Determination of the Equilibrium Constant of an Unknown

Topics: Mole, Sodium hydroxide, PH Pages: 5 (986 words) Published: November 13, 2007
Determination of the Equilibrium Constant of an Unknown
Ester Hydrolysis Reaction

The experiments to follow determined that the equilibrium concentrations of the reaction: ester + water ↔ alcohol + acid, are equal to 0.0363 moles of ester, 0.2852 moles of water, and 0.0268 moles each of alcohol and acid. Using this information the equilibrium constant was determined to be 0.06938.

1. Introduction

In this lab the equilibrium constant, Kc, for the acid catalyzed reaction between an unknown ester and water to form an unknown alcohol and an unknown carboxylic acid was determined. The equation for the reaction is the following: R1COOR2 + H2O ↔ R2OH + R1COOH (eq.1) The symbol R represents a general hydrocarbon residue of some specified structure. The equilibrium concentrations were determined through a series of experiments where a known base, Sodium Hydroxide (NaOH), was titrated into different reaction mixtures until an endpoint was reached. The equilibrium concentrations were then placed into the following equation to determine the constant: Kc = [R2OH] [ R1COOH]

[R1COOR2 ] [H2O] (eq.2) Equation one and two can be generalized to the following:
ester + water ↔ alcohol + acid (eq.3) Kc = [alcohol]eq [ acid]eq
[ester ]eq [water]eq (eq.4)

2. Results and Discussion

First, reaction mixtures were prepared according to the protocol in the first 5 columns of table 1, and stored for two weeks until they came to chemical equilibrium.
Table 1. Reaction Mixtures and Titration Values
Bottle3.00 M HCl
(ml)Ester #3 (ml)
Density: 0.9342 g/ml
Molar Mass: 74.08g/molAlcohol (ml)
Density: 0.7914 g/ml
Molar Mass: 32.04g/molNaOH

Next, 500ml of a 0.7 M NaOH solution was prepared and then standardized using potassium hydrogen phthalate (KHP). Table 2 summarizes the titration values for the standardization of NaOH. Table 2. Titration Values of Base

(L)M base

M1= 4.984g KHP X 1mole X 1 = 0.7069 M
204.23 g 0.0345 L

Using the same calculations for the other trials the average molarity of the NaOH solution was determined to be 0.7103. Using the equation:
∑differences X 1000 = part per thousand (ppt) precision
3 (average)
it was found that the ppt precision was 6, where 2ppt is ideal and anything above 10ppt should be repeated for accuracy.

Once the base was standardized and the mixtures were allowed to reach equilibrium, the base was titrated into the mixtures, using a phenolphthalein indicator, until a pale pink endpoint was reached. Column 6 in table 1 summarizes the results of the titrations. The values obtained for bottles 1 and 1a can be averaged to determine the amount of NaOH that the catalyst, HCl, required. This value, which was 16.87, should then be subtracted from the values obtained for bottles 2-5, the remainder of which would be from the acid that was produced in the reaction.

Next using the information from bottle 2, the initial amounts of the reactants was calculated.

5.00 mL Ester #3 x 0.9342g Ester #3 x 1 mole Ester #3 = 0.0631 moles of Ester # 3
mL Ester #3 74.08 g Ester #3

Since no water was added to bottle 2, it was only necessary to calculate to amount of water that came from the HCl solution.

5.00 mL HCl soln. x 1.217g HCl...
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