Title: Aqueous Solutions
1- Aqueous Solutions
- Chemical equilibrium - law of mass action
- For the general equation aA bB zZ
xX - species molar concentration of dissolved
species involved in the reaction - partial pressure of a gas in atmospheres of
gaseous reactants - 1 if the species is a pure liquid, pure
solid or solvent for a dilute solution - At equilibrium, the rate of the forward reaction
rate of reverse reaction - The position of the equilibrium is independent of
how it is established - Example for the equilibrium 1A 1B
2- Aqueous Solutions
- Chemical Equilibrium - Le Châteliers Principle
- When a stress is imposed on a system at
equilibrium the system will respond in such a
way as to relieve the stress. - Addition of a reactant
- Change in pressure or volume of a gas
- Change in temperature
- Types of equilibria
- Dissociation of water
- 2H2O H3O OH-
- H2O H OH-
3- Aqueous Solutions
- Types of equilibria
- Dissociation of water
- Example pure water
- 2H2O H3O OH-
- H3O OH-
- Example An aqueous solution 0.100 M NaOH
- 2H2O H3O OH-
- H3O
OH-0.100-H3O - Kw H3OOH- H3O(0.100-H3O)
- if H3O ltlt 0.100
- H3O x 0.100 1.00x10-14
- H3O 1.00x10-13
- OH- 0.100 1.00x10-13 0.100
4- Aqueous Solutions
- Types of equilibria
- Brønsted-Lowry acids and bases
- An acid is a substance that donates H to a
proton acceptor - A base is a H acceptor from a proton donor
- Conjugate acid - base pairs
- acid1 base1 acid2 base2
- base2 is the conjugate base of acid1
- Example
- acid1 base1 acid2 base2
- H2O NH3 NH4 OH-
conjugate pair
conjugate pair
OH- is the conj. Base of the acid H2O and NH4 is
the conj. acid of the base NH3 H2O is the conj.
Acid of the base OH- and NH3 is the conj. base of
NH4
5- Aqueous Solutions
- Types of equilibria
- Brønsted-Lowry acids and bases
- Example
- acid1 base1 acid2 base2
- HAc H2O H3O Ac-
- Ac- is the conj. base of Hac and H3O is the
conj. acid of H2O - H2O is the conj. base of H3O and HAc is the
conj acid of Ac- - Note H2O was an acid in the previous example
- Water is amphiprotic.
- Other amphiprotic substances
- HCO3- H2O H2CO3 OH-
- HCO3- H2O CO32- H3O-
6Aqueous Solutions Relationship between Ka and Kb
for conjugate a acid - base pair HAc H2O
H3O Ac- Ac- H2O HAc OH-
2H2O H3O OH- KwH3OOH-
The same result is obtained for a weak base in
water
7Aqueous Solutions Amine bases
8- Aqueous Solutions
- Calculate the H in an aqueous solution of a
weak acid or base - Example what is H in a 1.00 M NH4Cl solution
- NH4 H2O NH3 H3O
- Assume the 2nd equilibrium is suppressed
- by the 1st equilibrium
- NH3H3O
2H2O H3O OH-
KwH3OOH-
9- Aqueous Solutions
- Calculation of H for a solution of a weak acid
or base often leads to a quadratic equation as
in previous problem - The assumption that H ltlt CHA simplifies the
algebra and arithmetic - An error in the calculated H results
- See the table and figure showing errors in
calculated H as a function of Ka and CHA in
FAC7 - Sometimes the problem requires solution to a
complex equation in one unknown - ax3 bx1/2 c 0
- Use the method of successive approximations - or
Mathcad
10- Aqueous Solutions
- Types of equilibria
- Slightly soluble inorganic solids
- Assume dissolved solid is 100 dissociated into
ions - PbI2(s) Pb2(aq) 2I-(aq)
1
Stoichiometry demands I- 2Pb2 at
equilibrium Find the solubility of PbI2 in
concentration terms, how many moles PbI2
dissolves in 1 L of solution - find the C of
PbI2 in moles per liter of solution Solubility
S Pb2 since I- 2Pb2 at
equilibrium Pb2I-2 Pb2(2Pb2)2
7.1x10-9 4Pb23 7.1x10-9 Pb2 1.2x10-3
S
11- Aqueous Solutions
- Types of equilibria - slightly soluble inorganic
solids - Common ion effect - dissolve a solid in a
solution containing one of the ions common to
the solid - Example What is the solubility of PbI2 in a
solution that is 0.100 M in KI?
PbI2(s) Pb2(aq) 2I-(aq) Ksp
Pb2I-2 7.1 x 10-9 S Pb2 I- 0.100
2Pb2 Pb2(0.100 2Pb2)2 7.1 x
10-9 assume 2Pb2 ltlt 0.100 Pb2(0.0100) 7.1
x 10-9 Pb2 7.1 x 10-7 S Check assumption
2 x Pb2 1.42 x 10-6 ltlt 0.100
12- Aqueous Solutions
- Types of equilibria
- Complex ion formation
- Ag 2NH3 Ag(NH3)2
- Fe3 SCN- FeSCN2
- Oxidation - reduction equilibria
- 5Fe2 MnO4- 8H 5Fe3 Mn2 4H2O