Question

A precipitation titration was performed by adding 0.20 M Ag+ (from standard silver nitrate solution) from a buret into 40.0 mL mixture of 0.10 M I- (as NaI) and 0.10 M Cl- (as NaCl). Calculate the concentration of Ag+, I-, Cl-, Na+, and NO3- when (a) 0.0 mL, (b) 10.0 mL, (c) 20.0 mL, (d) 30.0 mL, (d) 40.0 mL, and (d) 50.0 mL of Ag+ were added. (Given Ksp AgI = 8.3 x 10-17 and Ksp AgCl = 1.8 x 10-10 )

Answer 1

The no. of millimoles of Na⁺ = (40 mL * 0.1 mmol/mL) + (40 mL * 0.1 mmol/mL) = 8 mmol

The no. of millimoles of I^- = 40*0.1 = 4 mmol

The no. of millimoles of Cl^- = 40*0.1 = 4 mmol

(a) 0.0 mL

The no. of millimoles of Ag⁺ = 0

The no. of millimoles of NO₃^- = 0

Therefore, the concentrations of Ag+, I-, Cl-, Na+ and NO3- are 0 M, 0.1 M, 0.1 M, 0.2 M and 0 M, respectively.

(b) 10.0 mL

The no. of millimoles of Ag⁺ = 0.2*10 = 2 mmol

The no. of millimoles of NO₃^- = 2 mmol

Total voluem = 10 + 40 = 50 mL

Therefore, the concentrations of Ag+, I-, Cl-, Na+ and NO3- are 5.367*10^-7 M​​, 0.08 M, 0.04 M, 0.16 M and 0.04 M, respectively.

Explanation: 2 mmol of Ag+ reacts with 2 mmol of Cl- to form 2 mmol of AgI.

i.e. [Ag+]_{in AgCl} = [Cl-]_AgCl = 2 mmol/50 mL* (1.8*10^-10)^1/2 = 5.367*10^-7 M

i.e. [Cl-] = 5.367*10^-7 M + (4-2) mmol/50 mL ~ 0.04 M

[I-] = 4 mmol/50 mL = 0.08 M

[Na+] = 8 mmol/50 mL = 0.16 M

[NO3-] = 2 mmol/50 mL = 0.04 M

(c) 20.0 mL

The no. of millimoles of Ag⁺ = 0.2*20 = 4 mmol

The no. of millimoles of NO₃^- = 4 mmol

Total voluem = 20 + 40 = 60 mL

Therefore, the concentrations of Ag+, I-, Cl-, Na+ and NO3- are 8.944*10^-7 M​​​, 0.067 M​, 8.944*10^-7 M​​, 0.133 M and 0.067 M, respectively.

Explanation: 4 mmol of Ag+ reacts with 4 mmol of Cl- to form 4 mmol of AgCl.

i.e. [Ag+]_{in AgCl} = [I-]_AgCl = 4 mmol/60 mL* (1.8*10^-10)^1/2 = 8.944*10^-7 M

[I-] = 4 mmol/60 mL = 0.067 M

[Na+] = 8 mmol/60 mL = 0.133 M

[NO3-] = 4 mmol/60 mL = 0.067 M

(d) 30.0 mL

The no. of millimoles of Ag⁺ = 0.2*30 = 6 mmol

The no. of millimoles of NO₃^- = 6 mmol

Total voluem = 30 + 40 = 70 mL

Therefore, the concentrations of Ag+, I-, Cl-, Na+ and NO3- are 7.669*10^-7 M​​​, 0.029 M​, 7.666*10^-7 M​, 0.114 M and 0.086 M, respectively.

Explanation: 4 mmol of Ag+ reacts with 4 mmol of Cl- to form 4 mmol of AgCl and the remaining 2 mmol of Ag+ reacts with 2 mmol of Cl- to form 2 mmol of AgCl.

i.e. [Ag+]_{in AgCl} = [Cl-]_AgCl = 4 mmol/70 mL* (1.8*10^-10)^1/2 = 7.666*10^-7 M

[Ag+]_{in AgI} = [I-]_AgI = 2 mmol/70 mL* (8.3*10^-17)^1/2 = 0.003*10^-7 M

[Ag+]_total = 7.669*10^-7 M

[I-] = (4-2) mmol/70 mL = 0.029 M

[I-]_total = 0.029 M + 0.003*10^-7 M = 0.029 M

[Na+] = 8 mmol/70 mL = 0.114 M

[NO3-] = 6 mmol/70 mL = 0.086 M