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Titrations are lab techniques used to determine the concentration of an unknown acid or base by reacting it with a standardized solution (titrant) of known concentration. The titration curve plots pH vs. volume of titrant added, revealing key points like the equivalence point (where moles of acid = moles of base) and the half-equivalence point (where pH = pKa for weak acids). This topic is high-yield on the AP exam—expect multiple-choice questions on curve shapes, FRQs on calculations, and lab-based analysis. Real-world example: Antacids (like Tums) are titrated with stomach acid (HCl) to determine their effectiveness in neutralizing excess acidity.
Goal: Find the concentration of an unknown acid/base. Steps:1. Write the balanced neutralization reaction (e.g., HCl + NaOH-NaCl + H?O).2. Use the titrant’s molarity and volume to find moles of titrant added at the EP. - mol = M × L (convert mL to L first!).3. Use stoichiometry to find moles of analyte (1:1 ratio for monoprotic acids/bases).4. Calculate analyte concentration: M = mol / L (use analyte volume in L).
Example: 25.0 mL of NaOH (unknown M) is titrated with 0.100 M HCl. EP is reached at 30.0 mL HCl. Find [NaOH]. - Moles HCl = 0.100 M × 0.0300 L = 0.00300 mol. - Moles NaOH = 0.00300 mol (1:1 ratio). - [NaOH] = 0.00300 mol / 0.0250 L = 0.120 M.
Goal: Predict the shape of a pH vs. volume curve. Steps:1. Identify acid/base strengths (strong/weak?).2. Initial pH: - Strong acid: pH-1–2. - Weak acid: pH-3–5 (use Ka to calculate). - Strong base: pH-12–13. - Weak base: pH-9–11 (use Kb to calculate).3. Buffer region (weak acid/base only): Flat part before EP; pH-pKa at half-EP.4. Equivalence point pH: - Strong-strong: pH = 7. - Weak acid-strong base: pH > 7 (conjugate base hydrolyzes water). - Strong acid-weak base: pH < 7 (conjugate acid hydrolyzes water).5. After EP: pH approaches titrant’s pH (e.g., excess NaOH-pH ~13).
Example: Sketch the curve for 50.0 mL of 0.10 M CH?COOH (Ka = 1.8 × 10) titrated with 0.10 M NaOH. - Initial pH: Use Ka to find [H?] = ?(Ka × [HA]) = ?(1.8 × 10 × 0.10)-pH-2.87. - Half-EP (25 mL NaOH): pH = pKa = 4.74. - EP (50 mL NaOH): pH > 7 (CH?COO? hydrolyzes water). - Curve starts at pH ~2.87, flat buffer region, EP at pH ~8.7, then rises sharply.
Goal: Find pH at initial, half-EP, EP, and post-EP. Steps:1. Initial pH (weak acid/base): - Weak acid: Ka = x² / [HA]?-solve for x = [H?]. - Weak base: Kb = x² / [B]?-solve for x = [OH?], then pOH-pH.2. Half-EP (weak acid/base): - pH = pKa (Henderson-Hasselbalch: [A?] = [HA]-log(1) = 0).3. EP (weak acid-strong base): - Find [conjugate base] = moles HA / total volume. - Use Kb = Kw / Ka to find [OH?] = ?(Kb × [A?]). - Convert to pH.4. Post-EP: - Calculate excess titrant moles-[OH?] or [H?]-pH.
Example: Calculate pH at EP for 25.0 mL of 0.10 M HCN (Ka = 6.2 × 10?¹?) titrated with 0.10 M NaOH. - Moles HCN = 0.0250 L × 0.10 M = 0.0025 mol. - Volume at EP = 25.0 mL (1:1 ratio)-total volume = 50.0 mL. - [CN?] = 0.0025 mol / 0.0500 L = 0.050 M. - Kb = Kw / Ka = 1 × 10?¹? / 6.2 × 10?¹? = 1.6 × 10. - [OH?] = ?(Kb × [CN?]) = ?(1.6 × 10 × 0.050) = 8.9 × 10 M. - pOH = 3.05-pH = 10.95.
Correction: Only strong acid-strong base titrations have EP at pH 7. Weak acids/bases shift EP pH due to hydrolysis.
Mistake: Forgetting to convert mL to L in molarity calculations.
Correction: Always convert volumes to liters (e.g., 25 mL = 0.025 L) before using M = mol / L.
Mistake: Using Ka instead of Kb (or vice versa) for conjugate species.
Correction: At EP for weak acid-strong base, use Kb = Kw / Ka for the conjugate base.
Mistake: Ignoring the buffer region in weak acid/base titrations.
Correction: The half-EP is critical—pH = pKa here, and the curve flattens due to buffering.
Mistake: Rounding intermediate values (e.g., pH calculations).
Using Henderson-Hasselbalch to find pH in buffer regions.
Multiple-choice traps:
Dilution effects: Adding water doesn’t change moles of acid/base, but it lowers concentration (affects pH calculations).
Tricky distinction:
Endpoint vs. equivalence point: Endpoint is where the indicator changes color; EP is where moles are equal. They’re not always the same (but should be close!).
Lab-based questions:
A 20.0 mL sample of 0.10 M HNO? is titrated with 0.10 M KOH. What is the pH at the equivalence point? (A) 1.0 (B) 3.0 (C) 7.0 (D) 11.0
Answer: (C) 7.0. Strong acid-strong base titration-EP pH = 7.
A student titrates 30.0 mL of a weak acid (HA) with 0.20 M NaOH. The equivalence point is reached after adding 25.0 mL of NaOH. (a) Calculate the molarity of the weak acid. (b) At the half-equivalence point, the pH is 4.7. What is the Ka of the weak acid?
Answer: (a) Moles NaOH = 0.20 M × 0.0250 L = 0.0050 mol. [HA] = 0.0050 mol / 0.0300 L = 0.167 M. (b) At half-EP, pH = pKa = 4.7-Ka = 10·? = 2.0 × 10.
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