Should the relatively low solubility of a base be considered in the calculation of the pH?
The $K_\mathrm{sp}$ of  $\ce{Ba(OH)2}$ is $0.005$.  

Therefore, the solubility is great enough that 0.055M  $\ce{Ba(OH)2}$ can all be in solution.

Secondly, you are diluting the $0.055~\mathrm{M}$ solution, so the concentration will be even less.

Finally, even if there was too much  $\ce{Ba(OH)2}$ to dissolve in water, the $\ce{HCl}$ would react with the $\ce{OH-}$ that is in solution, causing more  $\ce{Ba(OH)2}$ to dissolve.