ChemLab BK

BITSPilani, Pilani Campus1

1


2

Titrations involving iodine (I2)

A direct titration with only 1 reaction:Analyte + titrant (I2) → product

Indirect titration with 2 reactions:1. Analyte + I- → I2

2. I2 + titrant (Na2S2O3) → product

Iodometry Iodimetry

Reducing analyteStandard solution: Iodine (I2)

Oxidizing analyteStandard solution: Sodium thisoufate


3

2CuSO4 + 4KI 2CuI2 + 2K2SO4

2CuI2 Cu2I2 + I2

I2 + 2S2O32- S4O6

2- + 2I-

(blue) (colorless)

Principle: In neutral or faintly acidic solutions, cupric salts react with

iodide ion to liberate iodine.

The liberated I2 is titrated against the sodium thiosulphate (Na2S2O3.5H2O) solution using starch as indicator.

ESTIMATION OF COPPER BY IODOMETRY


4

• Sodium thiosulphate (hypo) is efflorescent nature therefore is a secondary standard.

• Double titration method. • Standardization of thiosulphate solution with a standard

CuSO4 solution • Standardized thiosulphate is used to titrate the CuSO4

solution of unknown strength.

Standard Solutions

SecondaryPrimary

Experimental Procedure


5

Experimental Procedure

0.5 to 0.7 g of copper sulfate

Transfer CuSO4 into a standard flask using a funnel.

Dissolve CuSO4 Add Na2CO3

Add glacial acetic acid

Make up the volume up to the mark

Preparation of standard (N/40) CuSO4.5H2O solution


6

Standardization of sodium thiosulfate

KI hypo KSCNStarch

hypo


7

Sources of error

1. Aerial oxidation of acidic iodide solution

4I- + 4H+ + O2 2I2 +2H2O

Use Na2CO3to remove oxygen in the reaction vessel by carbon dioxide. 2. Volatility of I2 formed

I- + I2 I3-

Use excess iodide solution which captures liberated iodine to form tri-iodide ions.


8

Equivalent weight of CuSO4. 5H2O is 249.7 gm. 2CuSO4 ≡ I2 ≡ 2Na2S2O3 ≡ 2e

97.24W (N)

Strength of copper sulphate solution

Standardization of hypo solution: N2 = =

2

11

VVN

297.2425V

W

Strength of unknown copper sulphate solution

N1’ =

12

2

'97.24'25VVVW

Report the strength of the given CuSO4 solution in g/l.


9

In a sports relay race, a chemical kinetics specialist runs slowly, and his group loses the race.When the chemical kinetics specialist is asked why he ran slowly, his reply was “I always wanted to be the significant rate determining step”.

Kinetics


10

KINETICS OF THE IODINATION OF ACETONE

Objective: To study the kinetics of the iodination of acetone in acidic medium using photometry, estimate the differential rate law and calculate the rate constant.

+HI2CH3 C

O

CH3 CH3 C

O

CH2I + HI

-d[I2]

dt= k [I2]x [CH3COCH3]y [H+]z

Determine the orders of the reaction x, y z Calculate the rate constant, k


11

x: reaction rate is studied using a large excess of acetone, and a high concentration of H+.

-d[I2]

dt= k' [I2]x

where k' = k [CH3COCH3]y [H+]z is a "pseudo rate constantIf x = 0, then [I2] = [I2]0 –k't x = 1, [I2] = [I2]0 e-k't

y: Initial concentration of I2 and H+ fixed and varying the initial concentration of acetonez: the initial concentration of I2 and CH3COCH3 fixed and varying the initial concentration of H+

Determination of the order


12

According to the Beer - Lambert law Absorbance A = εcb ε is the molar absorption coefficientb is the length of sample

Photometry


13

connection between the total concentration of iodine in solution and the absorbance A is obtained

Calibration Curve

Concentration

Abso

rban

ce


14

Rate Measurements

Experiment Acetone HCl H2O I2

1 2.0 ml 2.0 ml 4.0 ml 2.0 ml

2 4.0 ml 2.0 ml 2.0 ml 2.0 ml

3 2.0 ml 4.0 ml 2.0 ml 2.0 ml

4 2.0 ml 2.0 ml 2.0 ml 4.0 ml

Order wrt I2

Order wrt CH3COCH3: Exp 1 and 2Order wrt H+ : Exp 1 and 3For each experiment, calculate the rate constant


15

Time (s)

Abso

rban

ce

Rate constant = slope = y/x

Rate Measurements


16

Theoretical and Percent Yield

1. Balance the chemical equation 2. Find the limiting reagent


17

Find the limiting reagent when 1.22g O2 reacts with 1.05g H2 to produce H2O. Convert mass to moles: 0.038 mol O2, 0.5 mol H2

Calculate H2 moles necessary to react with O2: 0.076 mol

H2. Compare 0.076 mol H2 to actual mol of H2 (0.5mol H2), Since 0.5 mol H2 is more than 0.076 mol H2, H2 is the excess reagent and O2 is the limiting

reagent.

Limiting reagent


18

3. Find the theoretical yield: The maximum amount of product that could be formed from given amounts of reactants.4. Find the actual yieldThe amount of product actually formed or recovered when the reaction is carried out in the laboratory.5. Find the percentage yieldThe ratio of actual yield to theoretical yield.

percent yield = actual yield (g) x 100 theoretical yield (g)

Percent Yield


19

Weight of the crude product: 4 gWeight of recrystallized product from 1 g of crude product: 0.82 gPercent yield of pure product: 0.82 x 4 x 100 theoretical yield

Exercise

Percent Yield


20

Thank You

ChemLab BK

Documents

Transcript of ChemLab BK