Titration Calculations
Section: 3. Stoichiometry | Syllabus: Cambridge AS Level Physics 9702
What is a Titration? Titration is a technique used to find the exact volume of one solution needed to completely react with another solution. Titrations are commonly used to: Find the concentration of an unknown acid or alkali Find the exact volume needed for complete neutralization Determine purity of substances Titration Equipment Equipment Purpose Accuracy Burette Delivers variable volumes accurately ±0.05 cm³ Pipette Measures exact fixed volume ±0.05 cm³ Conical flask Contains solution being titrated Approximate Indicator Shows when reaction is complete - Key Terms End point: When the indicator changes color, showing the reaction is complete Titre: The volume of solution added from the burette Concordant results: Titres within 0.10 cm³ of each other (these are averaged) Acid-Base Titration Method Step 1: Use a pipette to measure a fixed volume of alkali into a conical flask Step 2: Add a few drops of indicator (e.g., phenolphthalein or methyl orange) Step 3: Fill the burette with acid and record the initial reading Step 4: Add acid slowly, swirling constantly, until the indicator just changes color Step 5: Record the final burette reading Step 6: Calculate the titre (final - initial reading) Step 7: Repeat until you get concordant results Titration Calculation Method Step 1: Write the balanced equation Step 2: Calculate moles of the known solution (n = C × V) Step 3: Use the mole ratio to find moles of unknown Step 4: Calculate concentration of unknown (C = n ÷ V) Remember: Volume → Moles → Use ratio → Moles → Concentration Example 1: Finding Unknown Concentration Question: 25.0 cm³ of sodium hydroxide solution is neutralized by 20.0 cm³ of 0.1 mol/dm³ hydrochloric acid.
Calculate the concentration of the sodium hydroxide. NaOH + HCl → NaCl + H₂O Step 1: Equation is balanced ✓ Step 2: Calculate moles of HCl (known) Volume = 20.0 cm³ = 0.020 dm³ Concentration = 0.1 mol/dm³ Moles = C × V = 0.1 × 0.020 = 0.002 mol Step 3: Use mole ratio From equation: 1 mole NaOH reacts with 1 mole HCl Ratio is 1:1 So 0.002 mol HCl reacted with 0.002 mol NaOH Step 4: Calculate concentration of NaOH Volume of NaOH = 25.0 cm³ = 0.025 dm³ C = n ÷ V = 0.002 ÷ 0.025 = 0.08 mol/dm³ Answer: Concentration of NaOH = 0.08 mol/dm³ Example 2: Different Mole Ratios Question: 25.0 cm³ of sulfuric acid is neutralized by 30.0 cm³ of 0.2 mol/dm³ sodium hydroxide.
Calculate the concentration of the acid. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O Step 1: Equation is balanced ✓ Step 2: Calculate moles of NaOH (known) Volume = 30.0 cm³ = 0.030 dm³ Moles = 0.2 × 0.030 = 0.006 mol Step 3: Use mole ratio From equation: 1 mole H₂SO₄ reacts with 2 moles NaOH Ratio is 1:2 So moles of H₂SO₄ = 0.006 ÷ 2 = 0.003 mol Step 4: Calculate concentration Volume of H₂SO₄ = 25.0 cm³ = 0.025 dm³ C = 0.003 ÷ 0.025 = 0.12 mol/dm³ Answer: Concentration of H₂SO₄ = 0.12 mol/dm³ Example 3: Finding Volume Question: What volume of 0.1 mol/dm³ HCl is needed to neutralize 20.0 cm³ of 0.15 mol/dm³ NaOH?
HCl + NaOH → NaCl + H₂O Step 1: Calculate moles of NaOH Moles = 0.15 × 0.020 = 0.003 mol Step 2: Use mole ratio (1:1) Moles of HCl needed = 0.003 mol Step 3: Calculate volume V = n ÷ C = 0.003 ÷ 0.1 = 0.03 dm³ = 30 cm³ Answer: 30.0 cm³ of HCl is needed Averaging Concordant Titres In practical titrations, you repeat the experiment and average the concordant results: Example data: Trial Titre (cm³) Use?
Rough 24.50 No (rough trial) 1 24.15 Yes 2 24.20 Yes 3 24.18 Yes Average: (24.15 + 24.20 + 24.18) ÷ 3 = 24.18 cm³ These are concordant because they're all within 0.10 cm³ of each other. Finding Mass from Titration You can find the mass of solute in a solution using titration data: Example 4: 25.0 cm³ of NaOH solution neutralizes 22.5 cm³ of 0.1 mol/dm³ HCl.
What mass of NaOH was in the 25.0 cm³? (Mr NaOH = 40) NaOH + HCl → NaCl + H₂O Step 1: Moles of HCl n = 0.1 × 0.0225 = 0.00225 mol Step 2: Moles of NaOH (ratio 1:1) n = 0.00225 mol Step 3: Mass of NaOH m = n × Mr = 0.00225 × 40 = 0.09 g Answer: 0.09 g of NaOH Back Titration Sometimes the substance can't be titrated directly, so we use back titration : Method: Add a known excess of reagent A to react with the unknown Titrate the leftover (unreacted) A with reagent B Calculate how much A reacted with the unknown Calculate the amount of unknown Common Indicators Indicator Acid color Alkali color Best for Phenolphthalein Colorless Pink Strong acid + strong alkali Methyl orange Red Yellow Strong acid + weak alkali Litmus Red Blue Not accurate for titrations Improving Accuracy Use a white tile: Makes color change easier to see Add acid dropwise near the end: Prevents overshooting Swirl constantly: Ensures complete mixing Repeat titrations: Get concordant results Read burette at eye level: Avoids parallax error Remove funnel from burette: Prevents drips affecting volume Calculation Summary Table To find Method Moles from volume n = C × V (convert cm³ to dm³!) Moles of unknown Use mole ratio from equation Concentration C = n ÷ V Volume V = n…
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