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Experiment 2: Acid-Base Titration

Experiment 2: Acid-Base Titration

Determination of the Concentration of Hydrochloric Acid Solution

Objectives

  • Prepare a standard solution of oxalic acid.
  • Standardise 0.2 M NaOH solution.
  • Determine the concentration of HCl solution.
  • Acquire the correct techniques of titration.

Introduction

Titration is a laboratory technique used to determine the concentration of a solution using another solution with a known concentration. A primary standard solution is prepared by dissolving an accurately weighed pure solid of a known molar mass in a known volume of distilled water.

Oxalic acid, H2C2O4, is a common primary standard used to determine the concentration of bases like NaOH. Solid NaOH is hygroscopic, making direct weighing inaccurate.

Chemical Equations

(1) H2C2O4(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(l)
(2) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Apparatus & Reagents

Reagents:
  • Hydrated oxalic acid (H2C2O4.2H2O)
  • 0.2 M NaOH (to be standardized)
  • HCl solution
  • Phenolphthalein
Apparatus:
  • Burette, Pipette, Volumetric flask
  • Conical flask, Analytical balance

Part A: Standard Solution Preparation

Part B: Standardisation of 0.2 M NaOH

TrialInitial Vol (mL)Final Vol (mL)Vol NaOH Used (mL)
Gross-
I-
II-
III-
Average Volume NaOH (I, II, III):0.00 mL

Part C: Determination of HCl Concentration

TrialInitial Vol (mL)Final Vol (mL)Vol NaOH Used (mL)
Gross-
I-
II-
III-
Average Volume NaOH (I, II, III):0.00 mL

Quantitative Analysis Results

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Lecturer: -

Part A: Standard Solution (Oxalic Acid)

Mass weighed0.0000 g
Moles (n = mass / Mm)0.0000 mol
Molarity (M = n / 0.250 L)0.0000 M
Molar mass H₂C₂O₄·2H₂O = 126.07 g/mol

Part B: Standardisation of NaOH

Average Volume NaOH used (trials I‑III): 0.00 mL

⚖️ Calculation (1 mol acid : 2 mol NaOH)

MNaOH = (2 × Moxalic × Vacid) / Vbase

= (2 × 0.0000 M × 0.025 L) / (0.00 L)

📊 Molarity of NaOH 0.0000 M

Part C: Determination of HCl

Average Volume NaOH used (trials I‑III): 0.00 mL

⚖️ Calculation (1:1 ratio)

MHCl = (MNaOH × VNaOH) / VHCl

= (0.0000 M × 0.00 L) / 0.025 L

🧪 Concentration of HCl 0.0000 M
Action Successful