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    Titration Curve of Weak Acid with Strong Base

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    A 30 mL sample of 0.35 M of lactic acid is titrated with a 0.25 M NaOH solution. Calculate the pH before NaOH is added. After 20 mL of NaOH is added. After 42 mL. After 45 mL. After 60 mL. After 80 mL. Graph the titration curve.

    © BrainMass Inc. brainmass.com December 24, 2021, 8:05 pm ad1c9bdddf
    https://brainmass.com/chemistry/titration/titration-curve-weak-acid-strong-base-245959

    SOLUTION This solution is FREE courtesy of BrainMass!

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    This site gives the Ka of lactic acid: http://www.mpcfaculty.net/mark_bishop/weak_acid_equilibrium.htm

    To simplify the reaction, I will use HA for lactic acid:

    HA + NaOH -> H2O + NaA

    This problem will use the Ka expression as in problem 9.

    Initial pH:
    (please see the attached file)

    Now you can use the pH equation:
    (please see the attached file)

    After 42 mL

    When NaOH is added, it neutralizes the acid. You need to know the moles of each:

    0.030 L * 0.35 M = 0.0105 mol lactic acid
    0.020 L * 0.25 M = 0.005 mol NaOH

    0.005 mol of lactic acid will be neutralized and 0.005 mol of the conjugate base will be added.

    0.0105 mol - 0.005 mol = 0.0055 mol / 0.050 L = 0.11 M HA
    0.005 mol / 0.050 L = 0.1 M A-
    (please see the attached file)

    Now you can use the pH equation:
    (please see the attached file)

    After 45 mL

    When NaOH is added, it neutralizes the acid. You need to know the moles of each:

    0.030 L * 0.35 M = 0.0105 mol lactic acid
    0.045 L * 0.25 M = 0.01125 mol NaOH

    0.01125 mol of lactic acid will be neutralized and 0.01125 mol of the conjugate base will be added. Some NaOH will be left as well.

    0.0105 mol - 0.01125 mol = 0 M HA
    0.01125 mol / 0.075 L = 0.15 M A-

    Now you will need to use the Kb expression:
    (please see the attached file)

    Because the Kb is so small, you can use the [OH-] to find pOH and pH. Remember that not all of the NaOH was used up:

    0.00075 mol / 0.075 L = 0.01 M NaOH

    Now you can use the pOH equation:
    (please see the attached file)

    The pH is 14 - pOH.

    14 - 2 = 12

    After 60 mL

    When NaOH is added, it neutralizes the acid. You need to know the moles of each:

    0.030 L * 0.35 M = 0.0105 mol lactic acid
    0.060 L * 0.25 M = 0.015 mol NaOH

    The acid will neutralize 0.0105 mol of NaOH:

    0.015 - 0.0105 = 0.0045 mol NaOH
    0.0045 mol / 0.090 L = 0.05 M NaOH

    Because the Kb is so small, you can use the [OH-] to find pOH and pH. Now you can use the pOH equation:
    (please see the attached file)

    The pH is 14 - pOH.

    14 - 1.30 = 12.7

    After 80 mL

    When NaOH is added, it neutralizes the acid. You need to know the moles of each:

    0.030 L * 0.35 M = 0.0105 mol lactic acid
    0.080 L * 0.25 M = 0.020 mol NaOH

    The acid will neutralize 0.0105 mol of NaOH:

    0.020 - 0.0105 = 0.0095 mol NaOH
    0.0095 mol / 0.110 L = 0.086 M NaOH

    Because the Kb is so small, you can use the [OH-] to find pOH and pH. Now you can use the pOH equation:
    (please see the attached file)

    The pH is 14 - pOH.

    14 - 1.07 = 12.93

    I have also included an Excel sheet with the values.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 8:05 pm ad1c9bdddf>
    https://brainmass.com/chemistry/titration/titration-curve-weak-acid-strong-base-245959

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