Calculate E for the titration of 50.00 mL of 0.100 M Ag+ when 25.00 mL of 0.150 M Cl−
is added. Silver wire is the indicator electrode and the saturated Ag-AgCl electrode is the
reference electrode. E = 0.197 V for the saturated Ag-AgCl electrode,E = 0.7993 V and Ksp =
1.8 × 10−8 for AgCl.
Answer: 0.497 V
1
Expert's answer
2019-06-28T02:46:45-0400
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