Q1. A sodium hydroxide solution is standardized by titrating
0.8592 g of primary standard potassium acid phthalate to a
phenolphthalein end point, requiring 32.67 mL. What is the
molarity of the base solution?
• Q2. Calculate the
• (i) Volume of HCl required to reach the equivalence point.
• (ii) pH at 0.0, 10.0, 25.0, and 30.0 mL of titrant in the titration
of 50.0 mL of 0.100 M NaOH with 0.200 M HCl.
• (iii) Using your titration data, plot a titration curve. (i.e. pH Vs
mL of titrant).
Q1.
Moles of potassium acid phthalate = 0.8592 g / 204.23 g/mol = 0.00421 mol
Moles of NaOH used = Moles of potassium acid phthalate = 0.00421 mol
Volume of NaOH used = 32.67 mL = 0.03267 L
Molarity of NaOH solution = Moles of NaOH / Volume of NaOH used
= 0.00421 mol / 0.03267 L
= 0.1289 M
Therefore, the molarity of the NaOH base solution is 0.1289 M.
Q2.
(i) Volume of HCl required to reach the equivalence point:
Since NaOH is in excess, the volume of HCl required to reach the equivalence point is equal to the initial volume of NaOH used, which is 50.0 mL.
(ii) pH at different points in the titration:
- At 0.0 mL of titrant: pH = 14 (since NaOH is strong base)
- At 10.0 mL of titrant: pH = 13.3 (close to the equivalence point, excess OH- ions)
- At 25.0 mL of titrant: pH = 7.0 (equivalence point, neutral solution)
- At 30.0 mL of titrant: pH = 1.5 (excess H+ ions from HCl)
(iii) Titration curve:
A titration curve can be plotted with pH on the y-axis and mL of titrant on the x-axis. The curve will start at a high pH (around 14), decrease as the HCl is added, reach a midpoint at pH 7 (equivalence point), and then decrease sharply as excess HCl is added beyond the equivalence point.