Question

An aqueous solution containing 1.00 g of oxobutanedioc acid (FM 132.07) per 100 mL was titrated with 0.09432 M NaOH.

1. Calculate the pH at th following volumes of added base: 0.5V_e1, V_e1, 1.5V_e2, V_e2, 1.05V_e2
2. Sketch the titration curve, using the values calculated above.
3. Which equivalence point would be best to use in this titration (which one is not blurred)?

Ka1 = 2.56

Ka2 = 4.37

Ve1 = equivalence point 1

Ve2 = equivalence point 2

The correct answers for part 1 are 2.56, 3.46. 4.37, 8.6, and 11.45 respectively. However, would you please show your work for all of them (specifically 1.05Ve2). Thanks.

Answer 1

* Answer - Let say oxobutanedoic acid = HA it is weak dipnotic acid HA & NaoH - NaHA +H₂O NaHA & NaoH > Na2O + H₂O J As we know, at first half equivalence point a buffer solution is formed with maximum buffer Capacity, whose pH is given as PH = 0. 5pkal = 0.5 x 2.56 = TPH₂ 1.281 TPH 2.56 2 At first equivalence point a solution of an amphiprotic salt NaHA is formed 4 Pkal + Pka2 2 pH - 2.56 + 4.37 3.465 2 :.PH. 3.465 3] At second equivalence point, A Solution of Salt NaA is formed . Whose pH as given by PH PKW + Pkaz tloge - 2 Where Cegh concenstation of Salt - mole of limiting reagent / total volume at second eqwratence point = 0.029 pH 14 + 4.37 + 10g (0.029) TpH = 8.6 at second half equivalence point a buffer Solution is formed with maximum buffer Capacity PH-0.5x pka 0.5 x 4.37 = 2.185 (p1.4571

at 1:05p acidic buffer (NaHA & Na₃A) is formed 1 pkol of log [A-] OTHA PH₂ 11.45