[iseeu]

An excess amount of silver chloride is added to a beaker of 2 M sodium cyanide. Calculate the concentration of all ions, other than OH- and H+, that are in solution once the system reaches equilibrium. (Hint, there should be 5 other ions present)

Please explain through your answer!! I'm really lost

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

What ions do you see immediately? What reactions (in water solution) can you expect? Try to write reaction equations.

[iseeu]

I already have the answers, I just have no idea why, and just wanted someone to explain it to me.

I know the ions are Na+, CN-, Ag+, Cl-, and the 5th one is Ag(CN)-2. but I don't understand why NaCl wouldn't be one of the ions to find the concentration of.

AgCl(s) --> Ag+(aq) + Cl- (aq)

Ag+ + 2CN- ---> Ag(CN)-2

AgCl(s) + 2CN- ---> Ag(CN)-2 + Cl-

I figured  [Cl-]= 1M and [Ag(CN)-2]= 1M. because sodium cyanide is 2M, and the mole ratio between sodium cyanide and Cl- and Ag(CN)-2 is 2:1.

But then again that assumption doesnt really make sense.

Please just go through the problem and explain why everything is done the way it is, because this problem makes absolutely no sense to me!

[Borek]

I know the ions are Na+, CN-, Ag+, Cl-, and the 5th one is Ag(CN)-2

There are definitely more, but this is a reasonable list.

but I don't understand why NaCl wouldn't be one of the ions to find the concentration of.

First of all - NaCl is not an ion. Just like HCN (and HCl and NaOH).

Ignoring the presence of HCN in such solution is a huge mistake IMHO.

AgCl(s) --> Ag+(aq) + Cl- (aq)

Ag+ + 2CN- ---> Ag(CN)-2

AgCl(s) + 2CN- ---> Ag(CN)-2 + Cl-

I figured  [Cl-]= 1M and [Ag(CN)-2]= 1M. because sodium cyanide is 2M, and the mole ratio between sodium cyanide and Cl- and Ag(CN)-2 is 2:1.

This is tricky. AgCN has even lower Ksp than AgCl, so if the mixture is left till it reaches equilibrium, practically all AgCl will be converted to AgCN. That in turn means very low concentration of Ag(CN)₂^- (as practically all CN^- is present in the precipitate).

At least looks like you are right about Cl^-.