acid base titration problem

[Mag]

consider the titration of 50.0mL of 0.10 M methyl ammonia with 0.10 M HCl. Calculate the pH after
15mL of titrant= 11.05
25mL of titrant= 10.68
50mL of titrant= 8.98
60mL of titrant= ?
At 60 mL there is an excess of 1 mmol of HCl. At this point do I say that the pH is 1 (-log of 0.10)? I'm at a loss.


Mag

[chem_tr]

You need to know the reaction first:

MeNH_2+HCl \longrightarrow MeNH_3^+Cl^-

Now you can place the millimoles of reactants to see which is excessive. You can find the millimole amounts by multiplying milliliters with molarity, but I think you know how to do this calculation.

Please consider the volume and millimoles together when trying to find the pH of the solution, 0.1 millimoles of HCl is in excess in 110 mL of total solution, where the contribution of MeNH_3^+ can be easily omitted.

[Mag]

This is what you were trying to tell me, correct?
(0.10M HCl)(60mL)=6mmol
(0.10M MeNH_2)(50mL)=5mmol


MeNH_2+HCl \longrightarrow MeNH_3^+Cl^-

initial: 5mmolMeNH_2 6mmolHCl
\Delta: -5mmol MeNH_2 -5mmolHCl
final: 0mmolMeNH_2 1mmolHCl 5mmolMeNH_3^++Cl^-

\frac{1mmol}{50mL (analyte) + 60mL (titrant)}

[chem_tr]

This is it. Congrats. The negative logarithm of the result will be your pH value.