pH of ammonium acetate solution
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 $\pu{1 M}$).