What are the limitations of the Hendersson-Hasselbalch equation?
The key approximation made in deriving the Henderson-Hasselbalch equation is that the equilibrium constant can be written as $$K=\frac{c_{H^+}c_{A^-}}{c_{HA}}$$
that is, we assume activity coefficients are unity. If you take the base-10 logarithm of this equation and rearrange terms you arrive at the HH equation:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}\right)$$ Problems usually arise with approximations made beyond the HH equation, in the context of a polyprotic compound particularly  by ignoring one of the equilibria, including perhaps water autodissociation, when this shouldn't be done.  

In the case of the problem you outline, you might 


ignore the second dissociation equilibrium and 
assume that added base neutralizes an equimolar amount of remaining acid $\ce{H2SO4}$ to form additional $\ce{HSO4-}$. 


Then

$$\begin{align}c_{HA} = c_{HA,\,initial}-c_{added\,OH^-}&amp;=\pu{0.0705 M} - \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&amp;=\pu{0.0538 M}\end{align}$$
$$\begin{align}c_{A^-} = c_{A^-,\,initial}+c_{added\,OH^-}&amp;=\pu{0.0295 M} + \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&amp;=\pu{0.0462 M}\end{align}$$

Applying the HH equation we then have, as the approximate pH
$$\begin{align}pH &amp;=1.91 + \log_{10}\left(\frac{0.0462}{0.0538}\right)\\&amp;=1.84\end{align}$$ which is above the starting pH of 1.53.

A more accurate value (although still approximate) can be obtained starting from the condition of charge balance for the system, and by assuming that the $\ce{OH^-}$ concentration is negligible, from which one obtains pH=1.75.