Question

Kinetic Study of an Iodine Clock Reaction.

Flask A contains 5.00 mL of 6 M HCl (density 1.11 g/mL). Flask B contains 5.00 mL of the same 6M HCl solution used in flask A, 2.00 mL of ethyl acetate (density 0.893 g/mL), and 3.00 mL of water (density 1.00 g/mL). Flask A was titrated with 1.00 M NaOH, and it required 31.95 mL to reach a phenolphthalein end point. Flask B was sealed and allowed to sit for 7 days to allow the system to reach equilibrium. At the end of the 7 day period, flask B was titrated with 1.00 M NaOH, and it required 42.75 mL of NaOH solution to reach a phenolphthalein end point.
Using the information above answer the following questions for the system outlined below.
ethyl acetate + water <--> ethyl alcohol + acetic acid

1-Based on the titration information for flask A, determine the number of moles of HCl in 5.00 mL of the 6 M HCl solution used for the experiment. Note that the concentration of the HCl solution is approximately 6M, and the actual concentration may be slightly higher or lower. Do not include units in your answer.

2-Calculate the moles of water in a 5.00 sample of the 6M HCl solution used for the experiment. Do not include units in your answer.

3-Determine the initial moles of water in flask B immediately after mixing. Do not include units in your answer.

4-Determine the initial moles of ethyl acetate in flask B immediately after mixing. Do not include units with your answer.

5-Determine the moles of acetic acid in flask B at equilibrium based on the titration data for flask B. Note that the HCl that was added to the flask is still present at equilibrium. Do not include units in your answer.

6-Using the moles of acetic acid at equilibrium, the stoichiometry of the reaction, and the initial number of moles of water, determine the moles of water at equilibrium. Do not include units with your answer.

7-Using the moles of acetic acid at equilibrium, the stoichiometry of the reaction, and the initial number of moles of ethyl acetate, determine the moles of ethyl acetate at equilibrium. Do not include units with your answer.

8-Calculate the equilibrium constant for the system.

Answer 1

1. For an acid-base titration experiment, V_aN_a = V_bN_b, V - volume of solution and N-Normality (no. of moles of H+/1L solvet)

Since for HCl, no. of moles of H+ is equal to that of HCl, N_a = 1.00*31.95/5.00 = 6.39.

no. of moles of HCl = 0.03195

2.weight of 5.00 mL HCl = 5.00*1.11 = 5.55

weight of HCl in 5.00mL = (6.39*36.55)*(5/1000) = 1.1678

weight of water = 5.55-1.1678 = 4.3822 and moles of water = 4.3822/18 = 0.2435

3. wt. of water = 4.3822+(3.00*1.00) = 7.3822 and moles of water = 7.3822/18 = 0.410

4. no, of moles of ethyl acetate (initial) = (2*0.893)/76 = 0.0235

5. Total volume in flask B, before titration = 5+2+3 = 10.00

moles of H+ = (42.75*1.00)/10.00 = 4.275 and therefore the no. of moles of acetic acid + HCl = 0.04275

No. of moles of acetic acid at equilibrium = 0.04275 - 0.03195 = 0.0108

6. amount of acetic acid produced = amount of water consumed

therefore the no. of moles of water at equilibrium = 0.410 - 0.0108 = 0.3992

7. amount of acetic acid produced = amount of ethyl acetate consumed

therefore the no. of moles of ethylacetate at equilibrium = 0.0235 - 0.0108 = 0.0127