Question

284 Report Sheet . Determination of the Dissociation Constant of a Weak Acid 2.96 2.99 4.37 4.55 C. Determination of pKof Unknown Acid First determination Second determination ml NaOH pH ml NaOH pH 0.00 mL 0.00 mL 1.00 mL 2.00 mL 2.00 mL 4.00 mL 3.64 3.00 mL 6.00 ml 3.97 4.00 mL 7.00 ml 4.17 5.00 ml 8.00 mL 6.00 mL 9.00 ml. 7.00 mL 9.50 mL 4.61 8.00 ml 10.00 ml 4.71 9.00 mL 10.50 ml 4.85 10.00 mL 4.61 12.00 ml 1100 ml 14.00 ml 12.00 ml 16.00 ml 13.00 ml 18.00 mL -14.00-ml. 18.50 mL 15.00 ml 19.00 ml 7.91 16.00 ml 19.50 ml 8,81 17.00-ml. 20.00 ml 11.21 -18.00 ml 11.71 20.50 ml -19.00-ml. 21.00 ml 20.00 ml. 25.00 mL 1299 21.00 ml. 11.53 22.00 ml 12.03 23.00 ml 12.38 24.00 ml 12.56 25.00 ml 12.71 Third determination ml NaOH PH 0.00 mL 3.01 4.00 mL 3.71 8.00 mL 4.51 9.00 mL 4.61 9.50 mL 4.72 10.00 ml 4.81 10.50 ml 4.91 11.00 mL 5.13 19.00 ml 7.99 19.50 ml R39 20.00 ml 1139 20.50 ml 11.89 2500 ml 13.04 4.99 5.39 $161 11.99 10.98 Volume at equivalence point Volume at one half equivalence point pk K Average K, (show calculations) Standard deviation of K.

Answer 1

i phayainst ime at half. mino proti aud. Irun volume a equivalence = 20.00 ml " " half." 10.00 mL Now Pl corresponding to half-equivalena & volume equivalente : 4.11. PH PKA u bl ( pH = pka at half-equivalence] ka : 10-01 2.45x10-5. (Any) from and a un volume at equivalence = 20.00 volume at half equivalence = 10.00 mL PH at half. equivalence = 441 = pka Kaz 10-4.91 1.94 x 10-5 (Aus) equivalence = 20.00 mL m 21 run; volume at equivalence point = 20.00 m 2 Nolume at half equivalence = 10.00 mL PH at half-equivalence 4.91= pka Ka=104 :. Average ku = (1.44 + 9.45 +1.55)X 10 - 1.88x10-5 1.55010-5 Standard deviation I P 44-1-889+(1.98-1-55)* + (2.45-1-88)*3 *10-10 3-1 Joogaf 4.39410-0036 + 0.1089 + 0-3244) X10-10 x 0661x10- 6.61X106 - 3-305x10-6 (A) Hope it helps you. ☺