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After the solution reaches equilibrium, what concentration of Ag+ remains?
A 120.0 mL sample of a solution that is 2.8*10^-3 M in AgNO3 is mixed with a 225.0 mL sample of a solution that is 0.10 M in NaCN.
I pretty much know how to solve this problem, but there's a tiny detail I'm confused on:
How do you know that the complex ion [Ag(CN)2]+ forms, and not just the precipitate AgCN?
2 Answers
- Dr WLv 73 weeks agoFavorite Answer
Great question.
The answer is... you wouldn't know it. But now you do.
When silver reacts with cyanide, it precipitates out as AgCN. But the addition of XS cyanide forms a soluble [Ag(CN)2](-).... not + but -... complex.
This is one of those things you learn along the way. Like how to pronounce methylamine. No student will automatically know this. But it should be covered in your textbook.
- jacob sLv 73 weeks ago
AgNO3 (aq) + NaCN (aq) ------------------> AgCN(s) + NaNO3(aq)
120x 2.8x10-3 225x0.1
= 0.336 22.5 0 0 initial mmoles
0 22.164 0.336 - after reaction.
As AgCN is a sparingly soluble salt with Ksp = 1.2 x10-16
the concentration of Ag+ in solution should be only from the salt dissoication
AgCN(s) -------------> Ag+ + CN-
x x x + (22.164/345)= x +0.06424
As x is very small compared to 0.06424 , we take [CN-] = 0.06424
Now Ksp = x (0.06424) = 1.2 x10-16
or x = 1.867 x 10-15 M
The concentration of Ag+ that remains in solution = 1.867 x 10-15 M
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