Is this problem about acid-base titration wrong? I wouldn't post mere 'problems' here normally, but I will hopefully be starting to give chemistry education to some 10-11-12th graders and I want to make sure that it is not I that has a problem but rather the problem in the book itself. Here is the problem:

$100~\mathrm{mL}$ of $\ce{H2SO4}$ solution is titrated with $5\cdot 10^{-2}\ \mathrm{M}$ solution of $\ce{NaOH}$. Here are the information that are given in the titration graph:
Total volume of $\ce{NaOH}$ solution added: $0$ | pH: $x$
Total volume of $\ce{NaOH}$ solution added: $400$ | pH: $7$
Total volume of $\ce{NaOH}$ solution added: $600$ | pH: $12$
According to what has been provided above, what is the $\mathrm{pH}$ of the initial $\ce{H2SO4}$ solution ($x$) ? (Assume that $\ce{H2SO4}$ fully dissociates in water to give $\ce{2H+}$ and $\ce{SO4^2-}$ ions)
A) 1
B) 2
C) 3
D) 4
E) 5

My answer: $-\log (2\cdot 10^{-1})$
My steps:

The $\ce{H+}$ ions in the acid solution have been neutralized by $400\ \mathrm{mL}$ of $5\cdot 10^{-2}\ \mathrm{M}\ \ce{NaOH}$ solution. That means that the amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of that $\ce{NaOH}$ solution must equal the amount of $\ce{H+}$ ions in the acid solution.

The amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of $5\cdot 10^{-2}~\mathrm{M}\ \ce{NaOH}$ solution can be found by taking the product of the liter of solution and the molarity of the solution. And that gives us $0.4\times 5\cdot 10^{-2} = 2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions. (Yes, I didn't use the units, I don't have to, I explained with words what I was going to do)

Since we used $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions to neutralize the acid solution, that should mean that there were $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{H+}$ in the acid solution.

To find the pH of the acid solution, we need to find the concentration of $\ce{H+}$ ions in that solution. The volume of the solution is known, $100~\mathrm{mL}$. And the amount of $ce{H+}$ ions too is known, $2\cdot 10^{-2}\ \mathrm{mol}$. The concentration of $\ce{H+}$ ions in the acid solution should therefore be $2\cdot 10^{-2} / 0.1 = 2\cdot 10^{-1}\ \mathrm{M}$.

$-\log[\ce{H+}] = -\log (2\cdot 10^{-1}) =$ none of the choices above.
Your calculations are correct. I also did a second sanity check: Adding $200~\mathrm{ml}\ \ce{NaOH}$ to a neutral solution does indeed give a $\mathrm{pH}\ 12$ solution, so they also wanted you to assume that the neutral point is the equivalence point.

Which answer to select? Well, considering that:

$$-\log(2 \cdot 10^{-2}) \approx 1.699 \approx 2$$

I would probably go with $2$. Along with telling the pupils that this is a badly worded question and in their exam there will be clearly solveable ones.