Question

This problem deals with Acid-base titrations and pH scale for a weak acid and a weak base.

Calculate the pH of the solution that results from adding 7.5 [ml] of Ammonia (NH3) to a beaker that contains, 100 ml of distilled water and 15 ml of 0.1 M Acetic acid (HC2H3O2).  

A buret containing 50 mL of 0.1 M Ammonia (NH3) is being used as the titrant.

The beaker containing 100 ml of distilled water, and 15 ml of 0.1 M Acetic acid (HC2H3O2) is the analyte.

Ka Acetic Acid = 1.7 x 10-5 at 25 C

pKa Acetic Acid = 4.76

Kb Acetic Acid = 5.8 x 10-10

pKb Acetic Acid = 9.24

Kb Ammonia = 1.8 x 10-5 at 25 C

pKb Ammonia = 4.75

Ka Ammonia = 5.6 x 10-10 at 25 C

pKa Ammonia = 9.25

Kw = 1.0 x 10-14  at 25 C

A) What is the calculated pH of the solution in the beaker before any of the titrant (NH3) is added?

B) What is the calculated pH of the solution in the beaker after 12 [mL] of titrant (NH3) is added?

C) What is the calculated pH of the solution in the beaker after 15 [mL] of titrant (NH3) is added?

D) What is the calculated pH of the solution in the beaker after 20 [mL] of titrant (NH3) is added?

Answer 1

A) [HC2H3O2] = 15×0.1 mmol/(100+15) mL = 0.013 M

pH = 1/2 {pKa - Log[HC2H3O2]}

= 1/2 {4.76 - Log(0.013)}

= 1/2 (4.76 + 1.88)

= 3.32

B) No.of mmol of CH3COOH = (15-12) mL × 0.1 mmol/mL = 0.3 mmol

[HC2H3O2] = 0.3 mmol/(15+12+100) mL = 0.00236 M

Now, pH = 1/2 {pKa - Log[HC2H3O2]}

= 1/2 {4.76 - Log(0.00236)}

= 1/2 (4.76 + 2.63)

= 3.7

C) pH = 7 + 1/2 (pKa - pKb)

= 7 + 1/2 (4.76 - 4.75)

= 7 + 0.005

= 7.005

D) No.of mmol of NH3 = (20-15) mL × 0.1 mmol/mL = 0.5 mmol

[NH3] = 0.5 mmol/(20+15+100) mL = 0.0037 M

pH = 14 - 1/2 {pKb - Log[NH3]}

= 14 - 1/2 {4.75 - Log(0.0037)}

= 14 - 1/2 (4.75 + 2.43)

= 14 - 3.59

= 10.41