(NH4)2S is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.200 M (NH4)2S solution.

[NH4] = M

[S] = M

Well, well, well, we have a solution of (NH4)2S here! Let me juggle these ions for you and calculate their concentration.

First, let's talk about the (NH4)+ ion. Since we have two of those in (NH4)2S, we know that for every one (NH4)2S compound, we have two (NH4)+ ions. So, the concentration of (NH4)+ is twice the concentration of the (NH4)2S solution. Therefore, [NH4] = 2 * 0.200 M = 0.400 M.

Now, let's focus on the sulfide (S2-) ion. It's solo, no multiplying by any numbers. So, the concentration of [S] in the solution is simply 0.200 M.

So, to sum it up:

[NH4] = 0.400 M
[S] = 0.200 M

Hope that clears things up for you!

To determine the concentration of each individual ion in a solution of (NH4)2S, we need to consider the dissociation of the compound in water.

(NH4)2S dissociates into two ammonium ions (NH4+) and one sulfide ion (S2-).

According to the stoichiometry of the compound, for every 1 mole of (NH4)2S that dissolves, we get 2 moles of NH4+ ions and 1 mole of S2- ions.

Given that the initial concentration of (NH4)2S is 0.200 M, we can find the concentration of each ion by multiplying the initial concentration by the appropriate stoichiometric factor.

For the ammonium ion (NH4+):
[NH4+] = 2 × [NH4] = 2 × 0.200 M = 0.400 M

For the sulfide ion (S2-):
[S2-] = 1 × [S] = 1 × 0.200 M = 0.200 M

Therefore, in a 0.200 M (NH4)2S solution, the concentration of NH4+ ions is 0.400 M, and the concentration of S2- ions is 0.200 M.

ijmk

(NH4)2S ==> 2NH4^+ + S^2-

So (NH4)2S = 0.200
That means NH4^+ must be twice that and
S^2- is the same as (NH4)2S

Bio student