Double Titration , Iodometric Titration , Idiometric Titration

The method involves two indicator (Indicators are substances that change their colour when a reaction is complete) phenolphthalein and methyl orange. This is a titration of specific compounds.

Let us consider a solid mixture of NaOH , Na2CO3 and inert impurities weighing w g. You are asked to find out the % composition of mixture. You are also given a reagent that can react with the sample, say, HCl along with its concentration (M1).

We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely phenolphthalein and methyl orange.

Now, we titrate this solution with HCl. NaOH is a strong base while Na2CO3 is a weak base. So it is safe to assume that NaOH reacts with HCl first, completely and only then does Na2CO3 react.

NaOH + HCl → NaCl + H2O

Once NaOH has reacted, it is the turn of Na2CO3 to react. It reacts with HCl in two steps:

Na2CO3 + HCl → NaHCO3 + NaCl and then,

NaHCO3 + HCl → NaCl + CO2 + H2O

As can be seen, when we go on adding more and more of HCl, the pH of the solution keeps on falling. When Na2CO3 is converted to NaHCO3, completely, the solution is weakly basic due to the presence of NaHCO3 (which is a weaker base as compared to Na2CO3Na2CO3).

At this instant phenolphthalein changes colour since it requires this weakly basic solution to change its colour.

Therefore, remember that phenolphthalein changes colour only when the weakly basic NaHCO3 is present.

As we keep adding HCl , the pH again falls and when all the NaHCO3 reacts to form NaCl, CO2 and H2O the solution becomes weakly acidic due to the presence of the weak acid H2CO3(CO2 + H2O).

At this instance methyl orange changes colour since it requires this weakly acidic solution to do so.

Therefore, remember methyl orange changes colour only when H2CO3 is present.

Now, let us assume that the volume of HCl used up for the first and the second reaction,

i.e. NaOH + HCl → NaCl + H2O and Na2CO3 + HCl NaHCO3 + NaCl be V1 (this is the volume of HCl from the beginning of the titration up to the point when phenolphthalein changes colour).

Let the volume of HCl required for the last reaction, i.e.,

NaHCO3 + HCl →  NaCl + CO2 + H2O be V2 (this is the volume of HCl from the point where phenolphthalein had changed colour upto the point when methyl orange changes colour).

Then, moles of HCl used for reacting with NaHCO3 = moles of NaHCO3 reacted = M1V2

moles of NaHCO3 produced by the Na2CO3 = M1V2

moles of Na2CO3 that gave M1V2 moles of NaHCO3 = M1V2

Mass of Na2CO3 = M1V2 × 106

% Na2CO3 $\large = \frac{M_1 V_2 \times 106}{w} \times 100 $

moles of HCl used for the first two reactions = M1V1

moles of Na2CO3 = M1V2

moles of HCl used for reacting with Na2CO3 = M1V2

moles of HCl used for reacting with only NaOH = M1V1 – M1V2

∴ moles of NaOH = M1V1 – M1V2

Mass of NaOH = ( M1V1 – M1V2) × 40

% NaOH $\large  = \frac{(M_1 V_1 – M_1 V_2)\times 40}{w}\times 100 $

Iodometric and Idiometric titration

The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titration’s.

I2 + 2e– → 2I–  (reduction)

2I → I2 + 2e–  (oxidation)

These are divided into two types

(a) Iodometric titration: In iodometric titrations, an oxidising agent in allowed to react in neutral medium or in acidic medium with excess of potassium iodide to librate free iodine.

KI + oxidising agent → I2

Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate Halogen, dichromates, cupric ion, peroxides, etc can be estimated by this method.

I2 + 2Na2S2O3 → 2NaI + Na2S4O6

2CuSO4 + 4KI → Cu2I2 + 2K2SO4 + I2

K2Cr2O7 + 6KI + 7H2SO4 → Cr2(SO4)3 + 4K2SO4 + 7H2O + 3I2

(b) Iodiometric titration: These are the titrations in which free iodine is used as it is difficult to prepare the solution of iodine (volatile and less soluble in water) it is dissolved in KI solution.

KI + I2 → KI3 (Potassium triiodide)

This solution is first standardises before use with the standard solution of I2 substance such as sulphite, thiosulphate, arsenite etc. are estimated.

The iodimetric and iodometric titrations, starch solution is used as indicator. Starch solution gives blue or violet colour with free iodine. At the end point the blue or violet colour disappears when iodine is completely changed to iodide.

Also Read :

→ Mole Concept & Principle of Atom Conservation (POAC)
→ Gravimetric Analysis
→ Oxidation and Reduction
→ Balancing Redox Equations
→ Volumetric Analysis
→ Law of Equivalence
→ ‘ n ‘ Factor
→ Applications of the Law of Equivalence : Simple titration , Back titration
Volume of Strength of H2O2

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