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13 Dec 2019

Silver bromide, AgBr, is a slightly soluble solid. Which of the following substances will not affect (cause a shift in one direction or another) the equilibrium established when AgBr(s) is dissolved in water. (HINT: ammonia forms a complex ion with Ag+ (aq)).

a. AgNO3

b. NH3

c. KBr

d. N2

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Irving Heathcote
Irving HeathcoteLv2
17 Dec 2019

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Related questions

When sodium chloride, NaCl, is added to a solution of silver nitrate, AgNO3, a white precipitate of silver chloride, AgCl, forms. If concentrated ammonia, NH3, is added to this precipitate, a clear colorless solution forms containing diamminesilver(I) ions, [Ag(NH3)2]+. Potassium iodide, KBr, solution added to this colorless solution will cause pale yellow silver bromide, AgBr, to precipitate. List the species, Cl-, NH3, or Br- in order of decreasing stability of the compounds or complexes that they can form with silver and explain your reasons.

When NaCl is added to a solution of AgNO3, a white precipitate of AgCl forms. If concentrated ammonia, NH3, is added to this precipitate, a clear colorless solution forms containing [Ag(NH3)2]+. KBr solution added to this colorless solution will cause pale yellow AgBr to precipitate. List the species, Cl-, NH3, or Br- in order of decreasing stability of the compounds or complexes that they can form with silver and explain your reasons.

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.

sp

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?

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