Acid/Base Chemistry
Strong acids: HCl, HBr, HI, HClO4, HClO3, HNO3, H2SO4
Ionize ~100%
Ka is very large essentially undefined
Conjugate base is weaker than water so ineffective in solution
[strong acid] initial = [H+] final
Strong bases: readily soluble salts with hydroxide (OH-)
LiOH, NaOH, KOH, Ca(OH)2, Ba(OH)2
Ionize ~100%
Kb is very large essentially undefined
Conjugate acid is water
Weak acids: acids that are not strong are said to be weak (HA)
Ionize << 100% (Ka < 1)
As Ka increases the acid strength increases while the conjugate base
strength decreases
Conjugate base is stronger than water so effective but a weak base
Weak bases: bases that are not strong are said to be weak (A-)
Ionize << 100% (Kb < 1)
As Kb increases the base strength increases while the conjugate acid
strength decreases
Conjugate acid is stronger than water so effective but a weak acid
Concepts:
[H+] = [OH-] for neutral solutions or pH = pOH or pH = 7 (at 25°C)
[H+] > [OH-] for acidic solutions or pH < pOH or pH < 7 (at 25°C)
[H+] < [OH-] for basic solutions or pH > pOH or pH > 7 (at 25°C)
As a solution gets more acidic the [H+] increases causing the pH to
decrease
Equations:
pH = -log[H+] [H+] = 10-pH pOH = -log[OH-] [OH-] = 10-pOH
pH + pOH = pKw = 14 (at 25°C)
[H+][OH-] = Kw = 1 x 10-14
(at 25°C)
(Ka)(Kb) = Kw = 1 x 10-14 (at 25°C and only for conjugate acid/base pairs)
pKa = -logKa
pH = pKa + log([A-]/[HA]) (for buffer solutions)
pH = pKa + log (n A-/n HA) (for buffer solutions)
Reactions Scenarios:
1. Acids:
HA  H+ + Aa. If strong use stoichiometry to solve for [H+]
b. If weak use Ka to solve for x, work in M (x = [H+])
2. Weak base:
A- + H2O  HA + OHUse Kb to solve for x, work in M (x = [OH-])
3. Strong acid and strong base: H+ + OH-  H2O (100%)
Work in mols – determine limiting reactant
After reaction is complete covert excess reagent into M in order
to get the pH or pOH
4. Strong base and weak acid:
OH- + HA  H2O + A- (100%)
Work in mols – determine limiting reactant
After reaction is complete:
a. if you only have A- flip reaction, get M and use Kb to solve for x
b. if you have excess OH- get M and find pOH then pH
c. if you have HA and A- (buffer solution) use equation
pH = pKa + log([A-]/[HA]) or pH = pKa + log (n A-/n HA)
5. Strong acid and weak base:
H+ + A-  HA (100%)
Work in mols – determine limiting reactant
After reaction is complete:
a. if you only have HA flip reaction, get M and use Ka to solve for x
b. if you have excess H+ get M and find pH
c. if you have HA and A- (buffer solution) use equation
pH = pKa + log([A-]/[HA])
No Reaction Scenarios:
1. 2 acids
a. if 2 strong acids you need to get the [H+]total
b. if strong acid and weak acid the [H+] will be determined by the strong
acid only (weak acids ionize very little normally but even less in the
presence of the strong acid)
2. 2 bases
a. if 2 strong bases you need to get [OH-]total
b. if strong base and weak base the [OH-] will be determined by the strong
base only
3. Buffer solutions