Hence 0.056 = 0.050 Note that the molar solubility of AgCl equals the molar conc

Question

Hence 0.056 = 0.050 Note that the molar solubility of AgCl equals the molar concentration of silver in the Because of the stability of Ag(NH,), most of the silver in solution will be in the formof this complexion. Therefore. because requals the concentration of Ag(NH,) the molar solubility of AgCl equals 0.050 M. Answer Check Check that you have the correct overall equilibrium for the problem and that you have then done the arithmetic correctly for the equilibrium constant. Once you have the molar solubility of the compound, x, you can put x back into the equilibrium- constant expression and calculate the value of K, as a final check. Exercise 17.11 What is the molar solubility of AgBr in 1.0 M Na,S,O, (sodium thio- sulfate)? Silver ion forms the complex ion Ag(S,Ox) See Tables 17.1 and 17.2 for data. See Problems 17.6

Explanation / Answer

the AgBr will dissociate as

                             AgBr (s) ---> Ag+ (aq) + Br- (aq)
solubility product = Ksp = [Ag+][Br-] = 5.0 x10-13

Let the molar solubility of AgBr = x

5.0x10-13 = x2

x = 7.07 X 10-6 M

Now in 1M Na2S2O3

                    AgBr + 2 Na2S2O3    -----> Na3[Ag(S2O3)2] + NaBr

Iniital                           1                            0                     0

Change                       -2x                          x                   x

Equilibrium            1-2x                             x                     x

Kf = 2.9 X 1013

So with formation of each complex ion one Br- is released

[Ag(S2O3)2]-3 = [Br-] = x

x2 = Ksp * Kf * [S2O3-2] 2

x2 = Ksp * Kf * [1 -2x] 2

x2 = 5.0 x10-13 X 2.9 X 1013 [1 -2x] 2

x2 = 14.5 (1 + 4x2 - 4x)

14.5 + 57x2 - 58x = 0

x = 0.575 or 0.442

this is solubility of AgBr in Na2S2O3

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