Solutions Notes (Chapter 16)
I. The Solution Process (Solvation)
Solvation is the process by which a solute dissolves in a solvent.
Miscible: when solutes and solvents are soluble in each other (solvation occurs)
Immiscible: when solutes and solvents are not soluble in each other (solvation does not occur)
What happens when:
Ionic molecules are the solute / Polar covalent molecules are the solute / Nonpolar covalent molecules are the soluteand water is the solvent?
/ and water is the solvent? / and water is the solvent?Dissociation of ionic molecules occurs (ions separate). Water is then attracted to the positive and negative ions. When all molecules have been “surrounded” the molecule is called hydrated.
Miscible / Dissociation does NOT occur. Water is polar and its “oppositely charged poles” will be attracted to other polar molecules' “oppositely charged poles.” When a solution is made between two polar molecules it is called molecular solvation.
Miscible / A solution will NOT occur. Water and any nonpolar molecule will not mix! Think of putting water and oil together. Water is polar and oil is nonpolar. The polar water is not attracted to the oil, because the nonpolar oil does not have any oppositely charged poles! Immiscible
II. Like Dissolves Like
We don’t always use water as the solvent! Solutions can be made from various substances – a rule of thumb to follow when trying to determine if two substances will form a solution is “like dissolves like.”
Polar Molecules + Polar Molecules Nonpolar Molecules + Nonpolar Molecules
Ionic and Ionic Ionic + Polar Molecules
Polar Molecules + Nonpolar Molecules Ionic Molecules + Nonpolar Molecules
III. Solubility - Now that we can figure out what we can mix to make a solution, how do we know how much solvent and how much solute to use?
Solubility: the maximum amount of a substance that will dissolve in a solvent (at a specific temperature)
According to solubility, solutions can be either:
a. unsaturated – a solution that is able to dissolve more solute (not enough)
b. saturated – a solution that cannot dissolve any more solute (just enough)
c. supersaturated – a solution that contains more solute than can be dissolved (too much!!)
The solubility of substances varies widely. For example 0.189 grams of Ca(OH)2 dissolves in 100 grams of water at 0°C. 122 grams of AgNO3 dissolves in 100 grams of water at 0°C.
IV. Factors Effecting Rate and Solubility
A. Factors Effecting Rate:
1. Agitation – stirring or mixing the solution will increase the rate at which the solute dissolves,
2. Surface Area – increasing the surface area of the solute increases the rate at which it dissolves
3. Temperature – increasing temperature will increase the rate or how fast the solute
B. Factors Effecting Solubility:
1. Increasing Temperature - increases the solubility of a solid solute in a liquid solvent
2. Decreasing Temperature-increases the solubility of a gaseous solute in a liquid solvent.
3. Pressure – The solubility of a gas increases as the pressure of the gas above the liquid increases.
V. Electrolyte VS Nonelectrolyte
1. Electrolyte – compounds that conduct an electric current in an aqueous solution OR in the molten state. An electrolyte solution contains charged particles (ions), which can move. Any salt dissolved in water is an electrolyte: NaCl, KI, etc.
Types of Electrolytes
1. Strong electrolytes – a large portion of the solute exists as ions:
a. aqueous solutions of all ionic compounds
b. strong acids: an acid that ionizes (separates into ions) completely in water.
HClO4, H2SO4, HNO3, HI, HBr, HCl
c. strong bases: a base that is present entirely as ions (one of which is OH-).
LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
2. Weak electrolytes – these are solutions in which only a small portion of the solute exists as ions
a. weak acids: an acid that only partially ionizes in water (assume any acid not listed above as a strong acid is a weak acid) -most weak acids have less than 2 oxygen's per hydrogen
b. weak bases: a base that only partially ionizes in water
-most Group I and II hydroxides are strong bases and the rest are weak (except for Be(OH)2
2. Non-Electrolytes- compounds that do NOT conduct electricity in either aqueous solution or when melted:
a. distilled water
b. gases
c. molecular compounds (2 nonmetals)
d. organic compounds – alcohols, sugars, etc. (anything containing a Carbon)
Practice Problems: Tell whether each of the following aqueous solutions would be a STRONG, WEAK, or NON electrolyte.
1. NaCl
2. CH3Br (l)
3. HMnO4
4. HC2H3O2
5. LiOH
6. HC6H7O6
7. CO2 (l)
8. HF
VI. Concentration: the concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution.
1. Molarity: the number of moles of solute in one liter of solution.
M = amount of solute (moles)
OR grams of solute/molar mass of solute
volume of solution ( liters) L of solution
A bottle labeled 6M HCl is pronounced 6 molar HCl and it was prepared by mixing 6 moles of HCl with enough water to make 1 liter of solution. Molarity is always moles/liter. Ex: 6M is 6 moles/ liter
Molarity is probably the MOST IMPORTANT unit of concentration that we work with in chemistry. Knowing the correct technique for preparing molar solutions is extremely important. A volumetric flask MUST be used.
Practice Molarity Problems:
1. Calculate the molarity, M, of a solution prepared by dissolving 11.85 g of potassium permanganate in enough water to make 750. mL of solution.
2. Calculate the mass of NaCl needed to prepare 175 ml of 0.500 M saline solution.
3. Calculate the volume (in mL) needed to prepare a 2.48 M sodium hydroxide solution containing 31.52 g of the dissolved solid.
2. Molarity of Ions in solution – Most ionic solids when dissolved in water, ionize (break apart into ions)
Ex: CaCl2 (aq) à Ca2+ + 2 Cl-
Na3PO4 à 3 Na+ + PO43-
Mg(OH)2 à Mg2+ + 2 OH-
Example Molarity of Ions Problems: Calculate the molarity of the ions in the following solutions:
1. 0.25 M calcium phosphide 2. 2.0 M Chromium (III) chloride
3. 0.25 M barium hydroxide 4. 0.55 M aluminum nitride
3. Dilutions – you need to make 800.0 mL of a 0.25 M solution of HCl. The only available HCl is concentrated (12 M). How would you do this? Being able to prepare dilutions is a very common application of chemistry. Our department buys concentrated acids, but normally uses more dilute solutions of these acids in our labs. Therefore, it is important to know how to correctly dilute.
M1V1 = M2V2 where M1 = concentrated solution V1 = volume of concentrated solution
M2 = dilute solution V2 = volume of dilute solution
In the above problem, M1 = 12.0 M M2 = 0.25 M Vl = ? V2 = 800.0 ml therefore
(12.0 M) (V1) = (0.25 M)(800 mL) and V1 = 16.67 ml.
This 16.67 mL is the amount of concentrated acid we will take out, but we still have to make 800 mL of 0.25 M. The other 783.3 mL of solution must be water (800 – 16.67). Therefore we would place 16.67 mL of HCl into 783.3 mL of water and we would have our diluted solution.
Practice Dilutions Problems:
l. How would you prepare 485 mL of 0.39 M solution of NaCl when a l.0 M solution of NaCl is all you can find?
2. If 300.0 mL of a 2.5 M solution of nitric acid is added to 500.0 mL of water, what is the molarity of the dilute solution?
3. Prepare 500. mL of a dilute solution (0.50 M ) of nitric acid from the l5.0 M stock solution. You will need 600. mL of the dilute solution.
4. Percent Solutions (2 types):
A. Percent mass or %(m/m)– used when a solid solute is dissolved in liquid, usually water.
% (m/m) = grams of solute
grams of solution
Example: Prepare a 10.00 % NaCl solution using 50.0 g water and solid salt:
0.l0 = x / 50 + x and x = 5.67 g NaCl in 50 grams of water
1. How many grams of water must be added to 25.0 g salt in order to have a 2.00 % (by mass) salt solution?
2. Prepare 600.0 g of a 3.00 % saline solution (NaCl solution).
B. Percent volume or %(v/v) – used when a liquid solute is mixed with a liquid solvent. The units are mL or L, but are worked the same.
% (v/v) = mL solute
mL solution
Example: Prepare a 20.00 % alcohol solution using 400.0 mL of water:
0.20 = x / 400 + x and x = 100 mL alcohol + 400 mL water
1. What is the percent (v/v) of ethanol in the final solution when 90.0 mL of it are diluted to a volume of 300. mL with water?
5. Molality: is the concentration of a solution expressed in moles of solute per kilogram of solvent. Molality is represented by a lower case m.
molality = amount of solute (moles) OR molality = grams of solute/molar mass
mass of solvent (kg) kg of solvent
5m NaOH is pronounced 5 molal NaOH solution.
5m = 5 moles of NaOH
1 kg of water 5m or 5moles/kg
Example Molality Problems:
1. Calculate the molality of a solution prepared by dissolving 5.0l g sodium sulfate in 700.0 g water.
2. Prepare a solution that is 2.50 molal barium nitrate in 1500. grams of water.
3. A solution is prepared by dissolving 3.00 g of potassium chromate in 58.5 g of water. Calculate the molality of the solution.
4. How many kg of water must be added to 8.3 g of oxalic acid, H2C204, to prepare a 0.050 m solution?
VII. Colligative Properties of Solutions - Characteristics of a solution that do NOT depend on size or type of solute particles, but only upon the concentration of particles. The physical properties of solutions are different from those of the pure solvent. Colligative properties depend on the number of ionized particles dissolved in a given mass of solvent:
NaCl à Na+ + Cl- 2 particles
Na2SO4 à 2Na+ + SO42- 3 particles
(NH4)3PO4 à 3NH41+ + PO43- 4 particles
What do you notice about the solutes that contained polyatomic ions?
Organic and molecular molecules do not ionize so they stay as one particle.
Practice Problems: How many particles will the following solutions become?
1. magnesium chloride 2. calcium bromide 3. aluminum sulfate
4. iron (III) nitrate 5. ethanol (C2H5OH) 6. dinitrogen pentoxide
B. Types of Colligative Properties
1. Boiling Point Elevation (BPE) – some solutes increase the point at which a solution boils
Example: salt added to water increases the boiling point of the water
This occurs because in water there are fewer particles to “energize” or boil, once you add salt the number of particles increases and more energy must be added to break the intermolecular forces that hold the molecules together.
Another explanation: Solutions have a lower vapor pressure than the pure solvent. Because the vapor pressure is decreased, more kinetic energy is required to raise the vapor pressure to atmospheric pressure; therefore the boiling point of a solution is higher than the boiling point of the pure solvent.
2. Freezing Point Depression (FPD) – some solutes decrease the point at which a solution freezes
Example: salt added to water decreases the freezing point of the water
Occurs because in water there are fewer particles to freeze, once you add salt the number of particles increases and more energy must be given off (lost) until the solution freezes. (Intermolecular forces are allowed to become stronger as more energy is lost that is why energy is lost during freezing.)
Another explanation: When a solute is present, the orderly pattern that the pure solvent normally takes when freezing is disrupted, so more kinetic energy must be lost from the solution for it to solidify (freeze). This lowers the freezing point of the solution.
Factors that affect the amount which the boiling point or freezing point changes are:
1. Molality of the solution
2. Whether the solution is ionic or molecular (# of particles)
3. The boiling point and freezing point constants
Colligative Property Constants
Solvent / Boiling Point (°C) / Kb (° C/m) / Freezing Point ( °C) / Kf (°C/m)Water / 100 / 0.5l2 / 0 / l.86
Benzene / 80.l / 2.53 / 5.48 / 5.12
Acetic Acid / 118.l / 3.07 / 16.6 / 3.90
Nitrobenzene / 210.88 / 5.24 / 5.70 / 7.00
Phenol / 182 / 3.56 / 43 / 7.40
C. Formulas for solving boiling point elevation or freezing point depression problems:
1. BPE = (molality)(Kb)(# particles of solute)
2. FPD = (molality)(Kf)(# particles of solute) Both BPE and FPD are always positive.
Molality = moles of solute OR Molality = grams solute/molar mass of solute
kg of solvent kg of solvent
D. Formulas to find freezing point or the boiling point are:
1. BP solution = BP pure solvent +BPE
2. FP solution = FP pure solvent - FPD
Practice Problems:
1. What is the boiling point of a solution in which 75.00 grams of sodium sulfate are dissolved in 800.0 grams of water? Kb = 5.12 °C/m
1st solve for BPE: BPE = grams solute/molar mass of solute (Kb) (# particles)
kg of solvent
BPE = 75.00 g/142 g/mol (5.12°C/m) (3) = 10.1°C
0.8000 kg
2nd solve for the boiling point of the solution: BP solution = BP pure solvent +BPE