Unit 7B – Part 3 : Titrations Lecture Notes

Required Documents

• Periodic Table
• Mole Tools
• Helpful references to revisit while solving titration questions:
  • Conversions & Constants to Memorize (Unit 1)
  • Strategies for Solving Concentration Problems (Unit 6)

Introduction to Titration

• Experimental technique based on a known reaction (most commonly acid–base) used to determine an unknown quantity.
  • The unknown solution in the flask = analyte.
  • The solution delivered from the buret = titrant (standard solution; concentration precisely known).
• Fundamental idea:
  • Add titrant until the chemical reaction reaches the stoichiometric point where both reactants are completely consumed in the correct mole ratio (the equivalence point).
  • Measure volume of titrant used → perform stoichiometric calculations to find unknown concentration, amount, or mass.

Indicators & Endpoints

• Indicator = dye that is one color in acidic medium and another in basic medium.
  • Only a few drops added so that it does not appreciably affect stoichiometry.
• Endpoint = the experimental signal (color change) indicating the reaction has reached the equivalence point.
  • Practically, the endpoint is taken as the equivalence point; small indicator error is negligible for most purposes.
• At the endpoint:
  \frac{\text{moles actually reacted}}{\text{coefficients}} = \text{same ratio for each reactant}

General Stoichiometric Strategy for Titration Calculations

1. Convert titrant volume → moles of titrant using its molarity.
   \text{mol titrant} = M{\text{titrant}} \times V{\text{titrant}}
2. Use balanced equation to change moles of titrant → moles of analyte.
3. Compute desired unknown (molarity, mass, percentage, etc.) for the analyte.

Worked Example 1 – Determining M of (\text{Ca(OH)}_2)

Situation
50.00 mL of unknown (\text{Ca(OH)}2) is titrated with (6.000\,\text{M}) (\text{HNO}3). 77.54 mL of acid is required.

Balanced equation
\text{Ca(OH)}2(aq) + 2\,\text{HNO}3(aq) \rightarrow \text{Ca(NO}3)2(aq) + 2\,\text{H}_2\text{O}(l)

Step-by-Step

• Step 1: Convert acid volume → moles.
  V{\text{HNO}3}=77.54\,\text{mL}=0.07754\,\text{L}
  n{\text{HNO}3}=6.000\,\frac{\text{mol}}{\text{L}} \times 0.07754\,\text{L}=0.46524\,\text{mol}
• Step 2: Mole ratio (coefficients 2:1)
  n{\text{Ca(OH)}2}=\frac{1}{2}\,n{\text{HNO}3}=\tfrac{1}{2}\times0.46524=0.23262\,\text{mol}
• Step 3: Convert moles → molarity of base.
  V{\text{Ca(OH)}2}=50.00\,\text{mL}=0.05000\,\text{L}
  M{\text{Ca(OH)}2}=\frac{0.23262\,\text{mol}}{0.05000\,\text{L}}=4.652\,\text{M}

Result
The calcium hydroxide solution is \boxed{4.65\,\text{M}} (3 sig figs).

Conceptual notes / pitfalls

• Dihydroxide base provides two OH⁻ but we account for that automatically through the coefficients.
• Always state units and keep moles of solute distinct from volume of solution.

Worked Example 2 – Mass of HBr in a Sample

Situation
An unknown HBr solution is titrated with (0.178\,\text{M}) (\text{Ba(OH)}_2). 23.67 mL of base reaches the endpoint. Find grams of HBr in the entire (unspecified-volume) acid sample.

Balanced equation
\text{Ba(OH)}2(aq) + 2\,\text{HBr}(aq) \rightarrow \text{BaBr}2(aq) + 2\,\text{H}_2\text{O}(l)

Step-by-Step

• Step 1: Volume → moles of base.
  V{\text{Ba(OH)}2}=23.67\,\text{mL}=0.02367\,\text{L}
  n{\text{Ba(OH)}2}=0.178\,\frac{\text{mol}}{\text{L}}\times0.02367\,\text{L}=0.00422\,\text{mol}
• Step 2: Mole ratio 1:2 (base:acid).
  n_{\text{HBr}}=2\times0.00422=0.00844\,\text{mol}
• Step 3: Moles → mass.
  Mr(\text{HBr}) = 1.01+79.90 = 80.91\,\frac{\text{g}}{\text{mol}} m{\text{HBr}}=0.00844\,\text{mol}\times80.91\,\frac{\text{g}}{\text{mol}}=0.682\,\text{g}

Result
The acid sample contained \boxed{0.682\,\text{g HBr}}.

Why the sample volume is irrelevant
All of the HBr present was neutralized; its initial volume or concentration is unnecessary once the reaction is complete.

Worked Example 3 – M of (\text{Sr(OH)}_2)

Situation
25.00 mL of (1.197\,\text{M}) (\text{H}3\text{PO}4) requires 38.46 mL of (\text{Sr(OH)}_2) to reach the endpoint. Calculate molarity of the base.

Balanced equation
2\,\text{H}3\text{PO}4(aq) + 3\,\text{Sr(OH)}2(aq) \rightarrow \text{Sr}3(\text{PO}4)2(s) + 6\,\text{H}_2\text{O}(l)

Step-by-Step

• Step 1: Acid volume → moles.
  V{\text{acid}}=25.00\,\text{mL}=0.02500\,\text{L} n{\text{H}3\text{PO}4}=1.197\,\frac{\text{mol}}{\text{L}}\times0.02500\,\text{L}=0.029925\,\text{mol}
• Step 2: Mole ratio 2:3 (acid:base).
  n{\text{Sr(OH)}2}=\frac{3}{2}\,n_{\text{acid}}=\frac{3}{2}\times0.029925=0.0448875\,\text{mol}
• Step 3: Moles → molarity of base.
  V{\text{base}}=38.46\,\text{mL}=0.03846\,\text{L} M{\text{Sr(OH)}_2}=\frac{0.0448875\,\text{mol}}{0.03846\,\text{L}}=1.167\,\text{M}

Result
The strontium hydroxide solution is \boxed{1.17\,\text{M}} (3 sig figs).

Take-Away Concepts Illustrated

• Polyprotic acid ((\text{H}3\text{PO}4)) and dihydroxide base ((\text{Sr(OH)}_2)) require correctly balanced coefficients; mis-counting protons/ hydroxides will mis-scale the final answer by factors of 2 or 3.

Practical Connections & Tips

• Reactions are often fast and go to completion → titration delivers very precise stoichiometric data.
• Indicators vs pH meters: indicators give approximate endpoint; pH meter can detect the sharp pH jump more precisely.
• Common lab errors: mis-reading buret, parallax, forgetting to rinse the inside wall of the flask with DI water during titration, adding excess titrant past the endpoint.
• Ethical/ safety note: strong acids/bases are corrosive; always add acid to water if diluting, wear goggles/ gloves, neutralize spills.

Mathematical & Chemical Highlights

• Molarity definition: M = \dfrac{\text{mol solute}}{\text{L solution}}
• Mole–mole conversions rely solely on balanced chemical equations.
• Multiplying conversion factors line-up to “map” units from given → desired; be explicit with each ratio to avoid skipped steps.

Further Practice & Next Steps

• Recommended: Work Q&A Sheet #13, problems 1–10 for extra titration stoichiometry practice.
• Upcoming topic: Limiting & Excess Reactants (Unit 7B – Part 4).