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    pH during titration prcoess

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    Calculate Titration. See attached file for full problem description.

    19. Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 mL of titrant in the titration of 25.0 mL of 0.200 M HA with 0.100 M NaOH. Ka = 2.0 * 10^-5.

    20. Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 mL of titrant in the titration of 50.0 mL of 0.100 M NH3 with 0.100 M HCl.

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    https://brainmass.com/chemistry/titration/ph-during-titration-prcoess-105567

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    Calculate the pH at 0, 10.0, 25.0, 50.0 and 60.0 mL titrant in the titration of 25.0 mL of 0.200M HA with 0.100M NaOH. (Ka for HA is 2 x 10^-5).

    (a) Addition of 0.0 mL base NaOH.

    Initial Concentration 0.200 0 0
    Equilibrium, M 0.200-x x x
    Assume at the equilibrium is x M.
    Then
    Since the concentration of HAis 0.300M ,which is much greater than 100Ka, then x at the denominator can be neglected.
    So

    Or
    M.

    (b) Addition of 10 mL of NaOH

    Initial mmol 5.00 1.00 0 0
    Final, 4.00 0 1.00 1.00
    The initial moles of is
    And NaOH is
    So after reaching the equilibrium, there is still 4 mmol of left.
    This is a buffer solution,
    , where V is the final volume of the mixture in mL.

    Or you can deduct it by yourself as followings
    Assume at the equilibrium is x M.

    Initial 4.00 ...

    Solution Summary

    The solution shows how to calculate the pH during the titration process when different amount of titrant added. The solution is detailed and well presented.

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