AQA A Level Chemistry复习笔记4.1.1 Making a Volumetric Solution

REQUIRED PRACTICAL 1

Volumetric Analysis

• Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (a volumetric solution) to determine the concentration of another unknown solution
• The technique most commonly used is a titration
• The volumes are measured using two precise pieces of equipment, a volumetric or graduated pipette and a burette
• Before the titration can be done, the standard solution must be prepared
• Specific apparatus must be used both when preparing the standard solution and when completing the titration, to ensure that volumes are measured precisely

Some key pieces of apparatus used to prepare a volumetric solution and perform a simple titration

1. Beaker
2. Burette
3. Volumetric Pipette
4. Conical Flask
5. Volumetric Flask

Making a Volumetric Solution

• Chemists routinely prepare solutions needed for analysis, whose concentrations are known precisely
• These solutions are termed volumetric solutions or standard solutions
• They are made as accurately and precisely as possible using three decimal place balances and volumetric flasks to reduce the impact of measurement uncertainties
• The steps are:

Volumes & concentrations of solutions

• The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of  solution
  • The solute is the substance that dissolves in a solvent to form a solution
  • The solvent is often water

   

• A concentrated solution is a solution that has a high concentration of solute
• A dilute solution is a solution with a low concentration of solute
• Concentration is usually expressed in one of three ways:
  • moles per unit volume
  • mass per unit volume
  • parts per million

   

Answer:

Step 1: Find the number of moles of NaOH needed from the concentration and volume:

number of moles  = concentration (mol dm-3) x volume (dm3)

n = 0.200 mol dm-3 x 0.250 dm3

n = 0.0500 mol

Step 2: Find the molar mass of NaOH

M = 22.99 + 16.00 + 1.01 = 40.00 g mol-1

Step 3: Calculate the mass required

mass = moles x molar mass

mass =  0.0500 mol x 40.00 g mol-1   = 2.00 g