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23 August, 09:28

A 25.0 mL sample of a 0.100 M solution of acetic acid is titrated with a 0.125 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. The Ka for acetic acid is 1.76*10-5.

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Answers (2)
  1. 23 August, 09:30
    0
    Ater adding 10.0 mL the pH = 4.74

    After adding 20.0 mL the pH = 8.75

    After adding 30.0 mL the pH = 12.36

    Explanation:

    Step 1: Data given

    Volume of a 0.100 M acetic acid solution = 25.0 mL

    Molarity of NaOH solution = 0.125 M

    A) 25.0mL of HAc (acetic acid) and 10.0mL of NaOH

    Calculate moles

    Equation: NaOH + HAc - -> NaAc + H20

    Moles HAc = 0.025 L * 0.100 mol / L = 0.00250 moles

    Moles NaOH = 0.01 L * 0.125mol/L = 0.00125 moles

    Initial moles

    NaOH = 0.00125 moles

    HAc = 0.00250 moles

    NaAC = 0 moles

    Moles at the equilibrium

    NaOH = 0.00125 - 0.00125 = 0 moles

    HAc = 0.00250 - 0.00125 = 0.00125 moles

    NaAC = 0.00125 moles

    Calculate molarity

    [HAc] = [Ac-] = 0.00125 moles / 0.035 L = 0.0357M

    Concentration at equilibrium

    HAc Ac - + H + Ka = 1.76 * 10^-5

    [HAc] = 0.0357 - x M

    [Ac-] = 0.0357 + X M

    [H+] = XM

    Ka=1.76*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]

    Ka=1.76*10^-5 = [0.00357][x]/[0.0357] usually good approx.

    Ka = 1.76 * 10-5 = x = [H+]

    pH = - log [H+] = - log [1.76*10^-5] = 4.74

    B) 25.0mL HAc and 20.0mL NaOH

    Calculate moles

    Equation: NaOH + HAc - -> NaAc + H20

    Moles HAc = 0.025 L * 0.100 mol / L = 0.00250 moles

    Moles NaOH = 0.02 L * 0.125mol/L = 0.00250 moles

    Initial moles

    NaOH = 0.00250 moles

    HAc = 0.00250 moles

    NaAC = 0 moles

    Moles at the equilibrium

    NaOH = 0.00250 - 0.00250 = 0 moles

    HAc = 0.00250 - 0.00250 = 0 moles

    NaAC = 0.00250 moles

    Calculate molarity

    [NaAC] = 0.00250 moles / 0.045 L = 0.0556M

    Concentration at equilibrium

    Ac - + H20 HAc + OH - Ka = 1.76 * 10^-5

    [Ac] = 0.0556 - x M

    [OH-] = XM

    [HAc] = XM

    **Kw = 1*10^-14 at STP I'll assume this for this problem

    **Kw = Ka*Kb ⇒ Kb = Kw/Ka

    **Kb = (1*10^-14) / (1.76*10^-5) = 5.68*10^-10

    Kb = ([HAc][OH-]) / [Ac-] = ([x][x]) / [0.0556]

    Kb = x² / 0.0556

    5.68*10^-10 = x² / 0.0556

    so x = 5.62*10^-6

    pOH = - log [OH-] = 5.250

    pH = 14 - pOH

    pH 14 - 5.25 = 8.75

    C) 25.0mL HAc and 30.0mL NaOH

    NaOH + HAc - -> NaAc + H20

    Initial numbers of moles

    Moles NaOH = 0.125M * 0.03 L = 0.00375 moles

    Moles HAc = 0.00250 moles

    moles NaAC = 0 moles

    Moles at the equilibrium

    Moles NaOH = 0.00375 moles - 0.00250 = 0.00125 moles

    Moles HAc = 0.00250 moles - 0.00250 = 0 moles

    moles NaAC = 0.00250 moles

    After the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.

    Calculate NaOH molarity

    (0.00125moles/0.055 L = 0.0227M NaOH

    Strong Bases dissociate 100% so [OH-] = 0.0227M

    pOH = - log[0.0227] = 1.644

    pH = 14-pOH = 12.36
  2. 23 August, 09:49
    0
    10mil 76 the experllon is the most numbers
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Home » Chemistry » A 25.0 mL sample of a 0.100 M solution of acetic acid is titrated with a 0.125 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. The Ka for acetic acid is 1.76*10-5.
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