pH of Ammonium Sulfide Solution What is the pH of a 0.100M aqueous solution of (

Question

pH of Ammonium Sulfide Solution

What is the pH of a 0.100M aqueous solution of (NH4)2S?
An exact analytical solution to this question can be algebraically very complex, however we can use some chemical intuition to greatly simplify it.

Here is a method to approach this question.

Combine the above equations to get:

Now, since H2S is diprotic we need to briefly consider the second dissociation step:

Combine the above equations to get:

So let us consider only the first equilibrium between H2S and HS-

Since we have already guessed that the equlibrium position for this reaction will be to the right, let us set up our initial conditions correspondingly.

That is:
[H2S]0 = 0.00 M
[HS-]0 = 0.100 M
[NH3]0 = 0.100 M
[NH4+]0 = 0.100 M

I hope you can see that this is stoichiometrically equivalent to 0.100 M (NH4)2S?

Now, we expect the actual equilibrium position to be shifted slightly to the left. Let us say that the amount of this shift is x M.

Thus:
[H2S] = x
[NH3] = 0.100 + x
[HS-] = 0.100 - x
[NH4+] = 0.100 - x

The above reduces to a simple quadratic equation and can be solved accordingly.
Alternatively, if our assumptions were correct, then we might expect x << 0.100 M.
We could then approximate the expression as:

or

and

This expression already gives a pretty good approximation of x, but you can take the value and insert it back into the unsimplified expression and solve for x again to get a better estimate. You can do this repeatedly and will find that the solution for x rapidly converges on a value which is not much different from the initial estimate.

Now you can determine approximate final concentrations for the acid/base conjugate pairs involved. You can then take either of the individual acid or base dissociation reaction equilibria to solve for either [OH-] or for [H+]

If our assumptions were good then:
[H3O+][OH-] = 1.0×10-14 should hold approximately for the values determined above.

Try it and then determine the pH and enter the value below.

Explanation / Answer

[OH-] = K2[NH3]/ [NH4+] , [NH3] = 0.1+0.1/180 =0.100556

[NH4+] = 0.1-1/180= 0.0944

[OH-] = 1.8*10-5*0.1000556/0.0944=1.92*10-5

poH= -log (1.92/10^5)= 4.72

pH= 14-4.72= 9.28

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