Question

A 10.0 mL solution of 0.064 M AgNO3 was titrated with 0.032 M NaBr in the cell. S.C.E. || titration solution | Ag(s) Find the cell voltage for 0.5, 14.0, 20.0, and 32.0 mL of titrant. (The Ksp of AgBr = 5.0 ✕ 10−13.)

Answer 1

Titration

(a) 0.5 ml of 0.032 M NaBr added

moles of AgNO3 = 0.064 M x 0.01 L = 6.4 x 10^-4 mols

moles of NaBr = 0.032 M x 0.0005 L = 1.6 x 10^-5 mols

remaining AgNO3 = 6.24 x 10^-4 mols

[Ag+] = 6.24 x 10^-4/0.0105 = 0.06 M

E = Eo + 0.0592 log[Ag+] - 0.241

   = 0.7998 + 0.0592 log(0.06) - 0.241

   = 0.47 V

(b) 14 ml of 0.032 M NaBr added

moles of AgNO3 = 0.064 M x 0.01 L = 6.4 x 10^-4 mols

moles of NaBr = 0.032 M x 0.014 L = 4.48 x 10^-4 mols

remaining AgNO3 = 1.92 x 10^-4 mols

[Ag+] = 1.92 x 10^-4/0.024 = 8 x 10^-3 M

E = Eo + 0.0592 log[Ag+] - 0.241

   = 0.7998 + 0.0592 log(8 x 10^-3) - 0.241

   = 0.43 V

(c) 20 ml of 0.032 M NaBr added

moles of AgNO3 = 0.064 M x 0.01 L = 6.4 x 10^-4 mols

moles of NaBr = 0.032 M x 0.020 L = 6.4 x 10^-4 mols

[Ag+] = [Br-]

E = Eo + 0.0592 log[Ag+] - 0.241

   = 0.7998 + 0.0592 log(sq.rt.Ksp) - 0.241

   = 0.20 V

(d) 32 ml of 0.032 M NaBr added

excess [NaBr] = 0.032 M x 0.012 L/0.042 = 9.143 x 10^-3 M

[Ag+] = Ksp/[NaBr] = 5 x 10^-13/9.143 x 10^-3 = 5.47 x 10^-11 M

E = Eo + 0.0592 log[Ag+] - 0.241

   = 0.7998 + 0.0592 log(5.47 x 10^-11) - 0.241

   = -0.05 V