• In acid-base titrations/reactions, are dealing with neutralization reactions (Use NmN Tables→ water is not a reactant)

○ N: neutralization reaction

○ M: mmols

■ In ice table, subtract number of mmols of limiting reactant

○ N: numbers

• Millimoles:

○ Ex: 40 mL x 0.2 M = 8 mmol

■ To convert back into concentration, do mmol/mL

• Halfway point/halfway equivalence point: half of the analyte has been neutralized

○ At halfway point pH = pKa; molarity of weak acid and conjugate base is equal; buffer is most effective

• Buffer Zone: region between the initial and eq. point; includes the halfway point
• What if you combine equal amounts of weak acid and weak base?

○ Solution will be acidic if Ka (of conjugate acid) is larger than Kb (of conjugate base) b/c more H3O+ ions are produced than OH- ions

Strong Acid-Strong Base Titrations

• They both dissociate completely

○ Strong acids: molarity of acid = molarity of H+

○ Strong Base: molarity of base = molarity of OH-

• The net ionic reaction for a strong acid–strong base titration is:
• Don’t need to do double ICE table at equilibrium (only NmN)
• pH is equal to 7 at the eq. point (only for strong + strong)

Weak Acid-Strong Base Titration

• The weaker the acid being titrated, the smaller the vertical area around the equivalence point
• pH at equivalence is above 7
• At 0 mL base added: use WMX ICE table, use Ka to determine pH
• Weak acid before eq point:

○ NmN ICE table first

○ Then Henderson-Hasselbalch

• At eq point: Have to do double ICE tables (only at the eq point)

○ NmN ICE table first → will have conjugate base left over

○ Then, WMX ICE table, use Kb to find [OH-]

• pH after equivalence:

○ NmN ICE table first → leftover strong base → do mmol/mL to find moles of [OH-] → find pOH- → find pH

Weak Base-Strong Acid Titration

• Process is practically the same with some differences
• pH at equivalence is below 7
• At 0 mL acid added:

○ WMX ICE table, use Kb to determine pOH, then pH

• Weak base before equivalence point:

○ NmN ICE table first → then Henderson-Hasselbalch (will have to use Kb to solve for pKa)

• At equivalence point:

○ NmN ICE table first → will have conjugate acid left over

○ Then do WMX ICE table → use Ka to find [H+]

• pH after equivalence:

○ NmN ICE table first → leftover strong acid → do mmol/mL to find moles of [H+] → find pH

Titration Curves

Strong Acid-Strong Base Titration

• Before the eq point: → Have left over strong acid (H+) which is why the pH is below 7

• For strong + strong only, the pH at eq. point is exactly 7
• Everywhere beyond equivalence point: have excess OH- which is why the pH is above 7

Weak Acid-Strong Base Titration

Graph:

At eq point only the conjugate base is present which is why the eq point is above 7

○ Justify: b/c this reaction  takes place

Weak Base-Strong Acid Titration

• In buffer zone 25mL is halfway point pH = pKa

○

• At eq point only the conjugate acid is present which is why the eq point is below 7

○ Justify: because this reaction takes place

• Beyond eq point have left over H+

Sketching a Titration Curve

• Need to plot 3 data point:
1. Starting PH
2. pH at equivalence
3. pH at the halfway point
• If given pkB, can solve for pKa to find pH