Bogna Haddad – Unit 3 Review, 2007.
SOLUTIONS AND SOLUBILITY
1. Concentrations
When talking about concentration we are talking about the amount of solute (what is dissolved, ex. salt) dissolved in a solvent (what is doing the dissolving, ex. water), to form a solution.
There are several ways to express the concentration of solutions. Regardless of the details, concentration is always expressed as the quantity of solute in a quantity of solution. Some of the common ways to show concentration are:
a) percent
mass percent = mass of solute x 100%
mass of solution
- remember both masses must have the same units! ex. g
volume percent = volume of solute x 100%
volume of solution
- again, both volumes have to be in the same
units, ex. mL
mass/volume percent = mass of solute x 100%
volume of solution
b) ppm, ppb
- used to represent the concentrations of solutions with very small amounts of solute
i) parts per million, ppm
- 1 part in 106
ppm = mass of solute (g) x 106
mass of solution (g)
ii) parts per billion, ppb
- 1 part in 109
ppb = mass of solute (g) x 109
mass of solution (g)
c) molar concentration
- the number of moles of solute in litres (L) of solution
- also referred to as “molarity”, M
molar concentration = amount of solute (mol)
volume of solution (L)
2. Properties of Solutions
a) Types of solutions
There are nine possible types of solutions:
/ solid / liquid / gas
solid / alloys, ex. brass / ex. brine: salt in water / ex. moth balls: naphtalene in air
liquid / ex. dental amalgam / ex. antifreeze: ethylene glycol in water / ex. humidity: water in air
gas / hydrogen in platinum / ex. carbonated water: CO2 in water / ex. air: oxygen in nitrogen
(gases ALWAYS mix with each other)
b) Water as the Universal Solvent
- water is a polar solvent and is known as the universal solvent since it is able to dissolve many more substances than any other solvent
- water acts to dissolve both molecular and ionic substances through intermolecular dipole interactions: it can dissociate (ionize) other substances by overcoming their intramolecular forces of attraction, as well as being able to hydrogen-bond with many substances
c) Aqueous Solutions
- these are solutions that involve substances in water
- there are three most common types of aqueous solutions:
i) electrolytes
- usually formed of highly soluble ionic compounds (formula units) that dissociate in water
- conduct electricity
ii) non-electrolytes
- usually formed of covalent compounds (molecules) that do not dissociate in water or ionic non-soluble formula units
- do not conduct electricity
iii) acids/bases/neutral solutions
- solutions that contain different ratios of ions in them
- aqueous solutions can be
i) unsaturated
- a solution that will dissolve more solute
ii) saturated
- a solution that will not dissolve more solute
- a solution that is saturated with respect to one solute can dissolve other solutes
iii) supersaturated
- formed when a saturated solution is heated to solubilize (dissolve) more solute
d) Predicting Solubility
solubility
- a physical property that describes the amount of solute that will dissolve in a given quantity of solvent at a specific temperature
- determined by intermolecular forces between the solute and the solvent
- recall: intermolecular forces are forces of attraction between molecules: hydrogen bonds are the strongest, then dipole-dipole interactions and London dispersion forces
- in order for a substance to dissolve, the forces between particles in the solute must be broken and new ones must be formed between the solute and the solvent:
- the solubility of compounds will depend on their dipole: the more polar (or ionic) something is the more likely it will be soluble in polar substances.
Two other factors to consider are:
i) the number of charges: compounds that have more charges (and therefore are in a stronger ionic bond) tend to be less soluble.
ii) the size of the ions involved: since small ions bind closer together than large ions, they tend to be less soluble
The general rule of thumb is that like dissolves like:
- polar compounds will dissolve in polar solvents, ex. water and sugar
- dipole-dipole interactions and hydrogen bonding increase solubility
- nonpolar compounds will dissolve in nonpolar solvents, ex. esters in oil
This gives rise to miscible and immiscible mixtures:
- when two liquids dissolve in each other they are said to be miscible, ex. ethanol and water
- when two liquids do not dissolve in each other they are said to be immiscible, ex. oil and water
Other factors affecting solubility
i) molecular size
- smaller substances dissolve easier
ii) temperature
- increase in temperature gives more energy for intermolecular bonds to be broken, therefore the higher the temperature the more soluble a solid or liquid solute becomes
- since gases expand when heated, the higher the temperature the less soluble a gas solute becomes
iii) pressure
- affects gases more than solids or liquids: the higher the pressure the more soluble a gas solute becomes
3. Reactions in Solution
When a soluble ionic compound is dissolved in a solution, the compound dissociates into individual ions:
ex. MgSO4 (s) --> Mg2+ (aq) + SO4 2- (aq)
In a double displacement reaction two solutions of ionic compounds are mixed, causing the cations (metals) to displace one another and anions (nonmetals) to displace one another:
ex. in solution, silver nitrate reacts with sodium chloride to form sodium nitrate and silver chloride
Double displacements can result in
- formation of a precipitate
- formation of a gas
- formation of water
a) Double Displacements that Produce a Precipitate
Reactions where a precipitate (a solid) forms can be written out as total and net ionic equations. Recall:
- total ionic equations show all the ions present in and around the reaction
- net ionic equations show only the ions that participate in the reaction (result in the formation of a precipitate), the remaining ions are referred to as spectator ions
To write ionic equations
write the chemical formulas of the compounds involved
write the complete balanced chemical equation for the reaction, using solubility rules to identify if a precipitate forms
write the total ionic equation: write the soluble ionic compounds as dissociated ions
identify and cross out spectator ions (these are the same on both sides)
write the net ionic equation out of the remaining ions
ex. In solution, sodium sulfide is mixed with iron (II) sulphate. Write the balanced total and net ionic equation for this reaction.
Unknown substances can often be identified using double displacement reactions and the formation of precipitates in a qualitative analysis. Once a precipitate is obtained, its mass can be measured in a gravimetric analysis.
4. Stoichiometry in Solutions
Recall, stoichiometry involves calculating the amounts of reactants and products in chemical reactions. Thus, you can use stoichiometry to determine the amount of ions that react. You can use the same basic steps that you learned in Unit 2 to solve stoichiometric problems involving solutions!
General Guidelines for Stoichiometry in Solutions:
convert mass or concentrations to moles
write the net ionic equation or chemical equation
use coefficients of balanced equation to determine the number of moles of required substance
if this is a limiting reagent problem identify the limiting reagent
calculate the mass or concentration of required substance
example 1: Finding the minimum volume of precipitate
An aqueous solution that contains silver ions is usually treated with chloride ions to recover silver chloride. What is the minimum volume of 0.25 mol/L magnesium chloride, MgCl2(aq), needed to precipitate all the silver ions in 60 ml of 0.30 mol/L silver nitrate, AgNO3 (aq)? Assume that silver chloride is completely insoluble in water.
SOLUTION:
convert concentrations to moles of AgNO3
write a chemical equation for the formation for the reaction
use coefficients of balanced equation to determine number of moles of MgCl2 required
calculate the volume of MgCl2 needed
example 2: Limiting Reagent
Mercury salts have a number of important uses in industry and in chemical analysis. Because mercury compounds are poisonous, however, the mercury ions must be removed from the waste water. Suppose that 25.00 ml of 0.085mol/L aqueous sodium sulphide is added to 56.5 ml of 0.10 mol/L mercury (II) nitrate. What mass of mercury(II) sulphide, HgS(s) precipitates?
SOLUTION:
convert mass or concentrations to moles
write net ionic equation or chemical equation
use coefficients of balanced equation to determine number of moles of required substance
If limiting reactant problem: identify the limiting reactant
Calculate the mass or concentration of required substance
example 3: The concentration of ions
Calculate the concentration (mol/L) fo chloride ions in each solution.
- 19.8 g of potassium chloride dissolved in 100 ml of solution
- 26.5 g of calcium chloride dissolved in 150 ml of solution
- a mixture of the two solutions in parts (a) and (b), assuming that the volumes are additive.
SOLUTION:
convert mass into moles for each solution
write the equations for the dissociation of the substance
use coefficients of balanced equation to determine number of moles of Cl- present
Calculate the concentration of each solution
Calculate the concentration of Cl- ions in the final solution
Solution / KCl / CaCl2Molar mass / 39.10 + 35.45 = 74.55 g / 40.08 + (2 x 34.45) = 110.98 g
Amount (mol) / 19.8 g/ 74.55 g/mol = 0.266 mol / 26.5 g / 110.98 g/mol = 0.239 mol
Dissociation equation / KCl à K+ + Cl- / CaCl2 à Ca2+ + 2 Cl-
Amount of Cl- / 0.266 mol x 1 mol Cl- = 0.266 mol
1 mol KCl / 0.239 mol x 2 mol Cl- = 0.478 mol
1 mol CaCl2
Concentration of Cl- / 0.266 mol = 2.66 mol/L
0.100 L / 0.478 mol = 3.19 mol/L
0.150 L
The concentration of Cl- of each solution is 2.66 mol/L and 3.19 mol/L respectively. The concentration of Cl- when mixed is 2.98 mol/L
example 4: The mass Percent of Ions
The leaves of a rhubarb plant contain a relatively high concentration of oxalate ions, C2O42-. Oxalate ions are poisonous, causing respiratory failure. To determine the percent of oxalate ions, a student measured the mass of some leaves. Then the student ground up the leaves and added excess calcium chloride solution to precipitate calcium oxalate. The student tested 238.6 g of leaves. The dried mass of calcium oxalate was 0.566 g. What was the mass percent of oxalate ions in leaves?
SOLUTION::
Convert mass to moles of calcium oxalate
write net ionic equation for the formation of calcium oxalate
use coefficients of balanced equation to determine number of moles of oxalate ions
Calculate the mass of oxalate ions
Calculate mass percent of oxalate ions in rhubarb
5. Acids and Bases
Acids and bases are electrolytes, just like salts. Recall, an electrolyte is a substance that dissociates into ions in water.
a) Properties of acids and bases
ACIDS / BASES- sour taste
- good conductors of electricity
- pH less than 7
- turn blue litmus paper red
- react with metals to produce hydrogen gas / - bitter taste
- good conductors of electricity
- pH more than 7
- turn red litmus paper blue
- do not react with metals
- feel slippery
b) Acid/base Theories
There are two theories of acids and bases:
i. Arrhenius theory
acid: a substance that dissociates in water to produce hydrogen ions, H+
ex. HBr (aq) à H+ (aq) + Br- (aq)
base: a substance that dissociates in water to produce hydroxide ions, OH-
ex. LiOH (aq) à Li+ (aq) + OH-(aq)
limitations
a) does not explain all compounds, ex. NH3 is a base, Al(NO3)3 is an acid
b) only describes reactions in water
ii. Bronsted-Lowry theory (this is the preferred theory)
acid
- a proton (H+) donor
- for a substance to behave as an acid (ie. donate a proton) the proton must be involved in a significantly polar bond
ex. HCl (aq) + H2O (l) --> Cl- (aq) + H3O+ (aq)
- water, in this reaction a base, accepts a proton to form ammonium, a
conjugate acid
- HCl, in this reaction an acid, donates a proton to form chloride, a
conjugate base
base
- a proton (H+) acceptor
- for a substance to be a base it has to have an available pair of electrons to allow it to bind the proton
ex. NH3 (aq) + H2O (l) --> NH4+ (aq) + OH- (aq)
- ammonia, in this reaction a base, accepts a proton to form ammonium, a
conjugate acid
- water, in this reaction an acid, donates a proton to form hydroxide, a
conjugate base
In an acid-base reaction a base becomes a conjugate acid and an acid becomes a conjugate base.
amphoteric substance
- a substance that acts as a base in one reaction yet acts as an acid in another
- in the two reactions above water could be either a base or an acid