[Sis290025]

Calculate the solubility of AgCl in a 0.104 M NH3.

CORRECT ANSWER: 4.98E-3

According to my textbook, this is the reaction that occurs:

AgCl + 2NH3 <-> Ag(NH3)2(+) + Cl-, where K = 3.1E-3

Using this information, I set up the following expression:

K = [Cl-][Ag(NH3)2(+)]/[NH3]^2

3.1E-3 = x^2/[0.104-x]^2

Solving for x:

sqrt(3.1E-3) = x/(0.104 - x)


sqrt(3.1E-3)*(0.104 - x) = x

sqrt(3.1E-3)*(0.104) - sqrt(3.1E-3)x = x
sqrt(3.1E-3)*(0.104)/[1 + sqrt(3.1E-3)] = x = 5.49E-3 ?

What am I doing incorrectly?

Please help.

Thank you.

[Borek]

x = 5.49E-3 ?

IMHO this is correct answer, not the one you quoted earlier - assuming you have copied all data (0.104 and 3.1*10^-3) correctly.

[Sis290025]

I copied the problem correctly and the answer that was deemed "correct" justly, too. I got the number K from the product of equilibrium constants where K = K_sp*K_f = (1.7E7*1.8E-10). (This is correct?) I have no idea why my value doesn't match the one in bold.

Thanks again.

[Borek]

I got the number K from the product of equilibrium constants where K = K_sp*K_f = (1.7E7*1.8E-10).

I don't get it, please elaborate. My tables shows that log(K) = 7.22.

[Sis290025]

Well, this part is from my textbook about complex ions:

For the formation of Ag(NH3)2(+), the net reaction is Ag+ + 2NH3 --> Ag(NH3)2(+), where K_f = 1.7E7

K_1 = [Ag(NH3)(+)]/[Ag+][NH3] = 2.1E3

K_2 = [Ag(NH3)2(+)]/[[Ag+][NH3]^2 = 8.1E3

These K values come from come from the 2 reactions to form the net:

Ag+ + NH3 <-> Ag(NH3)+
Ag(NH3)+ + NH3 <-> Ag(NH3)2(+)

K_f = K_1*K_2 = 1.7E7

The K constant for the net:

K = [Ag(NH3)2(+)][Cl-]/[NH3]^2 = K_f*K_sp = (1.7E7)*(1.8E-10) = 3.1E-3

I guess log(k) = 7.22 is the K_f value??

[Borek]

K = [Ag(NH3)2(+)][Cl-]/[NH3]^2 = K_f*K_sp = (1.7E7)*(1.8E-10) = 3.1E-3

OK, now I know what you mean. Note that you have neglected presence of AgNH₃⁺. Sounds reasonable.

I guess log(k) = 7.22 is the K_f value??

More precisely it is log of K₂.

However, I did some further analysis and I have found an error in your calculations - (.104-x) is wrong. Think about stoichiometry.

But even if corrected it still don't give the result you quoted as a correct answer.