Question

Hydrolysis of ethyl lactate pre-lab.

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): 0.6129

Concentration of stock HCl solution (M): 0.01016

Standardization titration volumes (of NaOH):

Run	Vi (mL)	Vf(mL)
1	2.92	17.96
2	17.96	32.88
3	32.88	47.88

Kinetics data:

Run	Reaction time (seconds)	Vi(mL)	Vf(mL)
0	0
1	370	3.68	6.54
2	686	6.54	10.82
3	1212	10.82	16.52
4	1802	16.52	23.18
5	2420	23.18	30.46

Rate constant = Answer

Answer 1

Standardization of NaOH. You have not mentioned what volume of HCl has taken!!! Lets consider 15ml ( you can change it)

volume used of NaoH

17.96-2.92 = 15.04 ml

32.88- 17.96 = 14.92 ml

47.88-32.88 = 15.00 ml

Mean 14.987 ml

Concernatation of NaOH = moles of HCl / volume of NaOH

0.01016 × 15 / 14.986 = 0.01017 M

Kinetic data

Lets take at 0 reaction second 0 ml of NaOH used.

Time	Vi	Vf	Vt	Vt-Vinfin	Vo-Vin/Vt-Vinf	Log10 (Vo-Vin)/(Vt-Vin)
370	3.68	6.54	2.86	-4.42	1.647058824	0.21670911
686	6.54	10.8	4.28	-3	2.426666667	0.385010125
1212	10.82	16.5	5.7	-1.58	4.607594937	0.663474292
1802	16.52	23.2	6.66	-0.62	11.74193548	1.06973969
2420	23.18	30.5	7.28

Now slope = Rate constant × Vinfinity / 2.303 = 0.0006

Rate constant = 0.00019 ml-1 sec-1