How Does Excess HCl Affect pH in Methyl Ammonia Titration?

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The discussion focuses on calculating the pH during the titration of 50.0 mL of 0.10 M methyl ammonia with 0.10 M HCl. After adding 60 mL of HCl, there is an excess of 1 mmol of HCl. The reaction between methyl ammonia (MeNH2) and HCl produces methyl ammonium chloride (MeNH3+Cl-). The initial amounts are 5 mmol of MeNH2 and 6 mmol of HCl. After the reaction, all MeNH2 is consumed, leaving 1 mmol of excess HCl. To find the pH, the concentration of HCl in the total volume (110 mL) is calculated, resulting in a concentration of 0.00909 M. The pH is then determined using the formula -log(0.00909), yielding a pH of approximately 2.04. This indicates a strong acidic solution due to the excess HCl.
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
 
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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.
 
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)}
 
This is it. Congrats. The negative logarithm of the result will be your pH value.
 
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