Question

2. Calculate E for the titration of 48.00 mL of 0.110 M Ag when 50.00 mL of 0.140 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- for AgCl.
please answer the following question

Answer 1

2) According to the condition given in question moles of Agt = concentration x volume = 0.11M X 48,0 me = 5.28 mmol moles of cl - concentration X volume = 0.14 x 48 50 me s 7 mmol Excess of cl = 7 - 5.28 s 1.72 mmol Total volume of solution = 50+48 = 98 me [ci] - 172 x 1000 1000 98 = 0.0175 M Emf of a cell is given by

E = E-E- = [0.7993 – 0.0591 log+] -0.197 = 0.6023 +0.059) log [Agt ] we know that Ksp = [Ag+ ][cé ] [Aq+J = Ksp 118x10 ° 1 [cê 3 0.0175 [Agt j = 1.028 x106m F = 0.6093 + 0.0591 log(1.028X1D) E = 0.6023 -0.3538 E = 0.248 V

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