Should the relatively low solubility of a base be considered in the calculation of the pH? Hydrochloric acid, $V=75.0~\mathrm{mL}$, $c=0.250~\mathrm{M}$, is added to $225.0~\mathrm{mL}$ of $0.0550~\mathrm{M}$ $\ce{Ba(OH)2}$ solution. What is the concentration of the excess $\ce{H+}$ or $\ce{OH-}$ ions left in this solution? (source)


I know the balanced equation but when considering $\ce{Ba(OH)2}$ is 'marginally soluble' which
I guess could be considered insoluble, $\ce{OH-}$ wouldn't really be in the result solution. 
Am I right or at where am I wrong? thanks for reading
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.