Practical considerations

• pH adjustment is critical in EDTA titrations. The pH is monitored with a pH meter or pH test paper.
• The metal ion under investigation should ideally be approximately 0.25 mM in a volume of 50--150mL of solution. Dilution of the metal ion may be necessary to avoid end-point detection problems.
• Do not add excess indicator, as too intense a colour can lead to problems, e.g. masking of the colour change.
• It is sometimes difficult to detect the end-point because the colour change can be slow to develop. Stirring is recommended to assist colour transformation.
• The use of metal-ion indicators to indicate the end-point of complexometric titrations is based on a specific colour change. Some individuals may find it difficult to detect a particular colour change (e.g. those with colour blindness). Alternative approaches for end-point detection are available based on a colorimeter/spectrophotometer or electrochemical detection.

Example calculation
A solution of Ni2+ (25.00mL) was titrated with 0.1036mol L-I EDTA at pH 5 and required 20.25 mL for the metal-indicator to change colour. What is the concentration (g L-1) of Ni2 +? The atomic weight of nickel is 58.71 g mol ":

1. Write the balanced equation for the reaction between the standard and the test substance
   
   

   ⇒Equation [23.10]

   Ni2+ + H2Y2− → NiY2 + 2H+



2. Determine the equivalences of the reacting species: 1 mole of EDTA is equivalent to 1 mole of Ni²⁺
3. Calculate the number of moles of standard substance (EDTA) used to reach the end-point of the reaction:
   
   

   moles of EDTA

   =

   volume (L) × molarity (mol L−1)

   
   

   =

   20.25 × 10−3 L × 0.1036 mol L−1

   
   

   =

   2.098 × 10−3 mol



4. Calculate the corresponding number of moles of Ni²⁺ present in the 25mL of nickel solution.
   Since 1 mole of EDTA is equivalent to 1 mole of Ni²⁺:
   
   

   moles of Ni2+

   =

   2.098 × 10−3 mol

   
   


5. Calculate the concentration of the Ni²⁺ solution:
   
   

   concentration of Ni2+ solution (mol L−1)	=	1000 × 2.098 × 10−3 (mol)
   		25 (mL)

   
   

   =

   83.92 × 10−3 mol L−1

   
   In this instance, the Ni²⁺ concentration is required in g L⁻¹. The molecular weight of nickel is 58.71 g mol⁻¹:
   
   

   concentration of Ni2+ solution (g L−1)	=	molecular weight (g mol−1) × molarity (mol L−1)
   		25 (mL)

   
   

   =

   58.71 × 83.92 × 10−3

   
   

   =

   4.927 g L−1