Question

In photography, unexposed silver bromide is removed from film by soaking the film in a solution of sodium thiosulfate, Na₂S₂O₃. Silver ion forms a soluble complex with thiosulfate ion, S₂O₃^2-, that has the formula Ag(S₂O₃)₂^3-, and formation of the complex causes the AgBr in the film to dissolve. The Ag(S₂O₃)₂^3- complex has K_form= 2.0 x 10¹³. How many grams of AgBr (K_sp= 5.4 x 10^-13) will dissolve in 126 mL of 1.27 M Na₂S₂O₃ solution?

Answer 1

K_form= [Ag(S₂O₃)₂^3-]/ ([Ag+][S₂O₃^2-]²) = 2.0 x 10¹³

K_sp= [Ag+][Br-] = 5.4 x 10^-13

AgBr = Ag⁺ + Br-                            K_sp= [Ag+][Br-] = 5.4 x 10^-13

Ag+   + 2 S₂O₃^2- = Ag(S₂O₃)₂^3-        K_form= [Ag(S₂O₃)₂^3-]/ ([Ag+][S₂O₃^2-]²) = 2.0 x 10¹³

………………………………………………….

AgBr + 2 S₂O₃^2- = Ag(S₂O₃)₂^3- + Br^-     K = [Ag(S₂O₃)₂^3-][Br-]/[S₂O₃^2-]² =K_form x K_sp

K = K_form x K_sp = 10.8

[Ag(S₂O₃)₂^3-] = [Br-]   because the concentration of available Ag⁺ equals [Br-].

Thus K = [Br-]²/[S₂O₃^2-]²

0.126 L x 1.27 mol/L = 0.160 mol S₂O₃^2- available in 127 mL solution

(0.080 mol of Ag(S₂O₃)₂^3- may be formed , with a small correction for re-equilibration.

The concentration of the complex Ag(S₂O₃)₂^3- is close to 1.27 M/2 = 0.635 M

AgBr + 2 S₂O₃^2- = Ag(S₂O₃)₂^3- + Br^-            

0               0              0.635 M        0.635 M                initial state

x               2x            0.635 M         0.635-x               at equilibrium

K = 0.635(0.635-x)/ (2x²)     

Solve the quadratic equation or….                       

Assuming x<<0.635

10.8 = 0.635²/x²

x = 0.193 M

Reiterate the calculation because   x is not <<0.635

10.8 = 0.635x(0.635-0.193)/x²

x= 0.161

Reiterate

10.8 = 0.635x(0.635-0.161)/x²

x= 0.166

Reiterate

10.8 = 0.635x(0.635-0.166)/x²

x= 0.166      Good!

At equilibrium [Br-] = (0.635-0.166) = 0.469 M

In 126 mL solution:

0.469 mol/L x 0.126 L = 0.0591 mol Br^- from AgBr dissolution

0.0591 mol x 187.77 g AgBr/mol = 11.1 g AgBr will be dissolved