Principles of method
The Vohlard method is used to determine halide ions l,, Cl-, Br-. A known excess of silver nitrate standard solution is added to the sample and the Ag+ ions in excess are determined using back-titration with a thiocyanate standard solution.
With the Vohlard method the silver ions are titrated using a standard solution of thiocyanate ions:
Ag+ + SCN- ↔ AgSCN
The iron acts as an indicator. The solution turns red with the first excess of thiocyanate ions (see right).
In this case, the concentration of the indicator is not important.
Principles of method (cont.)
The reaction occurs in a highly acid environment to avoid the formation of ferric hydroxide. This represents an obvious advantage compared to other methods for analysing halides because there is no interference from ions like carbonate, oxalate and arsenate (which form slightly soluble salts in neutral environments but not in acid ones).
Standardisation of potassium thiocynate solution
- Prepare 100 ml of a 0.1 N solution of potassium thiocynate;
- Take exactly 5.0 ml 0f 0.10 N AgNO3 solution, add about 3 ml of 3M HNO3, about 1 ml of saturated ferric alum solution and bring it up to a volume of about 40 ml with distilled water .
- Titrate with the KSCN solution adding it slowly, drop by drop, and shaking it all the time (precipitation titration needs to be carried out more slowly than acid-base titration). Initially the formation of a white precipitate can be observed due to the formation of silver thiocyanate. The first drop of reagent in excess after the equivalence point turns the solution a persistent, pale red colour due to the formation of [Fe(SCN)]2+.
- Note the volume of KSCN added.
- Calculate the concentration of the potassium thiocyanate solution.
- Do at least three titrations under the same conditions.
- Calculate the average value of the concentration of the potassium thiocyanate.
Preparation of a 0.1 N solution of potassium thiocyanate
P.M. = P.E. KSCN = 97.18 g/mol AgNO3 = 169.87 g/mol
N = 0.1 g = 9.718 g in 1 L of water.
Imagine we used 4.45 ml of titrant
Vi * Ci = Vf * Cf
Vi = 5 ml
Ci = 0,10 N
Vf = 4.45 ml
Cf = X
Cf = 0.112 N
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