An equilibrium question involving mixing equal volumes of HCl and H2SO4

Question

Question: Calculate the concentrations of H3O(+) and SO4(2-) in a solution prepared by mixing equal volumes of 0.2M HCl and 0.6M H2SO4. (Ka1 for H2SO4 is very large, and Ka2 for H2SO4 is 1.2E-2)

Here is my thought process:

1. Since the Ka1 for H2SO4 is very large, it will completely dissociate into HSO4- and H3O+. I then used Ka2 to find the equilibrium concentration of HSO4-, H3O+, and SO4(2-) before mixing it with HCl.

2. When the solutions of equal volumes were mixed, the concentration of each species in both solutions were halved. The concentration of H3O+ is now the sum of the halved H3O+ concentrations from each solution. Since the concentrations have changed, I calculated Qc from the new concentrations and determined that the dissociation equation proceeds to the right.

3. Lastly, I used Ka2 again to calculate the equilibrium concentrations in the mixed solution.

The final answer I got was [H3O+]=0.40856M and [SO4(2-)]=0.00856M.

I do not have the answer key for this question, so I would greatly appreciate it if you could help verify my thought process/answer. Thank you!

Answer 1

Mixing equal volumes of 0.2M $\ce{HCl}$ and 0.6M $\ce{H2SO4}$ would yield a solution with nominal concentration of 0.1M $\ce{HCl}$ and 0.3M $\ce{H2SO4}$.

Also given that $K_\mathrm{a2}$ for $\ce{H2SO4}$ is $1.2\times10^{-2}$.

RANT- I have come to absolutlye hate problems that don't use significant figures consistently. Is the initial HCl concentration really supposed to be 0.1 molar, or perhaps 0.10 molar?

Since the $K_\mathrm{a2}$ is given to two significant figures, I'll assume two significant figures for the concentrations.

Now the $\ce{HCl}$ will ionize completely, and the $\ce{Cl^-}$ anion can assumed to be a spectator ion.

Now the first ionization of $\ce{H2SO4}$ will ionize completely, and the second only partially.

Thus if $x=\ce{[H+]}$ from the ionization of $\ce{HSO4-}$, then the total $\ce{[H+]} = 0.4+x$

That means at equilibrium $\ce{HSO4-} = 0.3 -x$, and $\ce{SO4^{2-}} = x$

So:

\begin{align} K_\mathrm{a2} &= \dfrac{\ce{[H+][SO4^{2-}]}}{\ce{[HSO4-]}} \\ 1.2\times10^{-2} &= \dfrac{(0.40+x)(x)}{0.30-x} \\ 0 &= x^2 + 0.412x - 3.6\times10^{-3} \\ \end{align}

Thus $x=0.008560$

$\ce{[H+]} = 0.40 + 0.008560 = 0.41$

$\ce{[HSO4-]} = 0.30 - 0.008560 = 0.29$

$\ce{[SO4^{2-}]} = 0.008560 = 0.0086$