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 K_sp of AgBr = 5.0 ✕ 10⁻¹³.)

Answer 1

(a) 0.5 M of 0.032 M NaBr added

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

moles of NaBr added = 1.6 x 10^-5 M

Ag+ remaining = 6.075 x 10^-4 M

[Ag+] = 6.24 x 10^-4/0.0105 = 0.058 M

E = [0.799 - 0.0592 log(1/0.058)] - 0.241 = 0.485 V

(b) when 14 ml of 0.032 M NaBr added

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

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

Ag+ remaining = 1.92 x 10^-4 M

[Ag+] = 1.92 x 10^-4/0.024 = 0.008 M

E = [0.799 - 0.0592 log(1/0.008)] - 0.241 = 0.434 V

(c) 20 ml of 0.032 M NaBr added

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

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

Equivalence point,

E = 0.799 - 0.241 = 0.558 V

(d) 32 ml of 0.032 M NaBr added

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

moles of NaBr added = 0.032 x 0.032 = 1.024 x 10^-3 M

excess NaBr = 3.84 x 10^-4 mols

[Br-] = 9.143 x 10^-3 M

[Ag+] = Ksp/[Br-] = 5.47 x 10^-11 M

E = [0.799 - 0.0592 log1/5.47 x 10^-11] - 0.241 = -0.050