Based on the titration curve above, what volume of titrant was needed to reach t

Question

Based on the titration curve above, what volume of titrant was needed to reach the 15 mL 30 mL 4 mL 45 mL Based on the titration curve above, what would be the estimated pH at the equivalent 12.4 10.0 2.4 4.4 Based on the titration curve what would be the estimated pK_a. for the 12.4 10.0 4.4 7.2 Given the following information: Cd(s) l Cd^2+(aq) || Ag^+(aq) | Ag(s) E_cell degree = +1.20 V CD^2+(aq) + 2e^- rightarrow Cd(s) E_red degree = -0.401 V Calculate E_red degree for: Ag^+(aq) + e^- rightarrow Ag(s) -1.999 V -0.799 V +1.999 V +1.502 V

Explanation / Answer

Q8.

find volume needed to rech the quivalence point

from the graph, the volume is approx --> 30 mL (Actually about 28 mL)

Q9.

find pH at equuivalence point

in V = 28 mL. then pH = 10 approx

Q10

find pKa from the list

the pH = pKa, when half point

the half point V = 28mL/2 = 14 mL

the pH At V = 14 mL

approx 7.5

Q11

E° = Ecathode E anode

1.20 = Esilver - (0.401)

Esilver = 1.2 + 0.401 = 0.799 V

best answer is not shown, probably seen in E. it is positive 0.799 V

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