Question

Part 4: Calculating pH values as a Titration Progresses - this is a tough question! Finally, consider a 20.00 mL aqueous solution of 0.20 M H2A, where "HA" is some arbitrary diprotic acid (Kai = 4.35 x 10-7. K., = 4.85 x 10-11. Estimate the pH values of the solution after the addition of 0.00, 10.00, 20.00, 30.00, 40.00, and 60.00 ml of 0.20 M NaOH. a) Calculation #1 -0.00 ml NaOH Added: H2A exists alone in solution only until it reaches dissociative equilibrium - thus the pH calculation for this stage requires a simple ICE table calculation (Don't forget that since Kai >> Kaz, we can neglect protons formed from the further dissociation of HA™ to A2-...)

Answer 1

• Molarity of Malta o.2 Moles Volume of Naon = 0 as The required reaction? . И, н а на на о) - x x x+y . На к92 р. + хох+0 Ү- Ичку Kaiz 4.35x10 -7 kar = 4.85x10-11 Ка, >> ] , м >> Kay (x) (x+y) , х* (1) • 2-х в 2- 5 ч. 3 x | 1 , x k = 21. 2 xls 5 ч. 3 5 x x x x 7 - 2 - 0:2 - Х 2 - 0 2 ) 1 - 2 ) - г х 5 X: -3 х 2. ач х! Ч . р. -Ion (x) = - 1 1 - 49 = 1, 53/