acid base - pH of ammonium acetate solution

I'm trying to calculate the pH of a 1M $\ce{NH_4CH_3COO}$. I know that I'll have these reactions:

$\ce{NH_4CH_3COO <=> NH_4^+ + CH_3COO^-}$
$\ce{NH_4^+ <=> NH_3 + H^+}$
$\ce{CH_3COOH <=> CH_3COO^- + H^+}$

I know the $K_a$s of the last two, so I'm able to calculate the $K$ of the first one (it's $\frac {K_{a1}}{K_{a2}}$), which gives me these equations:

$\frac {K_{a1}}{K_{a2}} = \frac {(x-y)(x-z)} {1 - x}$
${K_{a1}} = \frac {y(y+z)} {x-y}$
${K_{a2}} = \frac {z(y+z)} {x-z}$

But I have only 2 independent equations (the first one is just the ratio of the second and third) and three variables so I'm unable to solve for $[\ce{H^+}]$, which is $y+z$...

What do I do?

Answer

Ok, I will follow the assumption proposed by @Zhe above (possible incorrect as he states, but please do not be confused by that).

In order to solve this problem we need two acidity constants: pka(ammonium ion) = 9.25 and pka(acetic acid) = 4.76.

First we state the proton balance (the amount of protons taken up have to be equal to the amount of protons given off in the system): Initially we have H2O and CH3COONH4.

Proton balance: [H3O+] + [CH3COOH] = [OH-] + [NH3]

At pH = 7, [H3O+] = [OH-] =10^-7 M At this pH the proton balance can be simplified as [CH3COOH] = [NH3]. The simplified proton balance will be true only at a pH that is exactly in the middle of the two pka-values. We get pH = (4.76 + 9.25)/2 = 7.005 ≈ 7 (only one significant figure is given, since you have stated the concentration as 1 M).