Titration calculations Knowledge-Based Study

Study notes

1. Titration is a quantitative volumetric analysis technique used to determine the unknown concentration of an identified analyte in solution by reacting it completely with a standard solution of known concentration, called the titrant.

   Key Definitions

   • Analyte: The substance whose concentration or mass is being determined.
   • Titrant: The solution of accurately known concentration delivered from a burette.
   • Equivalence Point: The exact point in the reaction where chemically equivalent amounts of titrant and analyte have reacted according to the balanced stoichiometric equation.
   • End Point: The physical point observed in the laboratory (such as a distinct color change of an indicator) that signals the equivalence point has been reached.

   The fundamental quantitative relationship governing solutions in titration is the amount of substance in moles ($n$), defined by molar concentration ($c$) and volume in liters ($V$):

   $n = c \times V$

   Titration calculations link the volume and concentration of one reactant to an unknown quantity of another reactant via the mole ratio from a balanced chemical equation.

   General Reaction

   Consider a general neutralization reaction between acid $\text{A}$ and base $\text{B}$:

   $a\text{A} + b\text{B} \rightarrow \text{products}$

   Where $a$ and $b$ represent the stoichiometric coefficients.

   Step-by-Step Procedure

   1. Write and balance the chemical equation to identify the stoichiometric ratio between the two reactants: $\frac{n_{\text{A}}}{a} = \frac{n_{\text{B}}}{b}$.
   2. Convert all volumes to liters ($\text{L}$ or $\text{dm}^3$) by dividing values given in $\text{mL}$ or $\text{cm}^3$ by 1000.
   3. Calculate the moles of the standard solution (titrant) used in the titration: $n = c \times V$.
   4. Use the stoichiometric mole ratio to calculate the moles of the unknown substance (analyte).
   5. Determine the required quantity (molar concentration, mass, or molar mass) using the calculated moles and known volume or sample mass: $c = \frac{n}{V}$ or $m = n \times M$.

   Case 1: 1:1 Stoichiometric Reaction

   A $25.0\text{ mL}$ sample of hydrochloric acid ($\text{HCl}$) is neutralized by $18.5\text{ mL}$ of $0.100\text{ mol/L}$ sodium hydroxide ($\text{NaOH}$). Calculate the concentration of the $\text{HCl}$ solution.

   • Balanced equation: $\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$
   • Mole ratio: $1\text{ mol HCl} : 1\text{ mol NaOH}$
   • Moles of titrant ($\text{NaOH}$): $n = 0.100\text{ mol/L} \times 0.0185\text{ L} = 1.85 \times 10^{-3}\text{ mol}$
   • Moles of analyte ($\text{HCl}$): $n = 1.85 \times 10^{-3}\text{ mol}$ (from $1:1$ ratio)
   • Concentration of $\text{HCl}$: $c = \frac{1.85 \times 10^{-3}\text{ mol}}{0.0250\text{ L}} = 0.0740\text{ mol/L}$

   Case 2: 1:2 Stoichiometric Reaction (Diprotic Acid)

   A $20.0\text{ mL}$ sample of sulfuric acid ($\text{H}_2\text{SO}_4$) requires $32.4\text{ mL}$ of $0.150\text{ mol/L}$ potassium hydroxide ($\text{KOH}$) to reach equivalence.

   • Balanced equation: $\text{H}_2\text{SO}_4\text{(aq)} + 2\text{KOH(aq)} \rightarrow \text{K}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$
   • Moles of $\text{KOH}$: $n = 0.150\text{ mol/L} \times 0.0324\text{ L} = 4.86 \times 10^{-3}\text{ mol}$
   • Moles of $\text{H}_2\text{SO}_4$: $n = \frac{4.86 \times 10^{-3}\text{ mol}}{2} = 2.43 \times 10^{-3}\text{ mol}$
   • Concentration of $\text{H}_2\text{SO}_4$: $c = \frac{2.43 \times 10^{-3}\text{ mol}}{0.0200\text{ L}} = 0.122\text{ mol/L}$

   Handling Titration Data

   Accurate quantitative analysis requires reliable experimental measurements:

   • Titre Concordance: Only concordant titres—typically within $0.10\text{ cm}^3$ (or $0.10\text{ mL}$) of each other—should be included when calculating the mean titre. The initial rough trial must be discarded.
   • Mean Titre Calculation: Calculate the average of concordant volumes before inserting volume into the stoichiometric formula.

   Common Pitfalls to Avoid

   • Ignoring Stoichiometry: Blindly using formulas such as $c_1 V_1 = c_2 V_2$ without dividing by stoichiometric coefficients fails whenever the mole ratio is not $1:1$. Always use $\frac{c_{\text{A}} V_{\text{A}}}{a} = \frac{c_{\text{B}} V_{\text{B}}}{b}$.
   • Volume Unit Inconsistency: Volumes measured in milliliters or cubic centimeters must be converted to liters (or cubic decimeters) before using $n = c \times V$ when concentration is expressed in $\text{mol/L}$.
   • Dilution Factors: If the titrated aliquot was taken from a diluted stock flask (e.g., $25.0\text{ mL}$ taken from a $250.0\text{ mL}$ volumetric flask), remember to scale the calculated moles by the dilution factor (e.g., $\times 10$) to find the total substance in the initial sample.