Calculating the pH of a ammonium hydrogen sulfate solution

I am attempting to complete a question in which I need to use an ICE table to calculate the pH of a $\pu{2.61 mol L-1}$ $\ce{NH4HSO4}$ (ammonium hydrogen sulfate) solution. 


When $\ce{NH4HSO4}$ dissociates I know that it splits into $\ce{NH4+}$ and $\ce{HSO4-}$ ions. 


According to my textbook, $\ce{HSO4-}$ is actually an acid and not a base, so I set up my equilibrium expression like such:


$$\ce{NH4+ + HSO4- + 2 H2O <=> 2 H3O+ + SO4^2- + NH3}$$


$\ce{NH4+}$ has a $K\_\mathrm{a}$ of $\pu{5.56e-10}$. $\ce{HSO4-}$ has a $K\_\mathrm{a}$ of $\pu{1.2e-2}$.


According to my ICE table: 


$\ce{NH4+}$ and $\ce{HSO4-}$ have starting concentrations of $\pu{2.61 mol L-1}$.


Both lose $x~\pu{mol L-1}$, and have equilibrium concentrations of $2.61 - x$.


$\ce{H3O+}$ has a concentration of $2x$ at equilibrium. $\ce{SO4^2-}$ and $\ce{NH3}$ each have a equilibrium concentrations of $x$. 


To calculate the final answer I am using the $K\_\mathrm{a}$ of $\ce{NH4+}$ and plugging in the values from my ICE table for the equilibrium concentrations of the products and reactant. 


Since $K\_\mathrm{a}$ of $\ce{NH4+}$ is so small, I am disregarding the changes and substituting $2.61$ instead of $2.61 - x$. 


The final answer I am getting is not the same as the answer in the back of the book which is approximately $\mathrm{pH} = 0.76$.


What is wrong with my process?




The hydrogen sulphate ion is a much stronger acid than that of the ammonium ion. Thus, almost every hydronium ion in the solution comes from the dissociation of hydrogen sulphate ions.


This said, the assumption,



> 
> "$\ce{H3O+}$ has a concentration of $2x$ at equilibrium. $\ce{SO4^2-}$ and $\ce{NH3}$ each have a equilibrium concentrations of $x$"
> 
> 
> 


is wrong.


When you solve the quadratic equation


$$K\_\mathrm{a} = \frac{[\ce{H3O+}] [\ce{SO4^2-}]}{[\ce{HSO4^-}]} = \frac{x^2}{c - x}$$


you will get a pH near your text book's value.


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