Question

when nh4cn is dissolved in water the solution will be?​

Answer 1

Answer:

Ammonium cyanide, NH4CN, is a solid where the atoms are grouped into the same ions that are generated in solution: NH4+ and CN−. Although it can be sublimed with very mild warming, it is fairly unstable. In particular it reacts with atmospheric water vapor, evolving ammonia and HCN. Dangerous stuff!

The first part of the answer to 1. is the decomposition of NH4CN into ions in aqueous solution, which should be obvious; one subtlety is that you may want to add (s) and (aq) next to each species, especially if your professor makes a point of doing so. So that's the dissociation. The data in the problem don't allow us to compute the equilibrium constant of the dissociation but it's safe to say that it is extremely high, that is, dissociation is complete: there is no NH4CN as such in solution.

The acid-base equilibrium equations can be the ones you wrote. Again, you may want to add more information, like this:

NH4+(aq)+H2O↽−−⇀NH3(aq)+H3O+(aq)K=5.5×10−10

Here the equilibrium constant K is the acid dissociation constant of the acid NH4+, the protonated form (aka conjugate acid) of ammonia. I got its value by dividing Kw, the dissociation constant of water (1.0×10−14), by the Kb of ammonia, given in the problem.

I encourage you to add similar information for the equilibrium involving cyanide ion. Hint: if you keep the reaction as you wrote it, the appropriate K is not the Ka of HCN but the Kb of the deprotonated form (conjugate base), CN−. (But you can instead consider the alternate reaction HCN+H2O↽−−⇀CN−+H3O+, and then the equilibrium constant is the Ka of HCN.)

Explanation:

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Answer 2

Explanation:

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