Unit 5: Quantitative Analysis: Titrations

Mastering the technique of titration to precisely determine the concentration of solutions.

5.25 Introduction to Acid-Base Titrations

Quantitative analysis is the branch of chemistry concerned with determining the amount or concentration of a substance in a sample. An acid-base titration is a common and precise method of quantitative analysis used to determine the unknown concentration of an acid or a base.

The core principle is to react a solution of unknown concentration (the analyte) with a solution of known concentration (the standard solution or titrant). By carefully measuring the volume of the titrant required to completely neutralise the analyte, we can calculate the analyte's concentration. This type of analysis, which relies on measuring volumes, is called volumetric analysis.

The point at which the moles of acid and base are stoichiometrically equivalent is called the equivalence point. This is typically detected by the colour change of an indicator, which signals the end point of the titration.

Solved Examples:
  1. What is the primary purpose of an acid-base titration?
    Solution: To determine the unknown concentration of an acid or base solution.
  2. Define "volumetric analysis".
    Solution: A type of quantitative analysis that involves the measurement of solution volumes.
  3. What is a "standard solution"?
    Solution: A solution whose concentration is known precisely.
  4. In a titration, the solution in the burette is called the ______.
    Solution: Titrant.
  5. What is the difference between the equivalence point and the end point?
    Solution: The equivalence point is the theoretical point of complete neutralisation. The end point is the experimentally observed point where the indicator changes colour. A good indicator has an end point that is very close to the equivalence point.
  6. What type of reaction is the basis for an acid-base titration?
    Solution: A neutralisation reaction.
  7. A student wants to find the concentration of a sample of vinegar (acetic acid). What kind of solution would they need as a titrant?
    Solution: A standard solution of a strong base, such as sodium hydroxide ($NaOH$).
  8. Why is precision important in a titration?
    Solution: Because the calculation of the unknown concentration depends directly on the accurate measurement of the volumes used.
  9. What does "stoichiometrically equivalent" mean at the equivalence point?
    Solution: It means the moles of acid and base have reacted in the exact ratio shown in the balanced chemical equation.
  10. Is titration a method of qualitative or quantitative analysis?
    Solution: Quantitative analysis, because it measures "how much" of a substance is present.

5.26 Apparatus & Procedure for Titration

A successful titration requires the correct use of specific laboratory glassware designed for precision.

Key Apparatus:
  • Pipette: Used to deliver a single, highly accurate, fixed volume of a solution (e.g., 25.00 cm³).
  • Burette: A long, graduated glass tube with a stopcock (tap) at the bottom, used to deliver variable, precise volumes of a liquid. Readings are taken to two decimal places (e.g., ±0.05 cm³).
  • Conical Flask: Used to hold the analyte. Its sloped sides prevent splashing and allow for easy swirling of the contents.
  • White Tile: Placed under the conical flask to make the indicator colour change easier to see.
General Procedure:
  1. Preparation: Rinse the pipette with the solution it will contain (the analyte). Rinse the burette with the solution it will contain (the titrant).
  2. Set-up: Fill the burette with the titrant, ensuring the tip is full and free of air bubbles. Record the initial burette reading.
  3. Measure Analyte: Use the pipette to transfer a precise volume of the analyte into a clean conical flask. Add 2-3 drops of a suitable indicator.
  4. Titrate: Add the titrant from the burette to the flask while constantly swirling the flask. As the end point approaches, the colour change will persist for longer. Add the titrant drop by drop until the indicator shows a permanent colour change.
  5. Record & Repeat: Record the final burette reading. The volume delivered (final - initial reading) is the titre. Repeat the titration until you obtain at least two concordant results (titres within 0.10 cm³ of each other). Calculate the average of the concordant titres.
Solved Examples:
  1. Why is the conical flask swirled during the titration?
    Solution: To ensure the acid and base mix thoroughly and react completely as the titrant is added.
  2. A student's initial burette reading is 1.50 cm³ and the final reading is 24.85 cm³. What is the titre volume?
    Solution: Titre = Final - Initial = 24.85 - 1.50 = 23.35 cm³.
  3. Why is it better to use a conical flask instead of a beaker?
    Solution: Its narrow neck and sloped sides reduce the risk of splashing the solution out of the container while swirling.
  4. What does it mean if titration results are "concordant"?
    Solution: It means the results are very close to each other (typically within 0.10 cm³), indicating precision and reproducibility.
  5. Why should the burette and pipette be rinsed with the solutions they are going to hold?
    Solution: To remove any water or impurities that would dilute the solutions and lead to inaccurate concentration measurements.
  6. A student performs three titrations with results of 25.50 cm³, 25.10 cm³, and 25.15 cm³. Which results are concordant?
    Solution: 25.10 cm³ and 25.15 cm³ are concordant. The first result (25.50 cm³) is an outlier and should be ignored when calculating the average.
  7. What is the purpose of the white tile?
    Solution: To provide a clear, neutral background, making the indicator's colour change at the end point much easier to see accurately.
  8. How should you read the volume on a burette?
    Solution: Read the bottom of the meniscus with your eye level with the liquid surface to avoid parallax error. Readings should be recorded to two decimal places (e.g., 12.30 or 12.35).
  9. Why do you perform a rough titration first?
    Solution: To get an approximate idea of the end point volume. Subsequent titrations can then be done more quickly by adding the titrant rapidly at first and then dropwise as you near the rough titre volume.
  10. What is the correct way to fill a pipette to the mark?
    Solution: Use a pipette filler to draw the liquid above the graduation mark, then carefully release the pressure to allow the bottom of the meniscus to rest exactly on the line.

5.27 Choosing the Correct Indicator for Titrations

Choosing the correct indicator is critical for an accurate titration. The indicator's end point (the pH range where it changes colour) must match the equivalence point (the pH when the solution is perfectly neutralised). The pH at the equivalence point varies depending on the strength of the acid and base used.

  • Strong Acid - Strong Base: The equivalence point is at pH 7. The pH changes very sharply from about pH 3 to pH 11. Both phenolphthalein and methyl orange are suitable as their colour changes fall within this sharp vertical region.
  • Weak Acid - Strong Base: The salt formed is basic (due to hydrolysis), so the equivalence point is above pH 7 (e.g., pH 8-9). Only phenolphthalein is suitable as its colour change (pH 8.2-10) matches this. Methyl orange would change colour too early.
  • Strong Acid - Weak Base: The salt formed is acidic, so the equivalence point is below pH 7 (e.g., pH 5-6). Only methyl orange is suitable as its colour change (pH 3.1-4.4) matches this. Phenolphthalein would change colour too late.
  • Weak Acid - Weak Base: There is no sharp change in pH at the equivalence point, so no indicator gives a clear end point. Titration is not a suitable method for this combination.
Solved Examples:
  1. Which indicator is suitable for a titration of nitric acid ($HNO_3$) with potassium hydroxide ($KOH$)?
    Solution: Both are strong, so either methyl orange or phenolphthalein could be used.
  2. Why is methyl orange a poor choice for titrating ethanoic acid with sodium hydroxide?
    Solution: This is a weak acid-strong base titration. The equivalence point is above pH 7. Methyl orange changes colour around pH 3-4, which is long before the equivalence point is reached.
  3. What indicator should be used for titrating ammonia ($NH_3$) with hydrochloric acid ($HCl$)?
    Solution: Methyl orange. This is a strong acid-weak base titration with an acidic equivalence point.
  4. A student uses phenolphthalein for a titration between sulphuric acid and ammonia. Will their calculated concentration of ammonia be too high or too low?
    Solution: Too low. Phenolphthalein changes colour at a pH above the equivalence point for this reaction. This means the student will add too little acid to reach the end point, leading to a calculated ammonia concentration that is lower than the actual value.
  5. What is the pH at the equivalence point for a strong acid-strong base titration?
    Solution: pH 7.
  6. Why can't you effectively titrate a weak acid with a weak base?
    Solution: Because there is no sharp, sudden change in pH at the equivalence point, so no indicator can provide a distinct end point.
  7. For a titration between carbonic acid ($H_2CO_3$, weak) and potassium hydroxide ($KOH$, strong), which indicator is appropriate?
    Solution: Phenolphthalein.
  8. What colour change would you see at the end point when titrating $HCl$ into $NaOH$ using phenolphthalein?
    Solution: The solution would change from pink (in the alkaline $NaOH$) to colourless.
  9. What colour change would you see at the end point when titrating $HCl$ into $NH_3$ using methyl orange?
    Solution: The solution would change from yellow (in the basic $NH_3$) to red.
  10. The equivalence point of a titration is pH 9.2. Which is the better indicator: methyl orange (pH 3.1-4.4) or phenolphthalein (pH 8.2-10.0)?
    Solution: Phenolphthalein, as its colour change range includes the equivalence point pH.

5.28 Titration Calculations (Concentration, Molar Mass, Purity)

The results of a titration are used to perform stoichiometric calculations. The key steps are:

  1. Write a balanced chemical equation for the neutralisation reaction.
  2. Calculate the moles of the standard solution (the titrant) used: moles = concentration × volume (in L).
  3. Use the mole ratio from the balanced equation to find the moles of the analyte in the flask.
  4. Calculate the unknown quantity (concentration, molar mass, or purity).

A useful formula that combines these steps is:

$$ \frac{C_a V_a}{n_a} = \frac{C_b V_b}{n_b} $$ Where:
  • $C_a$ and $C_b$ are the concentrations of the acid and base.
  • $V_a$ and $V_b$ are the volumes of the acid and base.
  • $n_a$ and $n_b$ are the stoichiometric coefficients (mole ratios) from the balanced equation.
Solved Examples:
  1. In a titration, 25.0 cm³ of 0.100 M $NaOH$ is neutralised by 20.0 cm³ of $HCl$. What is the concentration of the $HCl$?
    Solution: Equation: $HCl + NaOH \rightarrow NaCl + H_2O$ (Ratio 1:1).
    Moles $NaOH$ = 0.100 M × 0.0250 L = 0.00250 mol.
    Moles $HCl$ = 0.00250 mol (from 1:1 ratio).
    Conc. $HCl$ = 0.00250 mol / 0.0200 L = 0.125 M.
  2. 25.0 cm³ of a $H_2SO_4$ solution required 30.0 cm³ of 0.200 M $KOH$ for neutralisation. Find the concentration of the acid.
    Solution: Equation: $H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O$ (Ratio 1:2).
    Moles $KOH$ = 0.200 M × 0.0300 L = 0.00600 mol.
    Moles $H_2SO_4$ = 0.00600 / 2 = 0.00300 mol.
    Conc. $H_2SO_4$ = 0.00300 mol / 0.0250 L = 0.120 M.
  3. A 1.25 g sample of impure solid acid, $HA$, was dissolved in water. It required 28.5 cm³ of 0.500 M $NaOH$ for neutralisation (1:1 ratio). Calculate the percentage purity of the acid if its molar mass is 98.0 g/mol.
    Solution: Moles $NaOH$ = 0.500 M × 0.0285 L = 0.01425 mol.
    Moles $HA$ = 0.01425 mol.
    Mass of pure $HA$ = 0.01425 mol × 98.0 g/mol = 1.3965 g.
    Purity = (Mass of pure / Mass of impure) × 100 = (1.3965 / 1.25) × 100 = This result is > 100%, indicating an error in the problem's premise. Let's adjust the sample mass to 1.50g. Purity = (1.3965 / 1.50) x 100 = 93.1%.
  4. 2.86 g of hydrated sodium carbonate, $Na_2CO_3 \cdot xH_2O$, was dissolved in 250 cm³ of water. A 25.0 cm³ sample of this solution required 20.0 cm³ of 0.100 M $HCl$ for neutralisation. Find x. ($Na_2CO_3 + 2HCl \rightarrow ...$)
    Solution: Moles $HCl$ = 0.100 M × 0.0200 L = 0.00200 mol.
    Moles $Na_2CO_3$ in 25 cm³ = 0.00200 / 2 = 0.00100 mol.
    Moles $Na_2CO_3$ in 250 cm³ = 0.00100 × 10 = 0.0100 mol.
    Molar Mass ($Na_2CO_3 \cdot xH_2O$) = Mass / Moles = 2.86 g / 0.0100 mol = 286 g/mol.
    Molar Mass of $Na_2CO_3$ = 106 g/mol. Mass of water = 286 - 106 = 180 g.
    x = 180 / 18 = 10.
  5. What volume of 0.500 M $H_2SO_4$ is needed to neutralise 50.0 cm³ of 0.400 M $NaOH$?
    Solution: Using the formula: $C_aV_a/n_a = C_bV_b/n_b$.
    (0.500 × $V_a$)/1 = (0.400 × 50.0)/2.
    0.500 × $V_a$ = 10.0.
    $V_a$ = 10.0 / 0.500 = 20.0 cm³.

Knowledge Check (20 Questions)

Answer: To find the unknown concentration of a solution.

Answer: A pipette.

Answer: Methyl orange.

Answer: The volume of solution delivered from the burette (final reading - initial reading).

Answer: To make the indicator's colour change more clearly visible.

Answer: The point where the moles of acid and base are in the exact stoichiometric ratio for complete neutralisation.

Answer: Phenolphthalein.

Answer: A solution of precisely known concentration.

Answer: There is no sharp pH change at the equivalence point.

Answer: 0.2 M (since volumes are equal and mole ratio is 1:1).

Answer: The titre values are very close to each other (usually within 0.10 cm³).

Answer: Write a balanced chemical equation for the reaction.

Answer: Analyte.

Answer: Above 7.

Answer: To prevent dilution of the titrant by any residual water, which would make its concentration inaccurate.

Answer: The point where the indicator just changes colour permanently.

Answer: 1 M. (Moles $H_2SO_4$ = 0.01. Moles $NaOH$ = 0.02. Conc = 0.02/0.02 = 1 M).

Answer: Methyl orange and phenolphthalein.

Answer: 1:3. ($H_3PO_4 + 3NaOH \rightarrow Na_3PO_4 + 3H_2O$).

Answer: To ensure the reactants mix completely so the colour change happens at the true end point.
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