Calculating the pH of a ammonium hydrogen sulfate solution

Question

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?

Answer 1

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.