Home 11th Class Class 11 Chemistry Notes Stocihiometry – Equivalent Concept and Titration

Class 11 Chemistry Notes Stocihiometry – Equivalent Concept and Titration

Anand Meena

May 5, 2014

Volumetric analysis:

This mainly involves titrations based chemistry. It can be divided into two major categories.

Non-redox system
Redox system

Non-redox system

This involves following kind of titrations:

Acid – Base titrations
Back titration
Precipitation titration
Double indicator acid base titration

Titrimetric Method of Analysis:

A titrimetric method of analysis is based on chemical reaction such as.

aA + tT → product.

Where ‘a’ molecules of “analysis”, A reacts with t molecules of reagent T.

T is called Titrant normally taken in biuret in form of solution of known concentration. The solution of titrant is called “standard solution”.

The addition of titrant is added till the amount of T, chemically equivalent to that of ‘A’ has been added. It is said equivalent point of titration has been reached. In order to know when to stop addition of titrant, a chemical substance is used called indicator, which responds to appearance of excess of titrant by changing color precisely at the equivalence point. The point in the titration where the indicator changes color is termed the ‘end point’. It is possible that end point be as close as possible to the equivalence point. The term titration refers to process of measuring the volume of titrant required to reach the end point.

Acid- Base titration.

To find out strength or concentration of unknown acid or base it is titrated against base or acid of known strength. At the equivalence point we can know amount of acid or base used and then with the help of law of equivalents we can find strength of unknown.

Meq of acid at equivalence point = Meq of base at equivalence point

Ex.1 One liter solution of alkali (NaOH) is prepared by dissolving impure solid of alkali which contains 5% Na₂CO₃ and 8% CaCO₃ and 10% NaCl. A 10 ml portion of this solution required 9.8 mL of a 0.5 M H₂SO₄ solution for neutralization. Calculate weight of alkali dissolved initially.

Sol. gm eq. of Na₂CO₃ + gm eq. of CaCO₃ + gm eq. of NaOH = gm eq. of H₂SO₄

Back titration

Back titration is used in volumetric analysis to find out excess of reagent added by titrating it with suitable reagent. It is also used to find out percentage purity of sample. For example in acid-base titration supposes we have added excess base in acid mixture. To find excess base we can titrate the solution with another acid of known strength.

Ex.2 20 g. of a sample of Ba(OH)₂ is dissolved in 50 ml. of 0.1 N HCl solution. The excess of HCl was titrated with 0.1 N NaOH. The volume of NaOH used was 20 cc. Calculate the percentage of Ba(OH)₂ in the sample.

Sol. Milli eq. of HCl initially =50×0.1=5

Milli eq. of NaOH consumed = Milli eq. of HCl in excess =20×0.1=2

∴ Milli eq. of HCl consumed = Milli eq. of Ba (OH)₂=5-2=3

∴ eq. of Ba(OH)₂ = 3 / 1000 = 3 × 10^-3

Mass of Ba (OH)₂ = 3 × 10^-3 × (171/2) = 0.2565 g

% Ba (OH)₂=(0.2565/20)×100=1.28% Ans.

Ex.3 4.0 g of monobasic, saturated carboxylic acid is dissolved in 100 mL water and its 10 mL portion required 8.0 mL 0.27 M NaOH to reach the equivalence point. In an another experiment, 5.0 g of the same acid is burnt completely and CO₂ produced is absorbed completely in 500 mL of a 2.0 N NaOH solution. A 10 mL portion of the resulting solution is treated with excess of BaCl₂ to precipitate all carbonate and finally titrated with 0.5 N H₂SO₄ solution. Determine the volume of the acid solution that would be required to make this solution neutral.

Sol. Meq. Of NaOH=8×0.27=Meq. Of acid for 10 ml of acid solution

Meq. Of acid for 100 ml of solution

= (8 × 0.27 × 1000)/10 = (4/M) × 10³

M (acid) = 4/(8 × 0.27 × 10) × 1000 = 185.2

Formula of acid. = C_nH_2nO₂

Þ M = 14n + 32 = 185.2

Þ n = 11

Now 5g acid will produce (5/185.2) × 11 = (55/185.2) mol CO₂ after complete combustion.

2NaOH + CO₂ → Na₂CO₃ + H₂O

Total mass of NaOH available = 500 × 2 × 10^-3 = 1.0 mole

Moles of NaOH left unreacted = 1-(2 × 55/185.2) = (76/185.2) in 500 mL

Þ Molarity of NaOH after precipitation of Na₂CO₃ = 0.812

Therefore, 0.812 × 10 = 0.5 × V

Þ V = 16.24 mL Ans.

Precipitation titration:

In ionic reaction we can know strength of unknown solution of salt by titrating it against a reagent with which it can form precipitate. For example NaCl strength can be known by titrating it against AgNO₃ solution with which it forms white ppt. of AgCl.

So Meq. Of NaCl at equivalence point =Meq. Of AgNO₃ used= Meq. Of AgCl formed

Ex.4 A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a standardized acid as follows:

Co (NH₃)_x Cl₃ (aq) + HCl → NH⁺₄(aq) + Co³⁺ (aq) + Cl^–(aq)

A 1.58 g complex required 23.63 mL 1.5 M HCl to reach the equivalence point. Determine formula. If the reaction mixture at equivalence point is titrated with excess of AgNO₃ solution, what mass of AgCl will precipitate out?

Sol. The balanced chemical reaction is:

Co(NH₃)_x Cl₃ + xHCl → xNH⁺₄ + Co³⁺ + (x + 3) Cl^–

Precipitation titration