Question

At 25 C, a titration of 15.00 mL of a 0.0440 M A^NO, solution with a 0.0220 M Nal solution is conducted within the cell SCE | titration solution | Ag(s) For the cell as written, what is the potential after the addition of each volume of Nal solution? The reduction potential for the saturated calomel electrode is E = 0.241 V. The standard reduction potential for the reaction Ag+ + e _» Ag(s) is E = 0.79993 V. The solubility constant of AgI is Kgp = 8.3 x 1017 0.100 mL V 17.90 mL E = E = 30.00 mL V 42.70 mL V E =

Answer 1

Using Nernst equatron Agt (99.) tee - Ag(s) Entoh Eed - 005929 long Eckbalk = 0·7993 V – 0-0592 V tog Eanode = 0.241 V Eall = Ecathade - Exuale Eault = 0.79993 v 179993 V – 0.0592V log I -0.241V Ecell = 0.55893 y + 0.0592X doo [agt] [Ag+] = ☺0.100m) Agt in inimale = 0.0440 MX 15:00 mL = 0.66 mual 1 in mimale = 0'0220 MX 0.100ML = 0.0022 mmel Agt (04.) + 10() AgI (8) I fuced) 0.66 0.0021 Ewwel) 06578 0 06578 mmel LE D'0436 M (15+0:104) mL Eall = 0.55893 + 0.0592 log (0.043c) Eall = 0.478 VJ 5 117.90 m2 Azt in mmol = 0.0440 Mxis.com = 0.66 munde 1 in mimale = 0.0220MX 1740 mL = 0.3938 mmell Agt (arn) AIⓇ (ar) AGI(8) I(ww.) O'660 3936 Ewwele) 0 2662 0 111 0 2662_well = 0.00809M (35+17:90) ML Ecell = 0'5S 8934 0.0592 log (0.00809) Eall = 0 4351

0 1300ML Agt in wandle = 0.0440 MX 15.00m2 = 0.66 manal It in minste = 0.0220 MX 30.00 mL = 0.66 usmell Ag+Cav.) +I®(av) = AgI (8) I(mmal) 0.66 0.66 Efummel) S5 Ksp = [Agt ] [11] 83x1077 = S? S = 9.11 X1019 = Cagt] - [9] Eall = 0.55893 v. to.os 92 log (9.11X10-9) 200829 V 42.70 ML] Agt in in male a 0.0440 MX 15:00 mL =0.66 munale to in muele = 0·0220 MX 42.7omL = 0.939 4 ugurely Agt (ori) + I (09) Ag. I (4) Дие4 ), 066 019394 Erumel) 0' 2794 0 2754 mmel s 0:00484 M (1) = s.co+42.70) me Ksp = [A34] [I] 83X10 97= (Agt ] (0.00484) [Agt) = 1.714x10=1422 Ecell = 0.55893 + 0.0592 dog (1:714X10-14) Eall = -0.256