Question

# 1. For the titration of 25.0 mL of 0.1 M HCl (aq) with 0.1 M NaOH...

1. For the titration of 25.0 mL of 0.1 M HCl (aq) with 0.1 M NaOH (aq), at what volume of NaOH (aq) should the equivalence point be reached and why? If an additional 3.0 mL of 0.1 M NaOH (aq) is then added, what is the expected pH of the final solution?

2. What is the initial pH expected for a 0.1 M solution of acetic acid? For the titration of 25.0 mL of 0.1 M acetic acid with 0.1 M NaOH (aq), at what volume of NaOH (aq) is the equivalence point reached? Is the pH at the equivalence point greater than, less than or the same as in problem #1 above? Explain.

3. What is the initial pH expected for a 0.1 M solution of phosphoric acid (H3PO4)? For 25.0 mL of 0.1M H3PO4 (aq), what volume of 0.1 M NaOH (aq) is required to fully titrate all three protons to their end points?

4. In the titration of a weak acid with a strong base, how is the half equivalence point determined and what is its significance? How are the pKa and Ka of the weak acid determined from the half equivalence point?

5. Carefully read and study the lecture text sections related to this experiment. 5. Prepare and set up your lab notebook. Be sure to leave spaces for recording observations.

1. )

at eqiuivalence point : moles of acid = moles of base

25.0 x 0.1 = 0.1 x V

V = 25.0 mL

volume of NaOH needed = 25.0 mL

millimoles of acid = 25 x 0.1 = 2.5

millimoles of base = 3 x 0.1 = 0.3

here acid > base . so solutions gets acidic nature pH < 7

[H+] = 2.5 - 0.3 / total volume = 2.2 / 25 + 3 = 0.0786 M

pH = -log [H+] = -log (0.0786)

pH = 1.10

2)

C = 0.1 M

Ka for acetic acid = 1.8 x 10^-5

pKa = -log Ka = -log (1.8 x 10^-5) = 4.74

pH = 1/2 [pKa -logC]

pH = 1/2 [4.74 -log 0.1]

pH = 2.87 ------------------------------> initial pH

at equivalence point moles of acid = moles of base

25 x 0.1 = 0.1 x V

so volume of NaOH = 25 mL

.pH at equivalence point is more than problem #1

weak acid + strong base salt is formed . its pH more than 7

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/2 [4.74 + log 0.1]

pH = 8.87