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:

$\ce{NH_4CH_3COO &lt;=&gt; NH_4^+ + CH_3COO^-}$
$\ce{NH_4^+ &lt;=&gt; NH_3 + H^+}$
$\ce{CH_3COOH &lt;=&gt; CH_3COO^- + H^+}$

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?
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:  $\mathrm{p}K_{\mathrm{a}(\ce{NH4+})} = 9.25$ and
$\mathrm{p}K_{\mathrm{a}(\ce{CH3COOH})} = 4.76$.
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 $\ce{H2O}$ and $\ce{CH3COONH4}$.
Proton balance:
$[\ce{H3O+}] + [\ce{CH3COOH}] = [\ce{OH-}] + [\ce{NH3}]$
At $\mathrm{pH} = 7, [\ce{H3O+}] = [\ce{OH-}] = \pu{10^{-7} M}$:
At this $\mathrm{pH}$ the proton balance can be simplified as $[\ce{CH3COOH}] = [\ce{NH3}]$.
The simplified proton balance will be true only at a $\mathrm{pH}$ that is exactly in the middle of the two $\mathrm{p}K_{\mathrm{a}}$-values. We get $\mathrm{pH} = (4.76 + 9.25)/2 = 7.005 \approx 7$ (only one significant figure is given, since you have stated the concentration as $\pu{1 M}$).