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Buffers - Titrations

  1. Mar 5, 2008 #1
    You have 35.00 mL of a 0.350 M aqueous solution of the weak base NH2OH (Kb = 6.6 x 10^-9). This solution will be titrated with 0.350 M HCl.
    (a)How many mL of acid must be added to reach the equivalence point?
    (b)What is the pH of the solution before any acid is added?
    (c)What is the pH of the solution after 10.00 mL of acid has been added?
    (d)What is the pH of the solution at the equivalence point of the titration?
    (e)What is the pH of the solution when 40.00 mL of acid has been added?

    (a) Since the moles of acid and base are the same at the equivalence point, there will be 35 mL of HCl added.

    (b)Kb = [NH3][OH-]/[NH2OH]
    =6.6 x 10^-9 = x^2/.350
    =4.8 x 10^-5
    pOH = -log(4.8 x 10^-5)
    =4.32
    pH = 9.68

    (c) moles NH2OH = 0.03500 L / 0.350 moles/L = 1.23 x 10^-2
    moles HCl = 0.01000 L / 0.350 moles/L = 3.50 x 10^-3

    NH2OH + HCl ---> NH3 + H20
    Initial 1.23 x10^-2 3.50 x 10^-3 0
    Change -3.50 x 10^-3 -3.50 x 10^-3 +3.50 x 10^-3
    Equilibrium 8.80 x 10^-3 0 3.50 x 10^-3

    NH2OH ---> NH3 + OH-

    Kb = [NH3][OH-]/[NH2OH]
    6.60 x 10^-9 = (3.50 x 10^-3)x/8.80 x 10^-3
    [OH-] = 1.70 x 10^-8

    pOH = -log[OH-]
    =7.780
    Therefore, pH = 6.220

    (d) moles NH2OH = 1.23 x 10^-2
    moles HCl = 1.23 x 10^-2

    NH2OH + HCl ---> NH3 + H20
    Initial 1.23 x 10^-2 1.23 x 10^-2 0
    Change -1.23 x 10^-2 -1.23 x 10^-2 +1.23 x 10^-2
    Equilibrium 0 0 1.23 x 10^-2

    [NH2-] = 1.23 x 10^-2/0.0700 L = 0.176 M

    NH3 + H2O ---> NH2- + H+
    Initial 0.176 0 0
    Change -x +x +x
    Equilibrium 0.176 - x x x

    Ka = Kw/Kb
    =1.52 x 10^-6

    Ka = [NH2-][H+]/[NH3]
    1.52 x 10^-6 = x^2/0.176
    [H+] = 5.17 x 10^-4

    pH = -log[H+]
    =3.286

    (e)moles NH2OH = 1.23 x 10^-2
    moles HCl = 0.04000 L / 0.350 moles/L = 1.40 x 10^-2

    NH2OH + HCl ---> NH3 + H2O
    Initial 1.23 x 10^-2 1.40 x 10^-2 0
    Change -1.23 x 10^-2 -1.23 x 10^-2 +1.23 x 10^-2
    Equilibrium 0 1.70 x 10^-3 1.23 x 10^-2

    [H+] = 1.70 x 10^-3/0.07500 L = 2.27 x 10^-2 M

    pH = -log[H+]
    =1.645

    Does this look right to anyone? If not could you please tell me where I'm going wrong? Also, do my significant digits look correct? I get confused on how they work with logs.
     
  2. jcsd
  3. Mar 6, 2008 #2
    Lol....you go to Brock right?
     
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