Question

Silver bromide, AgBr (s), is an essential reagent in black and white film developing. It is,

however, only sparingly soluble in water. AgBr (s) has K = 5.0 × 10-13, making it difficult to

rinse AgBr from the film negative with water.

Instead, excess AgBr is removed by an aqueous solution of sodium thiosulfate (Na2S2O3), which forms the complex ion Ag(S2O3)23-:

Ag+ (aq) + 2 S2O32- (aq) Ag(S2O3)23- (aq) Kf = 4.7 × 10+13 a) To see how this helps, first determine the molar solubility of AgBr (s) in water.

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b) The large formation constant of Ag(S2O3)23- (aq) means that almost all of the silver is complexed with the thiosulfate. To determine how much, calculate the [Ag+] at equilibrium in a 1.0 L solution that initially contains 0.001 M silver ions and 0.200 M sodium thiosulfate. Do so by setting up an ICE table and assuming that x is small compared to 0.200 (but not 0.001). Show that this results in a value of x = [Ag(S2O3)23-] = 0.001. In other words, all of the Ag+ complexes.

c) Your result from part b suggests that no Ag+ remains in solution. This obviously can’t be correct, since it would result in an infinitely large reaction quotient that wouldn’t be equal to an admittedly large equilibrium constant. To determine the correct [Ag+], use the equilibrium concentrations determined in part b as your initial concentrations in a new ICE table. Then use Kf to determine [Ag+]. [This approach is called a stoichiometric shift and is useful when a reaction starts with only reactants but goes almost to completion. Is essence, we are approaching the equilibrium from the direction of all Ag(S2O3)23- and no Ag+ as opposed to the direction of all Ag+ and no Ag(S2O3)23-.]

d) Before we determine the solubility of AgBr (s) in a thiosulfate solution, we need to know the appropriate equilibrium constant. To find it, determine the value of the equilibrium constant for the reaction:

AgBr (s) + 2 S2O32- (aq) <-->Ag(S2O3)23- (aq) + Br - (aq) Kc = ??

e) Finally, calculate the molar solubility of AgBr (s) in 1.0 M sodium thiosulfate. In other words, what is [Ag(S2O3)23-] for the reaction given in part d when the initial concentration of thiosulfate is 1.0 M?

Answer 1

(a) The ionization of AgBr in water follows

AgBr (s) <======> Ag⁺ (aq) + Br^- (aq); K = [Ag⁺]Br^-] (the concentration of the solid is neglected).

We note that [AgBr]:[Ag⁺]:[Br^-] = 1:1:1

Now, K = 5.0*10^-13 M². Therefore, if y is the solubility of Ag⁺, then [Ag⁺]=[Br^-] =y. Therefore,

5.0*10^-13 M² = (y)(y)

or, y = √(5.0*10^-13 M²)

or, y = 7.07*10^-7 M

The molar solubility of AgBr in water is 7.07*10^-7 M.

(b) For this part, we write the reaction and set up the respective ICE chart (we note that the volume is 1.0 L, so no need to change initial concentrations)

Ag⁺ (aq) + 2 S₂O₃^2- (aq) <======> Ag(S₂O₃)₂^3-; K =4.7*10¹³

initial            0.001        0.200                                           0

change             -x             -2x                                              x

equibm.    (0.001 –x)(0.200 – 2x)                                    x

We write the expression for the formation constant K as

K = [Ag(S₂O₃)₂^3-]/[Ag⁺][S₂O₃^2-]²

or, K = (x)/(0.001 – x)(0.200 – 2x)²

We are given the condition that x is much small compared to 0.200 M, so that (0.200 – 2x) ≈ 0.200.

Hence, 4.7*10¹³ = (x)/(0.001 – x)(0.200)²

or, 4.7*10¹³.(0.04) = (x)/(0.001 – x)

or, 1.88*10¹² = x/0.001 – x

or, 5.32*10^-13 = (0.001 – x)/x

or, 5.32*10^-13 = (0.001/x) – 1

or, 1 = 0.001/x

or, x = 0.001

As has been hypothesized, x =[Ag(S₂O₃)₂^3-] = 0.001 M, assuming complete complexation.