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# Titrating hydrochloric and acetic acids with NaOH

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I would like to see the details of the calculations.
For my lab experiment I have the following data:
Titration 1:
25.00mL of unknown concentration HCl
2 drops of phenolpthalein to show end-point
1.0M NaOH
What is the molarity of the HCl?
What is the Ka of the HCl?
What is the equivalence point?
What is the half-equivalence point?
See attachment for titration details.

Titration 2:
5.00mL unknown concentration of acetic acid with 20.00 mL of H2O added and 2 drops of phenolpthalein to show end-point.
1.0M NaOH
What is the molarity of the acetic acid?
What is the Ka of the acetic acid?
What is the equivalence point?
What is the half-equivalence point?
See attachment for titration details.

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#### Solution Preview

Please see the attached file.

Titration 1:
Volume of NaOH 1.0M used = 15 (mL) - just before solution turns pink
Number of mole of NaOH used =Molarity*volume=1*0.015= 0.015 (mol)
The reaction between HCl and NaOH is:
HCl + NaOH = NaCl + H2O
Hence number of mole of HCl = Number of mole of NaOH = 0.015 (mol)
Molarity of HCl =Mol/Volume=0.015/0.025 = 0.6 (M)

For a strong acid such as HCl in this case, we just have to check in the literature for Ka value. I have found the following good reference with Ka (HCl) = 106
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/ph.php
You ...

#### Solution Summary

The solution provides detailed explanations and calculation on the concepts of titration, weak acid dissociation and equilibrium.

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