Chemistry: A titration of ethylamine with HCl.

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SUMMARY

The discussion focuses on the titration of 50 mL of 0.150 M ethylamine (C2H5NH2) with 0.100 M HCl, specifically calculating the pH at two points: after adding 10 mL more HCl than required and after adding 75% of the required acid. The relevant equations include the Henderson-Hasselbalch equation, pH = pKa + log(base/acid), with a given Ka of 2.34 x 10^-4. The first part of the problem involves limiting reagent calculations, while the second part utilizes the Henderson-Hasselbalch equation for pH determination.

PREREQUISITES
  • Understanding of acid-base titration concepts
  • Familiarity with the Henderson-Hasselbalch equation
  • Knowledge of limiting reagent calculations
  • Basic chemistry knowledge regarding pKa and Ka values
NEXT STEPS
  • Study the application of the Henderson-Hasselbalch equation in various titration scenarios
  • Learn about calculating pH at different points in a titration curve
  • Explore limiting reagent concepts in acid-base reactions
  • Review the properties and behavior of weak bases like ethylamine in titrations
USEFUL FOR

Chemistry students, educators, and anyone involved in laboratory work related to acid-base titrations and pH calculations.

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Homework Statement


For the titration of 50 mL of 0.150 M ethylamine (C2H5NH2), with 0.100 M HCl, find the pH :
a) when 10 mL more of HCl has been added than is required
b) when 75% of the required acid has been added

Homework Equations


Maybe Henderson Hasselbalch?: pH = pKa + log (base/acid)
Ka = 2.34*10^-4

The Attempt at a Solution


I honestly do not know where to go with this.
 
Last edited:
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I managed to solve most of the problem by now, just can't get the two parts still in the question.
 

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