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13 August, 10:55

Consider the titration of 100.0 mL of 0.200 M acetic acid () by 0.100 M. Calculate the pH of the resulting solution after the following volumes of have been added. 0.0 mL pH = 50.0 mL pH = 100.0 mL pH = 120.0 mL pH = 200.0 mL pH = 260.0 mL pH =

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  1. 13 August, 11:07
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    a) 2.72

    b) 4.26

    c) 4.74

    d) 8.65

    e) 8.79

    f) 12.22

    Explanation:

    Considered the titration of 100.0 mL of 0.200 M acetic acid (Ka=1.8 x 10^-5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.

    Step 1: Data given

    Volume of 0.200 M acetic acid = 100.0 mL = 0.100 L

    Concentration of KOH = 0.100M

    Ka of acetic acid = 1.8 * 10^-5

    pKa = 4.74

    Step 2: The balanced equation

    KOH + CH3COOH → CH3COOK + H2O

    a) When 0.0 mL is added

    pH = - log * (√[HA]*Ka)

    pH = - log (√ (0.200 * 1.8 x 10^-5)

    pH = 2.72

    b) When 50.0 mL KOH is added

    moles acetic acid = 0.100 L * 0.200 M = 0.0200 moles

    moles OH - added = 0.0500 L * 0.100 M=0.005 moles

    moles acetic acid in excess = 0.0200 - 0.005=0.015 moles

    moles acetate = 0.005 moles

    pH = pKa + log ([acetate]/[acetic acid])

    pH = 4.74 + log 0.005/0.015 = 4.26

    c) When 100.0 mL KOH is added

    moles acetic acid = 0.100 L * 0.200 M = 0.0200 moles

    moles OH - added = 0.100 L * 0.100 M=0.010 moles

    moles acetic acid in excess = 0.0200 - 0.010=0.010 moles

    moles acetate = 0.010 moles

    CH3COO - + H2O CH3COOH + OH-

    pH = 4.74 + log 0.0100/0.0100 = 4.74

    d) When 120.0 mL KOH is added

    moles acetic acid = 0.100 L * 0.200 M = 0.0200 moles

    moles OH - added = 0.120 L * 0.100 M=0.012 moles

    moles acetic acid in excess = 0.0200 - 0.012=0.008 moles

    moles acetate = 0.008 moles

    total volume = 0.220 L

    concentration acetate = 0.008/0.220=0.0364 M

    CH3COO - + H2O CH3COOH + OH-

    Kb = 5.6 * 10^-10 = x^2 / 0.0364-x

    x = [OH-] = 4.51 * 10^-6 M

    pOH = 5.35

    pH = 8.65

    e) When 200.0 mL KOH is added

    moles acetic acid = 0.100 L * 0.200 M = 0.0200 moles

    moles OH - added = 0.200 L * 0.100 M=0.0200 moles

    moles acetic acid in excess = 0.0200 - 0.0200 = 0 moles

    moles acetate = 0.020 moles

    total volume = 0.300 L

    concentration acetate = 0.020/0.300 = 0.0667 M

    CH3COO - + H2O CH3COOH + OH-

    Kb = 5.6 * 10^-10 = x^2 / 0.0667-x

    x = [OH-]=6.11 * 10^-6 M

    pOH = 5.21

    pH = 8.79

    f) When 260.0 mL KOH is added

    moles acetic acid = 0.100 L * 0.200 M = 0.0200 moles

    moles OH - added = 0.260 L * 0.100 M=0.0260 moles

    moles OH - in excess = 0.0260 - 0.0200 = 0.0060 moles

    total volume = 0.360 L

    [OH-] = 0.0060 moles / 0.360 L

    [OH-] = 0.0167

    pOH = - log [0.0167]

    pOH = 1.78

    pH = 14 - 1.78 = 12.22
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Home » Chemistry » Consider the titration of 100.0 mL of 0.200 M acetic acid () by 0.100 M. Calculate the pH of the resulting solution after the following volumes of have been added. 0.0 mL pH = 50.0 mL pH = 100.0 mL pH = 120.0 mL pH = 200.0 mL pH = 260.0 mL pH =
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