572 of 10th edition, 9th edition page 533.
Table will be provided.

16. Intermolecular forces play a roll in determining solubility.
If the solute-solute interactions, solute-solvent interactions and the solvent-solvent interactions exhibit the same types of intermolecular forces and of similar strength, then the substance will be form a homogeneous mixture – a solution

17. At a specific temperature there is a limit to the amount of solute that will dissolve in a given amount of solvent.
When this condition occurs the solution is said to be saturated.

18. Dissolving a Solid in a Liquid
Solution is saturated.
Solution is unsaturated. Solute dissolves
34 g KCl
35 g KCl
30 g KCl
25 g KCl
5 g KCl
15 g KCl
10 g KCl
20 g KCl
No more KCl can dissolve.
100 g H2O
20oC

19. Dissolving a Solid in a Liquid
A solution that is saturated at one temperature may not be saturated at another temperature.
Raise temperature
35 g KCl
Solute dissolves The solution is unsaturated
It may increase or decrease depending on the solute-solvent system.
Solubility of a solute in a solvent changes with temperature.
100 g H2O
50oC
20oC

20. Dissolving a Solid in a Liquid
A stress to the system will cause the solute in excess of the saturation limit to come out of solution.
Whenever a solution contains solute in excess of its solubility limit the solution is supersaturated.
A supersaturated solution is unstable.
Cool Solution to 20oC
35 g KCl
At 20oC the solubility of KCl in water is 34 g/100 g H2O.
100 g H2O
The solution contains 1 gram of KCl in excess of the solubility limit.
No KCl precipitates.
20oC
50oC

21. What affects the rate of dissolution?
Particle Size
Decreasing particle size increases the rate of dissolution (more surface area)
Temperature
In most cases, the rate of dissolving of a solid increases with temperature.
This occurs because solvent molecules strike the surface of the solid more frequently and harder, causing the solid to dissolve more rapidly.

22. Agitation or stirring.
When a solid is first put into water, it comes in contact only with water. The rate of dissolving is then at maximum.
Stirring distributes the dissolved solute throughout the water, more water is in contact with the solid, causing it to dissolve more rapidly.
As the solid dissolves, the amount of dissolved solute around the solid increases and the rate of dissolving decreases.

23. Many solids must be put into solution to undergo appreciable chemical reactions.

24. NaCl(s) + AgNO3(s) → no reaction
If the reactants are in the solid phase no reaction occurs.

25. NaCl(s) + AgNO3(s) → no reaction
NaCl(s) and AgNO3(s) do not react because their ions are securely locked in their crystal lattices.

26. If the reactants are dissolved in water an immediate reaction occurs.
Na+
Cl-
Ag+
If the reactants are dissolved in water an immediate reaction occurs.

27. NaCl(aq) + AgNO3(aq) → AgCl(s) +NaNO3(aq)
Mobile Ag+ and Cl- ions come into contact and form insoluble AgCl which precipitates.

28. NaCl(aq) + AgNO3(aq) → AgCl(s) +NaNO3(aq)
Na+ and remain in solution.

29. Concentrations Solutions have variable composition
To describe a solution, you need to describe the components and their relative amounts
The terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution
A dilute solution contains a relatively small amount of dissolved solute.
A concentrated solution contains a relatively large amount of solute.
Concentration = amount of solute in a given amount of solution
occasionally amount of solvent

30. Solution Concentration
Mass percent is the mass of solute divided by the total mass of solution multiplied by 100 (to put the value in terms of percentage).

31. 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) = 8.00 g
grams of solvent (H2O) = 50.0 g

32. Mass-volume percent is the mass of solute (in grams) divided by the total volume of solution (in mL) multiplied by 100

33. Solve the mass/volume equation for grams of solute.
A 3.0% m/v H2O2 solution is commonly used as a topical antiseptic to prevent infection. What volume of this solution will contain 10 g of H2O2?
Solve the mass/volume equation for grams of solute.
m/v percent

34. The volume percent is the volume of solute divided by the total volume of solution multiplied by 100.

35. Parts per million (ppm) can be expressed as 1 mg of solute per Kg of solution.
Parts per billion (ppb) are 1 g of solute per kilogram of solution.

36. Remember: mass of solute must be in the same units as mass of solution
The concentration of sodium fluoride, NaF, in a town’s fluoridated tap water is found to be 32.3 mg of NaF per 20.0 Kg of tap water. Express this NaF concentration in (a) ppm (m/m) and (b) ppb (m/m).
Mass of solute
Mass of Solution
ppm (m/m) =
x 106
Remember: mass of solute must be in the same units as mass of solution
3.23 x 10-2 g
2.00 x 104 g
ppm (m/m) =
x 106
= 1.62 ppm (m/m)

37. Remember: mass of solute must be in the same units as mass of solution
The concentration of sodium fluoride, NaF, in a town’s fluoridated tap water is found to be 32.3 mg of NaF per 20.0 Kg of tap water. Express this NaF concentration in (a) ppm (m/m) and (b) ppb (m/m).
Mass of solute
Mass of Solution
ppb (m/m) =
x 109
Remember: mass of solute must be in the same units as mass of solution
3.23 x 10-2 g
2.00 x 104 g
ppb (m/m) =
x 109
=1615 ppb
=1620 ppb

38. Solution Concentration Molarity
Moles of solute per 1 liter of solution
Used because it describes how many molecules of solute in each liter of solution

39. Molarity can be calculated from grams of solute by converting to moles of solute.
Molar mass = g
mole
Molarity =
(g solute)
(1 mole)
(molar mass g)
Liter of solution
moles
liter =

40. Molarity =
(g solute)
(1 mole)
(molar mass)
Liter of solution
moles
liter =

41. What is the molarity of a solution containing 1
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.

42. Calculate the number of moles of nitric acid in 325 mL of 16 M HNO3?
moles = liters x M
Use the equation:
Substitute the data given in the problem and solve:

43. Always follow the moles:
Dilution is the process of adding more solvent to a solution to lower its concentration.
It is common practice to have a “stock” solution of a specific concentration and dilute samples of it to meet your needs.
Always follow the moles:
250.0 mL 5.00 M NaCl
20.00 mL sol’n removed
20.00 mL sol’n then diluted to mL
5.00 mol NaCl
1 L sol’n
* L sol’n
= mol NaCl removed from the stock sol’n to be diluted.
Mdil =
0.100 mol NaCl
0.250 L
Mdil = M

44. Na+
NaNO3 solution

45. Moles of solute remain the same.
Na+
Solution volume is doubled.
Solution concentration is halved.
Moles of solute remain the same.

46. Another way of understanding this is: MstockVstock = MdilVdil
or: M1V1 = M2V2 V1 M1 V2 M2
250.0 mL 5.00 M NaCl
20.00 mL sol’n removed
20.00 mL sol’n then diluted to mL
5.00 mol NaCl
1 L sol’n
* L sol’n
= mol NaCl removed from the stock sol’n to be diluted.
Mdil =
0.100 mol NaCl
0.250 L
Mdil = M

47. 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.
V1 = 100. mL
M1 = 0.20 M
V2 = 250. mL
M2 = unknown
(100. mL)(0.20 M) = (250. mL)M2

48. moles of solute
kilograms of solvent
molality (m) =

49. The molar mass of CH3OH = 32.04 g/mol
What is the molality (m) of a solution prepared by dissolving 2.70 g CH3OH in 25.0 g H2O?
The molar mass of CH3OH = g/mol