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Calculate [Ag1+] in solution at equilibrium when 2.39 x 10-2 M AgNO3 is added to 1.00...

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Chemistry

Calculate [Ag1+] in solution at equilibrium when 2.39x 10-2 M AgNO3 is added to 1.00 L of a 1.65 MNH3 solution, assuming no volume change? (Kffor Ag(NH3)21+ is 1.1 x107)

a) 3.8 x 10-2

b) 2.39 x 10-2

c) 4.9 x 10-11

d) 8.5 x 10-10

e) 4.1 x 10-3

Answer & Explanation Solved by verified expert
4.5 Ratings (766 Votes)

no of moles of AgNo3 = molarity * volume in L

                                   = 2.39*10-2 *1   = 2.39*10-2 moles = 0.0239moles

no of moles of NH3     = molarity * volume in L

                                   = 1.65*1 = 1.65 moles

                  Ag+   +    2NH3 ---------> [Ag(NH3)2]+

I            0.0239    1.65    0

C -0.0239    -2*0.0239           0.0239

E            0              1.6022               0.0239

      Kf =    [Ag(NH3)2]+/[Ag+][NH3]2

     1.1*107   = 0.0239/[Ag+]*0.0239

     [Ag+]      = 0.0239/1.1*107 *0.0239

                 = 9.1*10-8 M   >>>> answer
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