Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, 0.1M $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

$$ \begin{align} \ce{ M(OH)3 <=> M(OH)^{+}_{2} + {OH^{-}} } \qquad pK_{b1} = 0.5 \\ \ce{ M(OH)^{+}_{2} <=> M(OH)^{2+} + {OH^{-}} } \qquad pK_{b2} = 0.7 \\ \ce{ M(OH)^{2+} <=> M^{3+} + {OH^{-}} } \qquad pK_{b3} = 9.0 \end{align} $$ All of the species in above equilibria are water soluble.

Progress so far:

1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

$$\begin{align} \ce{ M^{3+} + {H2O} <=> {M(OH)^{2+}} + H+ } \qquad pK_{a1} = 14.0 - 9.0 = 5.0 \quad ---(a) \\ \ce{ {M(OH)^{2+}} + {H2O} <=> {M(OH)^{+}_{2}} + H+ } \qquad pK_{a2} = 14.0 - 0.7 = 13.3 \quad ---(b)\\ \ce{ {M(OH)^{+}_{2}} + {H2O} <=> {M(OH)_3} + H+ } \qquad pK_{a3} = 14.0 - 0.5 = 13.5 \quad ---(c) \end{align} $$

3. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
4. Looking at the $pK_a$ values of $(b)$ and $(c)$ it can be assumed that reaction $(a)$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Thanks a lot.