Argentometric Titrations: Principles, Methods, and Applications, Slides of Chemistry

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ARGENTOMETRIC

TITRATIONS

Argentometric Titrations

• Argentometry is derived from a latin word Argentum

which means silver.

• It is the most widely applicable precipitation titrations

involving the use of silver nitrate with chlorides,

bromides, iodides, and thiocyanate.

• Since silver is always there, precipitation titrations

are referred to as Argentometric titrations.

• This implies that this type of titration is relatively

limited.

• According to end point detection method, three main

procedures are widely used depending on the type of

application.

The Mohr Method

• This method utilizes chromate as an indicator. Chromate forms a

precipitate with Ag

. Therefore, AgCl is formed first and after all

Cl

is consumed, the first drop of Ag

(titrant) in excess will react

with the chromate indicator giving a reddish precipitate

2Ag

+ CrO

Ag

CrO

(s)

titrant indicator red precipitate

• The concentration of titrant rises sharply near the equivalence

point, and the solubility of Ag

CrO

is exceeded.

• The appearance of red precipitate marks the endpoint.

• The applicability of the Mohr method is limited compared to

either of the other chemical indicator methods: it can be used to

analyze for Cl

or Br

anions.

The Mohr Method

• In this method, neutral medium should be used

since, in alkaline solutions pH > 10 , silver will

react with the hydroxide ions forming AgOH.

• In acidic solutions, chromate will be converted to

HCrO

4

and the end point is delayed. In other

words silver chromate solubility grows due to the

protonation of chromate anions. Therefore, the

pH of solution should be kept at about 7

The Volhard Method

• The indicator in Volhard titrations is Fe3+, which reacts

with titrant to form a red colored complex:

Fe

3+

+ SCN− Fe(SCN)

2+

(aq)

indicator titrant red complex

• This is a good method for the analysis of halide ions (F-,

Cl

, Br

, I

) and anions such as phosphates and

chromates in acidic environments using silver ions. If the

Volhard titration is not performed in an acidic medium,

the iron ion will precipitate as hydrated oxide (red brown

gelatinous Fe(OH)

3

, Fe

2

O

3

.H

2

O).

[n

Cl

- + n

SCN

- = n

Ag+

]

The Fajans Method

• It’s a direct titration that uses an adsorption

indicator.

• Adsorption indicators function in an entirely

different manner than the chemical indicators and

they can be used in many precipitation titrations,

not just argentometric methods.

• Imagine that we wish to analyse Cl−^ in a sample

solution by titrating with Ag

; the titration

reaction

would be

Ag

+ Cl

−

AgCl(s)

The Fajans Method

• After the equivalence point, there will be

an excess of titrant Ag

, some of these will

adsorb to the solid AgCl particles, resulting

in a positive charge. The figure below

illustrates this concept.Before^ equivalence^ point^ After^ equivalence^ point

Excess Cl analyte Excess Ag titrant Surface Charge Negative^ Positive Titration Cl-^ AgCl Cl

Cl

Cl

• (^) AgCl Ag

Ag

Ag

Ag

Before the equivalence point (^) After the equivalence point

Adsorption indicators are dyes, such

as dichlorofluorescein (shown below,

greenish yellow colour), that usually

exist as anions in the titration

solution.

The Fajans Method

O COO O Cl O Cl Dichlorofluoroscein

The Fajans Method

• Titration of Cl-^ with Ag+^ using

dichlorofluorescein as the indicator. The pink

end point is reached when the anionic

indicator becomes adsorbed on the cationic

particles of the precipitate

Yellow Pink

Class Exercise

1. A 0.32 g sample containing KCl (mw = 74.6)

is dissolved in 50 mL of water and titrated to

the Ag

CrO

end point, requiring 16.9 mL of

0.1 M AgNO

. A blank titration requires 0.

mL of titrant to reach the same end point.

Calculate the %w/w KCl in the sample?

Ans: To find the moles of titrant reacting with the sample, we first need to correct for the reagent blank thus: VAg=16.9 mL−0.7 mL = 16.2 mL (0.1 M AgNO 3 )X(0.0162 L AgNO 3 )=1.62× − moles of Ag

= moles of KCl weight KCl = 1.62X

• X 74.6 = 0.12 g. % KCl = ( 0.12/ 0.32 ) X100 = 37.5.

Class Exercise

3. 400 mg of a butter was heated and some water was added. After shaking and filtration, 10 ml of 0.2 M Ag + solution , some HNO 3 , drops of Fe 3+ solution and some nitrobenzene were added to the Filtrate. The excess Ag + in the aqueous layer was titrated with 0.1 M SCN - standard solution. If the volume of SCN - at the equivalent point was 15 mL , calculate the percentage of NaCl (mw = 58.5) in the butter sample? Ans: #moles NaCl = #moles Cl

= total #moles Ag

• excess #moles Ag

= 10 X 0.2 - 15 X 0.1 = 0. %w/w NaCl = ((05 x 58.5)/400) * 100 = 7.

Class Exercise

4. The sulphide contents of 100 mL of a water sample was titrated with a standard solution of 0.01 M AgNO 3 according to the following reaction equation : 2Ag+^ + S2-^ ↔ Ag 2 S. If the volume of AgNO 3 solution at the equivalent point was 8.5 mL , calculate the concentration (ppm) of H 2 S (mw = 34) in the water sample? Ans: MAgVAg MS^2 - VS^2 - nAg n S 2 - MAgVAg nAg n S 2 - MS^2 - VS 2 -
5. 01 M x 8. 5 mL x 1 100 mL x 2 MS 2 - 000425 M 000425 mol/L x 34 g/mol
6. 01445 g/L (^) ppm = mg/Kg = mg/L
7. 45 ppm

Class Exercise

6. A 2.0 gram sample containing Cl

   ( aw = 35.50 ) and ClO 4 - (mw= 99.5 ) 

was dissolved in sufficient water to give 250 mL of solution. A 50 mL aliquot required 14.0 mL of 0.09 M AgNO 3 in a Mohr’s titration. A 50. mL aliquot was then treated with V 2 (SO 4 ) 3 to reduce the ClO 4

to Cl

, following which titration required 40 mL of the same silver nitrate solution. Calculate the percent of Cl-^ and ClO 4 -^ in the sample? Ans: MCl- MAg+VAg+ VCl- MCl-

1. 09 M x 14 mL 50 mL
2. 0252 M
3. 0252 M x 35. 5 g/mol
4. 8946 g/L MClO- MAg+VAg+ VClO- MClO-
5. 09 M x ( 40 - 14 ) mL 50 mL
6. 0468 M
7. 0468 M x 99. 5 g/mol
8. 6566 g/L %w/w of Cl

= 0.8946(g/L)/8 (g/L) x 100 = 11.18% %w/w of ClO

= 4.6566 (g/L)/ 8 (g/L) x 100 = 58.2%