Hariri High School II

Chemistry Final Exam

Name: ______Date: ,2011

Grade: 12 G.S. Time: 120 min.

First Exercise (7pts.)

Benzoic Acid and Sodium Benzoate

Benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa) are used as food preservatives especially in beverages that do not contain alcohol.

The aim of this exercise is to study the behavior of these two compounds in an aqueous solution.

Given:

-This study is performed at 25 ºC.

-Solubility of benzoic acid: s = 2.5 g.L-1.

-Molar mass in g.mol-1 : benzoic acid M1 = 122, sodium benzoate M2 = 144.

-Sodium benzoate is an ionic compound highly soluble in water.

- Ion product of water Kw = 10-14.

- pKa (C6H5COOH/C6H5COO – ) = 4.2.

I- Aqueous Solution of Sodium Benzoate

A mass m = 0.36 g of sodium benzoate is dissolved in distilled water to obtain a volume V of solution of concentration C = 1.0x10-2 mol.L-1.

The pH of this solution is equal to 8.1.

A. Choose, among the available materials of the list given below, the required set for this

preparation. (Calculation is necessary). (1pt.)

-Sensitive balance.

-50, 100 and 250 mL graduated cylinders.

-Funnel.

-50, 250 and 500 mL volumetric flasks.

-Spatula.

-Watch glass.

B.Write the equation of the reaction between benzoate ion and water. (0.5pts.)

C. Determine the degree of conversion of benzoate ion into benzoic acid. (1pt.)

II- Verification of the Solubility of Benzoic Acid

A mass m´ of benzoic acid is introduced into a beaker containing distilled water.

The content of the beaker is stirred; small particles remain in suspension. After filtration,

a solution (S) is obtained.

A volume Va = 20 mL of (S) is titrated with sodium hydroxide solution (Na++HO –) of concentration Cb = 3.0x10-2 mol.L-1, in the presence of a convenient indicator. The added volume to reach the equivalence point is VbE = 13.6 mL.

A. Write the equation of the titration reaction. (0.5pts.)

B. The following two indicators are available in the laboratory:

-Methyl orange of pH range: red 3.1 – orange – 4.4 yellow

-Phenolphthalein of pH range: colorless 8.2 – pink – 10 purple.

1. Choose, by justifying, the convenient indicator for this titration. (0.5pts.)

2. Explain how you detect the equivalence point of this titration.(0.5pts.)

C. Determine the concentration Ca of solution (S). (0.75pts.)

D. Verify the solubility of benzoic acid given above. (0.75pts.)

III- Preparing a Mixture of two Solutions: Benzoic Acid and Sodium Benzoate

A volume V1 of benzoic acid solution of concentration C1 = 5.0x10-3 mol.L-1 and a volume V2 of sodium benzoate solution of concentration C2 = 5.0x10-3 mol.L-1 are mixed. The volume of the obtained solution is V = 300 mL and its pH is equal to 4.0.

A. What do you call this solution? Give its characteristics. (0.5pts.)

B. Determine V1 and V2. (1pt.)

Second Exercise (6.5pts.)

Banana Aroma

Banana aroma is an ester called isoamyl acetate, or 3-methylbutylethanoate according to IUPAC rules. It can be synthesized by the esterification reaction between ethanoic acid and isoamyl alcohol.

1mol of alcohol is mixed with 1mol of acid. At certain intervals of time, this mixture is transferred into an Erlenmeyer flask and cooled down directly. The acid present in each sample is titrated with sodium hydroxide solution. The quantity of acid present is denoted by na and is tabulated below:

t (hrs) / 0 / 1 / 2 / 4 / 6 / 8 / 10 / 15 / 20 / 25
na(mol) / 1 / 0.82 / 0.7 / 0.54 / 0.46 / 0.41 / 0.38 / 0.35 / 0.34 / 0.34
nester (x10-2 mol ) / 30 / 59 / 65 / 66

A. Show that nester = 1-na. (0.5pts.)

B. Calculate the missing values of nester in the above table. (0.5pts.)

C. At what time does the above system reach equilibrium? Calculate the degree of

conversion α at that time. (0.75pts.)

D. Find Kc for this reaction. (0.5pts.)

E. The same experiment is repeated by adding H+ ions to the reaction mixture at start.

What role do these ions play in this reaction? What is their effect on the degree of

conversion α?(0.5pts.)

F. Plot the curve nester = f(t). Determine the instantaneous rate at t=8hrs.

Scale: 1cm → 2hrs, 1cm → 5x10-2mol.(1.5pts.)

G. Determine the half life of the reaction.(0.75pts.)

H. Graphically, interpret qualitatively the variation of the rate of the reaction with time.

What kinetic factor explains this variation?(0.75pts.)

I. What is the purpose of cooling the reaction mixture before titration?(0.25pts.)

J. Starting with the same amounts of ester and acid, the synthesis is carried at a higher

temperature. Draw, on the same graph paper, the sketch of the curve nester = f(t) at the

newtemperature.(0.5pts.)

Third Exercise (6.5pts.)

Degree of Fermentation– Organic Compounds

The two parts of the problem are independent.

I. Degree of Fermentation

The alcoholic fermentation of glucose syrup ceases when the alcoholic degree of the mixture attains 16o (16mL ethanol in 100mL solution). 200L of 1.94mol/L glucose syrup is to be fermented. After a certain period of time t, a sample of 25mL of the mixture is removed to study the degree of the solution. The mass of the glucose left is 3g.

Given:

the variation of the volume of the solution is negligible.
The density of ethanol is 0.79g/mL.
M(C6H12O6) =180g/mol, M(C2H5OH) = 46g/mol.

A. Write the equation of the fermentation reaction. (0.25pts.)

B. Find the volume of ethanol in the sample at t. (1.25pts.)

C. Calculate the alcoholic degree at time t. Does the fermentation cease at time t?

(0.75pts.)

II. Organic Compounds

542

Identify A, B, C, D, E and F. (1.5pts.) Write the corresponding reactions. (2.75pts.)