Question

In: Chemistry

Calculate the equilibrium concentration of Ag+ (aq) in a solution that is initially 0.130 M AgNO3...

Calculate the equilibrium concentration of Ag+ (aq) in a solution that is initially 0.130 M AgNO3 and 0.610 M NH3 and in which the following reaction takes place:

Ag+ (aq) + 2NH3(aq) <--> Ag(NH3)2+ (aq) Kf = 1.7 x 107

Solutions

Expert Solution

The formation constant for the given reaction is Kf = 1.7*107. Since Kf is high, it is assumed that all the Ag+ (added as AgNO3) has coupled with NH3 to form the complex, [Ag(NH3)2]+ and the equilibrium concentration of [Ag(NH3)2]+ is equal to the initial concentration of Ag+, i.e, 0.130 M.

Again, as per the stoichiometric equation,

1 M [Ag(NH3)2]+ = 2 M NH3.

Therefore,

0.130 M [Ag(NH3)2]+ = (2 M NH3)*(0.130 M [Ag(NH3)2]+)/(1 M [Ag(NH3)2]+)

= 0.260 M NH3.

Therefore, equilibrium concentration of NH3 = (0.610 – 0260) M = 0.350 M.

Write down the expression for the formation constant of [Ag(NH3)2]+ as

Kf = [Ag(NH3)2+]/[Ag+][NH3]2

where [..] denote concentrations in mol/L or M.

Plug in values and get

1.7*107 = (0.130 M)/[Ag+](0.350 M)2

======> [Ag+] = (0.130 M)/(1.7*107)(0.350 M)2

Ignore units and obtain

[Ag+] = (0.130)/(1.7*107)(0.1225)

======> [Ag+] = 6.242*10-8 ≈ 6.24*10-8 (ans, correct to 3 sig. figs).

The equilibrium concentration of Ag+ is 6.24*10-8 M (ans).


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