August 17, 2022, 11:45:24 PM
Forum Rules: Read This Before Posting

Topic: titration calculation  (Read 6594 times)

0 Members and 1 Guest are viewing this topic.


  • Guest
titration calculation
« on: October 28, 2004, 01:37:08 AM »
 I am having trouble with the simplicity of calculating molarity for a titration. Here is what I am given:
   25.49 mL of NaOH were required to neutralize 0.5208g of KHP (M.W.=204.33) dissolved in water. this served to standardize NaOH

The vinegar solution for titration was prepared in the following manner: 25.0 mL of vinegar were diluted to 250.0 nL in a volumetric flash, and 25.0 mL of this diluted solution required 22.62 mL of the above standardized NaOH to reach phenolphthalein endpoint.

what was the molarity of the NaOH solution
what is the percentage by mass of the acid in the original vinegar sample.

- So I just get really stuck on all the equation stuff. I understand the concept fine, I just don't know how to apply it to the calculation. If you could help me with some steps to take that would be great. Thanks.

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7982
  • Mole Snacks: +555/-93
  • Gender: Male
Re:titration calculation
« Reply #1 on: October 28, 2004, 04:22:17 AM »
NaOH + KHP = KNaP + H2O
MV(volume in Liters)= n (number of moles).
From reaction: moles of NaOH = moles of KHP
M(NaOH)=0.5208/(204.33 x 0.02549)=

From the second titration you can calculate molarity of diluted vinegar.
M1V1 = M2V2

Concentration of vinegar before dilution was 10 times greater  - 10 x M1 (25 ml diluted to 250 ml)

Molar concentration is defined as the content of substance in moles in 1 liter of solution.
Changing moles to grams (x60 in this case) and assuming density 1 g/ml (quite reliable for diluted solution of acetic acid) you can calculate mass percentage of undiluted vinegar
as  M1 x 60
(M1 x 10 x 100 / 1000)

Sponsored Links