[Dolphinsiu]

An experiment has been done by a student by the following precedure:

1. Add 6 g of benzoic acid to 400 ml of DI water in 500 ml concical flask
    to make saturated solution.

2. Pour 250 ml of this solution into another conical flask (have been   
    previously heated at 40 C)at constant-temperature 40 C

3. After 25 mins, withdraw 10 ml with a 60 C-tempeature pipette.

4. Wrap this tip with small piece of filter paper and fasten with rubber 
    band and warn with hot-air blower.

5. Remove filter paper and discharge sample quickly to weighing bottle and weigh.

6. Wash into 250 ml concial flask, add about 20 ml of water and titrate with 0.0199M NaOH using phenolophthelain indicator.

Results has been recorded as follows
                                      1st            2nd              3rd
Mass of benzoic acid (g)    9.75           9.21              10
Average = 9.65 g
                                         1st            2nd              3rd
Volume of NaOH added (ml)    23.42        21.95           24.13
Average = 23.17 ml

Calculate the solubility of benzoic acid in moles per 1 kg of water at 40 C.

My attempted answer:

C6H5COOH + NaOH    --->   C6H5COONa + H2O

No. of moles of NaOH = 0.0199 x 23.17/1000 = 4.61 x 10^-4 mol

No. of moles of C6H5COOH = 4.61 x 10^-4 mol

No. of moles of 10 ml C6H5COOH = 4.61 x 10^-4 x 30/10
                                            = 1.38 x 10^-3 mol

Solubility of C6H5COOH at 40 C = 1.38 x 10^-3 / 9.65 x 10-3
                                           = 0.143 mol/1 kg of water

Am I right?

[Dolphinsiu]

I believe my work is correct!

Then I ask another question.

ln (S2/S1) = Delta H/ RT (1/T1 - 1/T2)

S1 and S2 are the solubilites at tmeperatures T1 and T2.

However, I still don't know why the calculated S2 from this equation is smaller than that measured in experiment (Delta H has been determined by solubilities at 20 C and 30 C and now we use T1 = 20 C and T2 = 40 C).

Is it due to precipation of benzoic acid on filter paper, or other reasons?