Question

Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na₂S₂O₃, called hypo) to form the complex ion Ag(S₂O₃)₂^3-. The reaction with silver bromide is:

AgBr(s) + 2 S₂O₃^2- (aq) ⇌ Ag(S₂O₃)₂^3- (aq) + Br^- (aq)

The equilibria that need to be considered are:

Ag⁺ + 2 S₂O₃^2- (aq) ⇌ Ag(S₂O₃)₂^3- (aq) K_f = 4.7 x 10¹³

AgBr(s) ⇌ Ag⁺ (aq) + Br^- (aq) K_sp = 5.0 x 10^-13

What mass of Na₂S₂O₃ is required to prepare 1.00 L of a solution that will dissolved 1.00 of AgBr by the formation of Ag(S₂O₃)₂^3-?

Answer 1

Ag⁺ + 2 S₂O₃^2- (aq) ⇌ Ag(S₂O₃)₂^3- (aq) K_f = 4.7 x 10¹³

AgBr(s) ⇌ Ag⁺ (aq) + Br^- (aq) K_sp = 5.0 x 10^-13

Combining equation above 2 equations that cancel out silver ions : AgBr(s) + 2 S₂O₃^2- (aq) ⇌ Ag(S₂O₃)₂^3- (aq) + Br^- (aq)

Ksp = [Ag⁺] [Br^-] = 5.0 x 10^-13

Molar mass of AgBr = 187.77 g/mol

Molarity of AgBr . = [Br^-] = wt/molar mass)/vol in lts = (1 g/187.77 g/mol)/1 lt = 0.00533 M

Ksp = [Ag⁺] [Br^-] = 5.0 x 10^-13

[Ag⁺] = 5 x 10^-13 / 5.33 x 10^-3 = 0.94 x 10^-10 M

   Ag⁺ + 2 S₂O₃^2- (aq) ⇌ Ag(S₂O₃)₂^3- (aq)

At equilibrium , [Ag(S₂O₃)₂^3- ] = 5.33 x 10^-3 as AgBr is dissolved

K_f = 4.7 x 10¹³ = [Ag(S₂O₃)₂^3- ]/ [Ag⁺] [S₂O₃^2- ]² = 5.33 X 10^-3/0.94 x 10-¹⁰ [S₂O₃^2- ]²

[[S₂O₃^2- ] = 5.67 x 10⁷/4.7 X 10¹³ = 1.1 x 10^-3 M

Total no. of moles of thiosulphate ion = amount of thiosulphate that reacted with silver ions and free thiosulphate ions =

in [Ag(S₂O₃)₂^3- ] 2 thiosulphate ions + free thiosulphate ion = 2 x 0.00533 + 0.0011 = 0.01176 M

Volume is 1 litre

So, no. of moles of thiosulphate ions = 0.01176 moles

Molar mass of Na₂S₂O₃ = 158.11 g/mol

Mass of Na₂S₂O₃ required = Moles x molar mass = 0.01176 moles x 158.11 g/mol = 1.86 gms