Precipitation Titration: A Method for Forming Insoluble Solids in Chemistry, Study notes of Biochemistry

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Precipitation Titration

Precipitation Reactions

 Precipitation is the formation of a solid in a solution solid formed is called the precipitate A precipitation reaction occurs when water solutions of two different ionic compounds are mixed and an insoluble solid separates out of solution. KCl + AgNO 3 AgCl + KNO 3 The precipitate is itself ionic; the cation comes from one solution and the anion from another. Cl-^ solution Precipitating agent White precipitate

Argentometric titration:

• Titrations involving silver are termed argentometric, from the Latin name for silver, argentum.
• The major precipitation reaction used is that of silver with a range of anions. These anions include:

• Halides (Cl

   , Br - , I - ) 

• Pseudohalides (S 2 - , HS - , CN - , SCN - )
• The reaction rates for the silver salt precipitation is rapid.
• The reaction ratio is 1:1 and silver salts formed are generally quite insoluble.
• Argentometric methods involving precipitation titrimetry:
• Mohr’s Method
• Fajan’s Method
• Volhard’s Method 4

Mohr’s Method:

• This direct method uses potassium chromate ( chromate ions (CrO 4 2  ) ) as an indicator in the titration of ( Cl - , Br - , and CN - )ions (analyte) with a silver nitrate standard solution (titrant).
• After all the chloride has been precipitated as white silver chloride , the first excess of titrant results in the formation of a silver chromate precipitate ,
• which signals the end point ( 1 ). The reactions are: Ag + (aq) + Cl - (aq) → AgCl(s) K sp = 1.8 x 10 - 10 white precipitate
• End point determination by brick red color precipitate, Ag 2 CrO 4 (s)

2 Ag

(aq) + CrO 4 2 - (aq) → Ag 2 CrO 4 (s) K sp = 1.2 x 10

• 12
• AgCl is less soluble than Ag 2 CrO 4 so it will precipitate first 5

• This method uses a back titration with potassium thiocyanate and is suitable for the determination of chlorides, bromides and iodides in acidic solutions. - First, Cl - is precipitated by excess AgNO 3

Removing AgCl(s) by filtration / washing

• Excess Ag
  • is titrated with KSCN in the presence of Fe 3 +
• When Ag
  • has been consumed, a red complex forms as a result of: The Volhard titration can be used for any anion that forms an insoluble salt with silver

Volhard method:

Ag (aq) Cl (aq) AgCl (s)

   Ag (aq) SCN (aq) AgSCN (s)

   (aq) (aq) (aq)     3 - 2 Fe SCN FeSCN Red complex

• Conditions for Volhard’s method:
• The solution must be acidic , with a concentration of about 1 M in nitric acid to ensure the complex formed is stable , and to prevent the precipitation of Iron(III) as hydrated oxide.
• The indicator concentration should not be more than 0.2M.
• In case of I - , indicator should not be added until all the I - is precipitated with Ag + , since it would be oxidized by the Fe(III). 2 Fe 3 + + 2 I - 2 Fe 2 + + I 2 The AgX ↓precipitate must be filtered off, before titrating with SCN
  • to prevent any error, for example in the case of chloride ion, AgCl will react with the titrant (SCN
    • ) and cause a diffuse end point. AgCl + SCN - AgSCN + Cl - OR Use tartrazine as indicator instead of Iron(III). (^8)

The mechanism of indicators action:

• The best–known adsorption indicator is fluorescein , which is used to indicate the equivalence point in the titration of Cl - with Ag + . Fluorescein is a weak acid, which partially dissociates in water to form fluoresceinate anion.
• The fluoresceinate anion has a yellow–green colour in solution. 10

• When Cl - is titrated with Ag + in the presence of fluorescein, the negatively charged fluoresceinate anions are initially repelled by the negatively charged AgCl colloidal particles, with their primary **adsorption layer of Cl

ions**.

• Thus the fluorescein remains in a yellow–green colour prior to the equivalence point.
• At the equivalence point , the colloidal AgCl particles undergo an abrupt change from a negative charge to a positive charge by virtue of Ag + ions adsorbed in the primary adsorption layer.
• The fluoresceinate ions are strongly adsorbed in the counter–ion layer of the AgCl colloids, giving these particles a red colour and providing an end point colour change from yellow–green to red or pink. 11

Comparison of argentometric titration methods

Method Advantages Disadvantages Mohr Simple

• Alkaline solution only

• Not suitable for I

• Requires a blank Volhard
• Capable for direct Ag

and indirect halide analyses

• Very clear colour change
• Must use 1M of nitric acid solution
• Some problems with some ions Fajans
• Capability for different pH ranges and selectivity with different indicators
• Difficult with dilute solutions
• Should not be a high background ionic level

Titration Curves for Argentometric Methods

Plots of titration curves are normally sigmoidal curves consisting of pAg (or pAnalyte) versus volume of AgNO 3 solution added. Example: Titration of chloride with silver. The points on the curve can be calculated, given the analyte concentration, AgNO 3 concentration and the appropriate K sp . A useful relationship can be derived by taking the negative logarithm of both sides of a solubility-product expression. Thus, for silver chloride, 𝐾 𝑠𝑝 = 𝐴𝑔

[𝐶𝑙 − ] 𝑙𝑜𝑔𝐾 𝑠𝑝 = −log( 𝐴𝑔

𝐶𝑙 − ) 𝑙𝑜𝑔𝐾𝑠𝑝 = − log 𝐴𝑔

− 𝑙𝑜𝑔 𝐶𝑙 − 𝑝𝐾 𝑠𝑝 = 𝑝𝐴𝑔

+𝑝𝐶𝑙 −

Example Calculate pCl for the titration of 100.0 ml 0.100 M NaCl with 0.100 M AgNO 3 for the addition of 0.0, 20.0, 99.0, 99.5, 100.0 and 110.0 ml AgNO 3

K

sp AgCl is 1.0 x 10

• 10 Solution a) Addition of 0.0 ml Ag + **[Cl

] = 0.100 M pCl = - log [Cl

] = - log 0. = 1** b) Addition of 20.0 ml Ag + Initial mmol Cl

= 100.0 ml x 0.100 M = 10.0 mmol mmol added Ag + = 20.0 ml x 0.100 M = 2.0 mmol mmol Cl

left **= 8.0 mmol [Cl

] left = 8.0 = 0.0667 M ( 100 + 20 ) ml pCl = - log [Cl

] = - log 0.** = 1.

c) Addition of 99.0 ml Ag d) Addition of^ 100.0^ ml Initial mmol Cl

• = 100.0 ml x 0.100 M = 10.0 mmol mmol added Ag

= 100.0 ml x 0.100 M= 10.0 mmol Equivalence point is reached. The solution contain saturated AgCl solution **Ksp = [Ag

][Cl

] = 1.0 x 10

• 10 [Cl

] = √Ksp = √1.0 x 10

• 10 = 1.0 x 10
• 5 pCl = - log 1.0 x 10
• 5 = 5**