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

Calculate [Ag¹⁺] in solution at equilibrium when 2.39 x 10^-2 M AgNO₃ is added to 1.00 L of a 1.65 M NH₃ solution, assuming no volume change? (K_f for Ag(NH₃)₂¹⁺ is 1.1 x 10⁷)

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

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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