Cliffs AP Chemistry, 3rd Edition

(a) Restatement: Moles of HA in fifty-mL aliquots.

At the end of titration, moles of HA = moles of NaOH.

average volume of NaOH ...

3

=87 45 84 68 91 23++

= 87.79 mL

moles HA = moles NaOH = VNaOH×MNaOH

..
.

liter

liter

mole mole

1

0 08779

1

==##0 100 87810 -^3

(b) Restatement: Molecular weight of HA.

.

.'

.'

.

878 10

48 99

500 00

25 00

MW

mole HA

mL HA sol n

mL HA sol n

gHA

= # 3 #

)

= 279 g/mole

* = average

(c) Restatement: pH of fifty-mL aliquots (assume 100% ionization).

Average volume of fifty-mL aliquots = 48.99 mL

.

..
ln
H liters solutionmoles H
liter HA so

mole H M

0 04899

+ ==++878 10# -^3 =0 179

7A l

pH = −log[0.179] = 0.747

(d) Restatement: pOH of fifty-mL aliquot.

pOH = 14.000 −pH = 14.000 −0.747 = 13.253

(e) Restatement: Effects of following errors.

(1) Balance reading 0.010 gram too low.

• Student would think she or he had 25.000 grams when there were actually
  25.010 grams.


• In the calculation of molecular weight, grams/mole, grams would be too low,
  so the effect would be a lower MWthan expected.
  (2) An impurity in the sample of HA.


• Student would have less HA than expected.


• In the calculation of molecular weight, g/mole, there would be less HA avail-
  able than expected. Therefore, in the titration against NaOH, it would take
  less NaOH than expected to reach the equivalence point. This error would
  cause a larger MWthan expected, because the denominator (moles) would
  be smaller.


• These results assume that the impurity does not have more H+/mass of impu-
  rity than the HA.

Acids and Bases