Cliffs AP Chemistry, 3rd Edition

(d) Restatement: Equivalent mass of unknown acid.

GEM

moles of H furnished

grams of acid

= +

.

.

mole H

g acid

686 10

0 250

= # -+ 3

= 36.4 g/mole acid

2. A student wanted to determine the molecular weight of a monoprotic, solid acid,
   symbolized as HA. The student carefully measured out 25.000 grams of HA and
   dissolved it in distilled H 2 O to bring the volume of the solution to exactly 500.00 mL.
   The student next measured out several fifty-mL aliquots of the acid solution and then
   titrated it against standardized 0.100 M NaOH solution. The results of the three titrations
   are given in the table.

Trial Milliliters of HA Solution Milliliters of NaOH Solution

1 49.12 87.45

2 49.00 84.68

3 48.84 91.23

(a) Calculate the number of moles of HA in the fifty-mL aliquots.

(b) Calculate the molecular weight of the acid, HA.

(c) Calculate the pH of the fifty-mL aliquot solution (assume complete ionization).

(d) Calculate the pOH of the fifty-mL aliquot solution (assume complete ionization).

(e) Discuss how each of the following errors would affect the determination of the

molecular weight of the acid, HA.

(1) The balance that the student used in measuring out the 25.000 grams of HA

was reading 0.010 gram too low.

(2) There was an impurity in the acid, HA.

(3) The NaOH solution used in titration was actually 0.150 M instead of 0.100 M.

Answer:

2. Given: 25.000 grams of HA.
   Dissolved in H 2 O to make 500.00 mL of solution.
   Fifty-mL aliquots (samples) of acidic solution.
   0.100 M NaOH used for titration.
   See results of titrations in given table.

Part II: Specific Topics