Determine the value of the rate constant for the hydrolysis of ethyl lactate fro

Question

Determine the value of the rate constant for the hydrolysis of ethyl lactate from the following data. It was collected using exactly the same procedure as you will in the lab.

Mass of ethyl lactate (grams): 0.5554

Concentration of stock HCl solution (M): 0.00985

Standardization titration volumes (of NaOH):

Run

Vi (mL)

Vf(mL)

1

3.22

18.20

2

18.20

33.12

3

33.12

48.02

Kinetics data:

Run

Reaction time (seconds)

Vi(mL)

Vf(mL)

0

0

1

304

1.34

3.36

2

707

3.36

7.08

3

1318

7.08

12.30

4

1866

12.30

18.34

5

2450

18.34

24.98

Rate constant =

Lab Preocedure (Information that may be of use)

A. Preparation and standardization of 0.01M of ethyl lactate

1. Use a florence flask to prepare 500ml of 0.01M NaOH by quantitive dilution of the approximate 0.1M NaOH already prepared for you

2. From the side bench obtainn 200ml of the stock 0.01M HCl

3. Standardize the 0.01M of NaOH solution by titrating 15.00ml of 0.01M HCl with your NaOH solution, using 3 drops of bromothymol blue indicator

B.Preparation of Ethyl Lactate solution

1. To prepare 500ml of 0.01M solution of ethyl lacate deliver 0.59g(~25drops) of ethyl lacate into a weighed 20 ml beaker and weigh again

2. Transfer quantatively to the volumetric flask and fill to mark

C.Kinetic studies

1.Into each of the 5, 250ml erlenmeyer flasks deliver 75ml of distilled water, 10.00ml of 0.01M ethyl lactate and 4 drops of broothymol blue indicator

2. To start the reacion, add 10.00 ml of 0.01M NaOH. Start the stopwatch when half of the pipetful has been delivered. Allow it to stand. Record room temp

3.After 5 mins, cool in an ice bath for 45s swirling the solution, deliver 10.0ml of 0.01M HCl to "stop" the reaction. Record the stop time when half the pipetfull has been delivered

4. Add crushed ice to the solution in the flask. Back titrate immediately with 0.01M NaOH while swirling until it turns green. Then continue the titration until it turns blue

Run

Vi (mL)

Vf(mL)

1

3.22

18.20

2

18.20

33.12

3

33.12

48.02

Explanation / Answer

6        2450              18.34    24.98 24.98- 24.98 = 0 18.34 - 24.98 = -6.64 -    -

Vt = Final titre value

Slope = 0.0016 /min

Slope = KVt / 2.303

Slope = K X 24.98 / 2.303

K = 0.0016 X 2.303 / 24.98

K = 0.00015

For 10 ml ethyl lactate rate constant = 0.00015 min-1/ml

Run Reaction Time Vi (ml) Vf (ml) Vf - Vt Vi - Vt Vi - Vt /
Vf - Vt
log (Vi - Vt /
Vt - Vf) 1 0 2 304 1.34 3.36 3.36 - 24.98 = - 21.62 1.34 - 24.98 = -23.64 1.09 0.037 3 707 3.36 7.08 7.08 - 24.98 = -17.9 3.36 - 24.98 = -21.62 1.20 0.079 4 1318 7.08 12.30 12.30 - 24.98 = -12.68 7.08 - 24.98 = -17.9 1.41 0.149 5 1866 12.30 18.34 18.34 - 24.98 = -6.64 12.30 - 24.98 = -12.68 1.91 0.281

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