Mada za sehemu hiiChemical AnalysisMada 3
- Volumetric Analysis
- Qualitative Analysis
- Physical Chemistry Analysis
Volumetric analysis in chemistry
Volumetric analysis, commonly referred to as titration, is a quantitative chemical analysis method used to determine the concentration of an unknown solution by adding a volume of a solution of known concentration. The process involves measuring the volume of a titrant (standard solution) needed to react with a given volume of the analyte (unknown solution) to reach the equivalence point, where the reaction is complete.
- Titrant: A solution of known concentration used to react with the analyte.
- Analyte: The unknown solution whose concentration is to be determined.
- Equivalence Point: The point at which the moles of titrant equal the moles of analyte.
- End Point: The point at which an indicator changes color, signaling the completion of the titration.
- Indicator: A substance that changes color to signal the end of the titration.
This is the most common type of titration. It involves the neutralization reaction between an acid and a base. The goal is to determine the concentration of one of the solutions by using the volume of the titrant required to reach the equivalence point.
Reaction Equation: For example, when titrating hydrochloric acid (HCl) with sodium hydroxide (NaOH):
Example 1: Calculation for acid-base titration
Titrant: 0.1 M NaOH Analyte: HCl Volume of NaOH used: 25.0 mL Volume of HCl: 50.0 mL
Step-by-step calculation
- Write the balanced equation:
- Mole ratio between NaOH and HCl is 1:1.
- Use the formula:
- Substitute the values:
- Solve for : M
Result: The concentration of HCl is 0.05 M.
Redox titration involves the transfer of electrons between the titrant and analyte. The most common example is the titration of oxalic acid with potassium permanganate (KMnO₄).
Reaction Equation: The reaction between potassium permanganate and oxalic acid is:
Example 2: Calculation for redox titration
Titrant: 0.02 M KMnO₄ Analyte: Oxalic acid (H₂C₂O₄) Volume of KMnO₄ used: 32.5 mL Volume of oxalic acid solution: 25.0 mL
Step-by-step calculation
- Mole ratio from the reaction: KMnO₄ : Oxalic acid = 2:5
- Use the formula:
- Substitute the values:
- Solve for : M
Result: The concentration of oxalic acid is 0.13 M.
In precipitation titration, the titrant reacts with the analyte to form an insoluble precipitate. A typical example is the titration of chloride ions with silver nitrate.
Reaction Equation:
Example 3: Calculation for precipitation titration
Titrant: 0.1 M AgNO₃ Analyte: NaCl Volume of AgNO₃ used: 45.0 mL Volume of NaCl solution: 50.0 mL
Step-by-step calculation
- Mole ratio from the reaction: Ag⁺ : Cl⁻ = 1:1
- Use the formula:
- Substitute the values:
- Solve for : M
Result: The concentration of NaCl is 0.09 M.
Complexometric titrations involve the formation of a complex between a metal ion and a ligand. A common example is the titration of calcium ions (Ca²⁺) with EDTA (ethylenediaminetetraacetic acid).
Reaction Equation:
Example 4: Calculation for complexometric titration
Titrant: 0.05 M EDTA Analyte: Ca²⁺ solution Volume of EDTA used: 25.0 mL Volume of Ca²⁺ solution: 50.0 mL
Step-by-step calculation
- Mole ratio from the reaction: Ca²⁺ : EDTA = 1:1
- Use the formula:
- Substitute the values:
- Solve for : M
Result: The concentration of Ca²⁺ is 0.025 M.
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