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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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    Attachments

    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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