In: Chemistry
A 10.0 mL solution of 0.030 M AgNO3 was titrated with 0.015 M NaBr in the cell. S.C.E. || titration solution | Ag(s) Find the cell voltage for 0.4, 10.0, 20.0, and 27.0 mL of titrant. (The Ksp of AgBr = 5.0 ✕ 10−13.) 0.4 V 10.0 V 20.0 V 27.0 V
(a) 0.4 ml of 0.015 M NaBr added
moles of AgNO3 = 0.03 M x 0.01 L = 3 x 10^-4 mols
moles of NaBr = 0.015 M x 0.004 L = 6 x 10^-5 mols
excess AgNO3 = 2.4 x 10^-4 mols
[Ag+] = 2.4 x 10^-3/0.014 = 0.017 M
Ecell = 0.7994 - 0.241 = 0.558 V
E = 0.558 + 0.0592 log [Ag+] = 0.4 V
(b) 10 ml of 0.015 M NaBr added
moles of AgNO3 = 0.03 M x 0.01 L = 3 x 10^-4 mols
moles of NaBr = 0.015 M x 0.010 L = 1.5 x 10^-4 mols
excess AgNO3 = 1.5 x 10^-4 mols
[Ag+] = 1.5 x 10^-4/0.02 = 7.5 x 10^-3 M
Ecell = 0.7994 - 0.241 = 0.558 V
E = 0.558 + 0.0592 log [Ag+] = 0.432 V
(c) 20 ml of 0.015 M NaBr added
moles of NaBr = 0.015 x 0.02 = 3 x 10^-4 M
[Ag+] = sq.rt.(5 x 10^-13) = 7.07 x 10^-7 M
E = 0.558 + 0.0592 log [Ag+] = 0.194 V
(d) 27 ml of 0.015 M NaBr added
excess moles of NaBr = 0.015 x 0.007 = 1.05 x 10^-4 mols
[NaBr] = 1.05 x 10^-3/0.037 = 2.84 x 10^-3 M
[Ag+] = Ksp/[NaBr] = 5 x 10^-13/2.84 x 10^-3 = 1.762 x 10^-10 M
E = 0.558 + 0.0592 log [Ag+] = -0.019 V