Volumetic Analysis - calculate concentration

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SUMMARY

The discussion focuses on calculating the concentration of ammonia in a 'Cloudy Ammonia' solution through volumetric analysis. A 25.00 mL sample was diluted to 250.00 mL, and a 25.00 mL aliquot was titrated with 0.200 M hydrochloric acid, requiring 26.75 mL to reach the endpoint. The chemical reaction involved is NH3 + HCl → NH4Cl. Participants emphasized the importance of understanding acid-base titration calculations and encouraged users to consult course notes or online resources for further assistance.

PREREQUISITES
  • Understanding of acid-base titration principles
  • Familiarity with volumetric flasks and their use
  • Knowledge of molarity calculations
  • Basic chemistry equations involving NH3 and HCl
NEXT STEPS
  • Study the process of acid-base titration in detail
  • Learn how to calculate molarity and concentration from titration data
  • Explore examples of volumetric analysis in chemistry textbooks
  • Review the stoichiometry of the reaction between ammonia and hydrochloric acid
USEFUL FOR

Chemistry students, educators, and anyone involved in laboratory work requiring volumetric analysis and titration techniques.

Mr Mooch
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Please Help! Volumetic Analysis - calculate concentration

Dear Physics Forum Fiends,

I was wondering if anyone out there who needs to earn some extra points to get into heaven perhaps or is simply feeling extraordinarily generous, might like to give this problem a go?
I am woefully lost in chemistry!

Homework Statement



A 25.00mL sample of 'Cloudy Ammonia' was diluted to a volume of 250.00mL in a volumetric flask. A 25.00mL aliquot of the diluted solution was titrated with 0.200 M hydrochloric acid. To reach the end point, 26.75 mL of acid was required.
Calculate the concentration of ammonia in the cloudy ammonia.


Homework Equations



NH3 + HCl > NH4Cl
 
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Welcome to PF:
This is an acid-base titration calculation ... from the look of it, this is part of a course so you should find examples in your course notes. Failing that - there are many many online.

Give the calculation an honest try and we'll help you out where you get stuck.
 


Hello,

Thank you for your reply. I just got some help from somewhere else which has helped me to understand what's going on a lot better.

Thanks,
 


...but I really appreciate your offer of assistance!

Best wishes,
 

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