Question

The following data shows the titration of an unknown weak acid (called HA now) created by dissolving 1.78 g of this acid in distilled water.
Volume of 0.600 M NaOH added (1st measurement) pH of titrated solution (2nd #)
0.00 mL - ?
10.0 - 4.40
20.0 - ?
30.0 - 5.35
40.0 - 8.55
50.0 - 12.25

(a) The titration equivalence point occurred at the 40.0 mL data point. How does the data verify that HA is a weak acid rather than a strong one?
(b) Write a proper net ionic equation for the titrated of HA with NaOH .
(c) How many mole of HA were titrated in order to reach the equivalence point?
(d) What would the molar mass be for HA?
(e) Write the dissociation reaction for HA in water.
(f) If the initial [HA] = 0.500 M, find the pH of this weak acid given a Ka of 1.3x10^-5
(g) Find the pH at the 20.0 mL titration data point.

Answer 1

1. At half of the EP, i.e., at 20 mL , pH is about 4.97 (average value). The pH variation between 10 …30 mL added NaOH is small and indicates a buffer HA/A^- during titration.
2. HA + HO^- = A^- + H₂O     (net ionic equation)
3. 0.600 mol/L x 0.020 L = 0.012 mol   (added NaOH and consumed HA)
4. 1.78g/0.012 mol = 148 g/mol
5. HA + H₂O = A^- + H₃O⁺
6. pKa = 4.9  

      pH = 0.5pKa-0.5logC_acid =

            = 2.45 + 0.15 = 2.60

g. pH = pKa = 4.9     (here is the half EP where [HA]=[A-])