How Does Excess HCl Affect pH in Methyl Ammonia Titration?

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Discussion Overview

The discussion centers on the effect of excess hydrochloric acid (HCl) on the pH during the titration of methyl ammonia. Participants explore calculations related to pH at various volumes of titrant added, particularly focusing on the scenario where excess HCl is present after reaching a certain volume.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant calculates the pH at different volumes of HCl added, providing specific values for 15 mL, 25 mL, and 50 mL of titrant.
  • Another participant emphasizes the importance of understanding the reaction between methyl ammonia and HCl, suggesting that the millimoles of reactants should be considered to determine which is in excess.
  • A participant confirms the calculations of millimoles for both methyl ammonia and HCl, illustrating the stoichiometry of the reaction and the final amounts of each species present after the reaction.
  • There is a suggestion that the pH can be calculated using the negative logarithm of the concentration of excess HCl in the final solution volume.

Areas of Agreement / Disagreement

Participants generally agree on the stoichiometry of the reaction and the method for calculating pH, but there is uncertainty regarding the exact pH value at 60 mL of titrant, particularly in how to handle the excess HCl.

Contextual Notes

Participants discuss the calculations based on specific volumes and concentrations but do not resolve the final pH value at 60 mL of titrant, leaving some assumptions about the contributions of other species in the solution unaddressed.

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:

[tex]MeNH_2+HCl \longrightarrow MeNH_3^+Cl^-[/tex]

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 [itex]MeNH_3^+[/itex] can be easily omitted.
 
This is what you were trying to tell me, correct?
(0.10M HCl)(60mL)=6mmol
(0.10M [tex]MeNH_2[/tex])(50mL)=5mmol


[tex]MeNH_2+HCl \longrightarrow MeNH_3^+Cl^-[/tex]

initial: 5mmol[tex]MeNH_2[/tex] 6mmol[tex]HCl[/tex]
[tex]\Delta[/tex]: -5mmol [tex]MeNH_2[/tex] -5mmol[tex]HCl[/tex]
final: 0mmol[tex]MeNH_2[/tex] 1mmol[tex]HCl[/tex] 5mmol[tex]MeNH_3^++Cl^-[/tex]

[tex]\frac{1mmol}{50mL (analyte) + 60mL (titrant)}[/tex]
 
This is it. Congrats. The negative logarithm of the result will be your pH value.
 

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