- 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:
- Starting PH
- pH at equivalence
- pH at the halfway point
- If given pkB, can solve for pKa to find pH