Q1. A sodium hydroxide solution is standardized by titrating

Question

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)

Answer 1

Q1:
First, calculate the moles of potassium acid phthalate:
0.8592 g / 204.22 g/mol = 0.0042 mol

Since the stoichiometry between potassium acid phthalate and sodium hydroxide is 1:1, the moles of sodium hydroxide is also 0.0042 mol.

Next, calculate the molarity of the base solution:
Molarity = moles / volume
Molarity = 0.0042 mol / 0.03267 L = 0.1285 M

Therefore, the molarity of the base solution is 0.1285 M.

Q2:
(i) The volume of HCl required to reach the equivalence point can be calculated using the balanced chemical equation of the reaction between NaOH and HCl. Since the stoichiometry is 1:1, the volume of HCl required to reach the equivalence point is 50.0 mL.

(ii) To calculate the pH at different points in the titration, you need to consider the reactions that occur with NaOH and HCl. At 0.0 mL of titrant, the solution is pure NaOH, so the pH can be calculated using the initial concentration of NaOH. At 10.0, 25.0, and 30.0 mL of titrant, the pH will depend on the amount of HCl added.

(iii) To plot a titration curve, you can plot the pH versus the volume of titrant added. The pH will change as the titrant is added due to the neutralization reaction between NaOH and HCl. You can calculate the pH at different points in the titration using the concentrations of NaOH and HCl at those points.