Titrimetry

1

• Titrimetry
• Definition Volumetric analysis is a method in
  which the final step is the determination of
  the volume of one solution containing a reactant
  completely reacts with a reactant in another
  solution.
• Definition A titration is the procedure by which
  the volume of a standard solution required to
  react with an analyte is carried out usually
  using a buret.
• A standard solution is a solution whose
  concentration is accurately known.
• Std. Solns must be stable so that the
  concentration remains unchanged
• Std. Solns should react rapidly with the analyte
  
• Std. Solns should react completely with the
  analyte in order to produce sharp endpoints
• Std. Solns should undergo selective reactions
  with an analyte thus giving a balanced chemical
  equation for its reaction
• The concentration of a standard solution can be
  determined by
• Direct weight weigh the standard reagent and
  dilute to a known volume
• Indirect titrate against a known amount of a
  primary standard

2

• Titrimetry
• The equivalence point in a titration is the point
  in the titration when a chemically equivalent
  amount of titrant has been added to the
  analytical solution
• Example In the titration of HCl with Ba(OH)2
• 2HCl Ba(OH)2 BaCl2 2H2O
• the equivalence point occurs when 1 mol Ba(OH)2
  has been added to 2 mol HCl or the mole ratio
  of Ba(OH)2HCl is 12
• The endpoint in a titration is the point in the
  titration when a physical change occurs
• Color change of an indicator
• Change in electrical resistance of the titration
  mixture
• Change in electrical potential of an electrode
  inserted in the titration mixture
• Primary Standards should
• Be obtainable in high purity
• Be stable to air and light
• Be non-hygroscopic and non-efflorescent
• Be readily available and inexpensive
• Have a high molecular weight
• Have good solubility in the solvent used for the
  analysis

3

• Titrimetry
• Concentration units for solutions
• The usual unit of concentration of a species in
  solution, cA, is molarity
• Example What is if 30.00 mg H2SO4 is
  contained in 10.35 mL solution?
• Analytical molar concentration this is an
  operational definition of concentration of an
  analyte, regardless of the state of its
  dissociation
• The Equilibrium concentration is the actual
  concentration of each species in solution at
  equilibrium
• Example Dissolving 1.000 mol H2SO4 in enough
  water to make 1.000 L solution gives
  1.000 M
• Since H2SO4 is 100 dissociated, H2SO40.000 M
• HSO4- is 1 dissociated
• HSO4-0.99 M
• H1.01 M
• SO42-0.01 M

4

• Titrimetry
• Algebraic relationships
• (1)
• (2)
• (3)
• Example Describe how to prepare 250.0 mL of
  0.1000 M AgNO3.

5

• Titrimetry
• Example Describe how to make 100.0 mL of 0.02500
  M AgNO3 from 0.1000 M AgNO3.
• In diluting a solution, the number of mmol solute
  remains unchanged
• Therefore, equation (3) can be enhanced
• 0.02500 M x 100.0 mL0.1000 x Vsoln 2
• Vsoln 225.00 mL
• To prepare the solution, place 25.00 mL 0.1000 M
  AgNO3 in a 100.0 mL volumetric flask and dilute
  to volume

6
Titrimetry Calculations from titration data four
step process 1 Derive a stoichiometric
relationship between the number of mmol analyte
and number of mmole of standard This usually
requires a balanced chemical equation! 2
Calculate the number of mmol of standard used
with Equation (1) or (3) 3 Convert the number of
mmol standard to mmol analyte using the results
of step 1 4 Solve for cA or weight of A using
Equations (1) and/or (2)
7

• Titrimetry
• Calculating the concentrations of standard
  solutions
• Example 528.0 mg KHP required 24.21 mL NaOH
  solution ot reach the phenolphthalein endpoint.
  What is cNaOH?
• 1
• 2
• 3
• 4

8

• Titrimetry
• Calculating the concentration of a secondary
  standard solution from the volume ratio of the
  two solutions
• Example 25.00 mL of 0.1 M HCl require 27.23 mL
  of 0.1068 M NaOH to reach the phenolphthalein
  endpoint. What is the molarity of the HCl
  solution?
• 1 HCl NaOH NaCl H2O
• 2 number mmol NaOH cNaOH x VNaOH
• 0.1068 M x 27.23 mL
• 2.908 mmol NaOH
• 3
• 4

9

• Titrimetry
• Calculation of the weight percent of analyte in a
  sample
• Example 1.5216 g impure KHP required 31.25 mL of
  0.1068 M NaOH for complete reaction. What is the
  wt. KHP in the sample?
• 1 KHP NaOH NaKP H2O
• 2 number mmol NaOH cNaOH x VNaOH
• 0.1068 M x
  31.25 mL 3.338 mmol NaOH
• 3
• 4 rearranging equation (1)
• 5