Calculate the equilibrium concentration of Ag+(aq ) in a solution that is initially 0.170 M AgNO3...

Calculate the equilibrium concentration of Ag+(aq ) in a solution that is initially 0.170 M AgNO3 and 0.990 M NH3, and in which the following reaction takes place: Ag+(aq)+2NH3(aq)=Ag(NH3)2+(aq) (Kf = 1.70x107)

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Expert Solution

Given that

[AgNO3] = 0.170 M

[NH3] = 0.990 M

Kf = 1.70x10⁷

0.170M will react with NH3 and there will be zero amount of Ag+ left after reaction

However the complex formed will generate more amount of Ag+ due to dissociation

Ag⁺(aq) + 2NH₃(aq) = Ag(NH₃)₂⁺(aq)

Initial 0.170 0.990 0

Change -0.170 -2(0.170) 0.170

Equilibrium +x 0.65 +2x 0.170 - x

Kf = [0.170-x] / [x][0.65+2x]^²

we may ignore x in denominator and numerator in addition and subtraction

Therefore

1.70x10⁷ = [0.170] / [x] (0.65)²

x = 2.367 X 10^-7 M

queen_honey_blossom answered 1 year ago

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