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(Solved): Part 1: Titration of a Polyprotic Acid A student titrates 20.00 mL of 0.450 M weak acid (H3A) with ...



Part 1: Titration of a Polyprotic Acid
A student titrates 20.00 mL of 0.450 M weak acid (H3A) with 0.750 M strong base (NaOH)

Part 1: Titration of a Polyprotic Acid A student titrates 20.00 mL of 0.450 M weak acid (H3A) with 0.750 M strong base (NaOH). The Ka values for the acid are as follows: Ka? = 2.5 x 10-3; Ka? = 3.5 x 10-5; Ka3 = 4.5 x 10-7 1. Determine the equivalence point volumes for all three equivalence points in this titration 2. Determine the Kb values for all three conjugates: Kb1 (A³), K2 (HA²), and K?3 (H?A) 3. Determine the initial pH of the titration solution (i.e., before base is added) 4. Determine the pH at each of the three equivalence points 5. Determine the pH after 4.00 mL of NaOH is added 6. Determine the pH after 18.00 mL of NaOH is added 7. Determine the pH after 37.00 mL of NaOH is added Part 2: Titration of a Polyprotic Base Sera titrates a 10.00 mL sample of a 0.0750 M weak base (A4) with 0.2500 M HCl, a strong monoprotic acid. The Kb values for this base are Kb1 = 2.00 x 10³, Kb2 = 3.00 x 106, Kb3 = 4.00 x 10¹?, and Kb4 = 5.00 x 10-¹¹. 8. Calculate the volume of HCL, in mL, needed to reach the first equivalence point of this titration. 9. Calculate the pH of the titration solution after 4.50 mL of HCI has been added. 10. Calculate the pH of the titration solution after 8.00 mL of HCI has been added. 11. Calculate the pH of the titration solution at the third equivalence point. 12. Calculate the pH at the fourth and final equivalence point in this titration.


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Part 1: 1. i) The first equivalence point corresponds to the point where 1 mole of NaOH has been added per mole of H3A in the solution being titrated, that is: H3A + HO- H2A- + H2O Moles of NaOH=moles of H3A MNaOHVNaOH=MH3AVH3A VNaOH=MH3AVH3A/MNaOH V
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