#Acid-Base Titrations: Your Ultimate Guide 🧪

Hey there, future AP Chem master! Let's tackle acid-base titrations together. It might seem like a beast, but we'll break it down step-by-step. Think of this as your go-to guide for the night before the exam. Let's get started!

#What's a Titration Anyway?

Titration is a lab technique used to find the unknown concentration of a solution (the analyte or titrand). We do this by adding a solution of known concentration (the titrant) using a burette. The key is to add the titrant until we reach the equivalence point.

#The Magic Equation

At the equivalence point, we use the equation: nMaVa = mMbVb

• n and m are the stoichiometric coefficients from the balanced equation.
• Ma and Mb are the molarities of the acid and base.
• Va and Vb are the volumes of the acid and base at the equivalence point.

#Titration Setup

Here's a visual of a typical acid-base titration setup:

• Analyte: The solution with the unknown concentration.
• Titrant: The solution with the known concentration, added from the burette.
• Equivalence Point: The point where the reaction is complete.

#Titration Curves: The pH Story 📈

As we add titrant, the pH of the solution changes. Plotting this change creates a titration curve.

#Strong Acid - Strong Base Titration

Let's look at the titration of 1M NaOH into 25 mL of 1M HCl:

#Pre-Titration

Before adding any NaOH, we have only HCl. The pH is calculated as: pH = -log[H+] = -log(1) = 0. #### Pre-Equivalence Point

As we add NaOH, the reaction is: H+ + OH- → H2O

• We still have excess H+, so the solution is still acidic.
• The pH slowly increases as we neutralize the H+.

#Equivalence Point

At the equivalence point, moles of HCl = moles of NaOH. For our example, this occurs at 25 mL of NaOH added.

• We have only water and salt (NaCl) at this point.
• The pH is 7 for strong acid-strong base titrations.

#Post-Equivalence Point

After the equivalence point, we have excess OH-, so the solution becomes basic, and the pH increases.

#Full Titration Curve

Here’s the full curve for a strong acid-strong base titration:

#Titrations with Weak Acids and Bases 🧪

Most acids and bases are weak, so let's look at those titrations. The process is similar, but with a few key differences.

#Buffers are Here!

Before the equivalence point, you'll have both the weak acid and its conjugate base (or weak base and its conjugate acid). This means we have a buffer! 💡

• The solution is less responsive to pH changes.
• We have a half-equivalence point at half the volume of the equivalence point where pH=pKa or pOH=pKb.

#Equivalence Point pH

Unlike strong acid-strong base titrations, the pH at the equivalence point is not 7. It will be:

• Acidic for a weak base titrated with a strong acid.
• Basic for a weak acid titrated with a strong base.

This is because, at the equivalence point, you form a conjugate base or acid that affects the pH.

#Titration Curves for Weak Acids/Bases

Here are some examples:

#Reading a Burette 📏

Accurately reading a burette is crucial for titrations. Always read from the bottom of the meniscus at eye level.

#Example Problems 📝

Let's work through some examples to solidify our understanding.

#Finding Molarity at Equivalence Point

Problem: Find the concentration of HF when titrating with NaOH. The equivalence point occurs when 20mL of 0.1M NaOH is added to 10mL of HF.

Solution:

Using MaVa = MbVb:

Ma(10mL) = (0.1M)(20mL)

Ma = (0.1)(20)/10 = 0.2M

Therefore, the concentration of HF is 0.2M.

#Weak Acid/Strong Base Titration

Problem: Find the pH of the solution formed from the titration of 25mL of 0.1M CH3COOH with 10mL 0.1M KOH (Ka = 1.8 * 10^-5).

Solution:

1. Reaction: CH3COOH + OH- <=> CH3COO- + H2O
2. Initial mmols:
   • CH3COOH: 25mL * 0.1M = 2.5 mmol
   • OH-: 10mL * 0.1M = 1.0 mmol
3. Stoichiometry:
   • After reaction: 1.5 mmol CH3COOH, 1.0 mmol CH3COO-
4. Henderson-Hasselbalch:
   • pH = pKa + log([CH3COO-]/[CH3COOH])
   • pH = -log(1.8 * 10^-5) + log(1.0/1.5) = 4.74 + log(1/1.5) = 4.56

#Weak Base/Strong Acid Titration

Problem: Find the pH of the solution formed from the titration of 30mL of 0.5M NH3 with 10mL 0.1M HCl (Kb = 1.8 * 10^-5).

Solution:

1. Reaction: NH3 + H+ <=> NH4+
2. Initial mmols:
   • NH3: 30mL * 0.5M = 15 mmol
   • H+: 10mL * 0.1M = 1 mmol
3. Stoichiometry:
   • After reaction: 14 mmol NH3, 1 mmol NH4+
4. Henderson-Hasselbalch (for bases):
   • pOH = pKb + log([NH4+]/[NH3])
   • pOH = -log(1.8 * 10^-5) + log(1/14) = 4.74 + log(1/14) = 3.59
5. pH:
   • pH = 14 - pOH = 14 - 3.59 = 10.41

#Final Exam Focus 🎯

• Key Concepts: Equivalence point, half-equivalence point, buffer region, titration curves (shape and interpretation), Henderson-Hasselbalch equation.
• High-Value Topics: Titrations of weak acids/bases, calculating pH at different points in the titration, understanding the impact of conjugate acids/bases.
• Common Question Types:
  • MCQs: Identifying equivalence points, choosing indicators, calculating pH.
  • FRQs: Drawing and interpreting titration curves, performing stoichiometry calculations, using Henderson-Hasselbalch.

Titration curves are a high-value topic. Make sure you can interpret and draw them.

#Last-Minute Tips

• Time Management: Quickly identify the type of titration and the key points to calculate.
• Common Pitfalls: Forgetting to use pOH for bases, not considering the stoichiometry of the reaction, misreading the burette.
• Strategies:
  • Always write out the balanced equation and net ionic equation.
  • Use the Henderson-Hasselbalch equation when buffers are present.
  • Double-check your calculations and units.

#Practice Questions

*   Set up the ICE table for HA <=> H+ + A-
*   Ka = [H+][A-]/[HA] = x^2/(0.100-x) ≈ x^2/0.100 = 1.0 x 10^-5
*   x = [H+] = 0.001 M
*   pH = -log(0.001) = 3.0

*   Calculate moles of HA and OH-: moles HA = 0.020 L * 0.100 M = 0.002 mol; moles OH- = 0.010 L * 0.100 M = 0.001 mol
*   Use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = pKa + log(0.001/0.001) = pKa
*   pKa = -log(1.0 x 10^-5) = 5.0, so pH = 5.0

*   Using MaVa = MbVb: (0.100 M)(20.0 mL) = (0.100 M)Vb
*   Vb = 20.0 mL

*   At equivalence point, all HA is converted to A-.  Moles of A- = 0.002 mol, total volume = 40 mL = 0.040 L
*   [A-] = 0.002 mol/0.040 L = 0.050 M
*   A- + H2O <=> HA + OH-  Kb = Kw/Ka = 1.0 x 10^-14 / 1.0 x 10^-5 = 1.0 x 10^-9
*   Kb = [HA][OH-]/[A-] = x^2/0.050 = 1.0 x 10^-9
*   x = [OH-] = 7.07 x 10^-6 M; pOH = 5.15; pH = 14 - 5.15 = 8.85

You've got this! Remember, understanding the concepts is key. Good luck on your exam! 🎉