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Help calculating theoretical pH

Discussion in 'The Aquarium' started by mudplayerx, Mar 29, 2007.

  1. mudplayerx

    mudplayerx New Member

    Hey guys I need help.
    Suppose I have a 5ml 0.25M solution of CH3COOH. I manually measured the pH with an electric pH meter and got pH 2.80.
    I then diluted the solution to 50.0ml for a molarity of 2.5x10^-4M. The measured pH of this solution was 3.23.
    How do I find the "theoretical pH" of these two solutions? I have no clue.
    The Ka of CH3COOH is 1.8x10^-5
    Thanks
  2. reefreak29

    reefreak29 New Member

    CH3COOH or Acetic acid
    OAc- is the shortcut for CH3COO- or the Acetate ion, the conjugate base of Acetic acid
    1) VHCl = 0 mL (before addn of HCl)
    Change of Concentrations Due to Dilution (M1V1 = M2V2)
    [OAc-] = 0.100 M
    50 mL
    [OAc-] = (0.100)(50)/95 = 0.0526 M
    [HOAc] = 0.100 M
    45 mL
    [HOAc] = (0.100)(45)/95 = 0.0474 M
    Vtotal = 95 mL
    You took 30.00 mL of the buffer:
    [OAc-] = 0.0526M
    30 mL
    [OAc-] =
    (0.0526)(30.00)
    =
    0.0522 M
    30.25 + 0
    [HOAc] = 0.0474 M
    30 mL
    [HOAc] =
    (0.0474)(30.00)
    =
    0.0470 M
    30.25 + 0
    Plug these values in the ICE Table:
    HOAc + H2O < = > H3O+ + OAc-]
    Initial 0.0470 M 10-7 0.0522 M
    Change -x +x +x
    Equilibrium 0.0470 - x x 0.0522 + x
    Then into the equilibrium expression:
    Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0.0522 + x)
    [HOAc] 0.0470 - x
    5% of 0.0474 = 2.37 x 10-3
    assume x is less than this value
    1.76 x 10-5 = x (0.0522 )
    0.0470
    Solving for x = [H3O+]
    [H3O+] = 1.58 x 10-5
    pH = - log [H3O+] = 4.80 M
    2) Addn of 0.100 M HCl
    Calculate these concentrations for each volume of acid added corresponding to the increments added in Table 12.2 in the manual:
    [OAc-] = 0.0526M
    30 mL
    [OAc-] =
    (0.0526)(30.00)
    30.25 + VHCl added
    [HOAc] = 0.0474 M
    30 mL
    [HOAc] =
    (0.0474)(30.00)
    30.25 + VHCl added
    Remember that the volumes are additive (i.e., the volume of HCl added after the second addition of the acid is 0.50 + 1.00 = 1.50 mL, third time is 1.50 + 1.50 = 3.0 mL and so on )
    You are adding an acid to the buffer, calculate [HCl] for each volume of acid added:
    [HCl]
    =
    0.100 M
    x
    V HCl added
    30.25 + VHCl added
    You will be adding this value to [HOAc] and subtracting it from [OAc-] in the ICE table:
    Example: After addition of 0.50 mL of HCl, [HCl] = 0.00163
    HOAc + H2O < = > H3O+ + OAc-]
    Initial 0. 0462 + 0.00163 10-7 0.0513 - 0.00163
    Change -x +x +x
    Equilibrium 0.0478 - x x 0.0497 + x
    Then into the equilibrium expression:
    Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0.0497 + x)
    [HOAc] 0.0478 - x
    5% of 0.0478 = 2.39 x 10-3
    assume x is less than this value
    1.76 x 10-5 = x (0.0497 )
    0.0478
    Solving for x = [H3O+]
    [H3O+] = 1.69 x 10-5 < 2.39 x 10-3
    pH = - log [H3O+] = 4.77 M
    Notice that the pH did not change that much compared to the original pH of 4.80 unlike the degassed water where the pH dropped from 7.0 to 2.1 with the addition of 0.25 mL of HCl.
    Also, notice that the addition of an acid decreases the pH.
    Part C: Addition of NaOH to Degassed Water
    1) Without the base, the pH should be close to 7.0, [OH-] = 10-7
  3. reefreak29

    reefreak29 New Member

    1) Without the base, the pH should be close to 7.0, [OH-] = 10-7
    2) Calculate [NaOH] for each volume corresponding to each increment of base added in Table 12.3 on p.74 of the manual:
    [NaOH]
    =
    0.100 M
    x
    V NaOH added
    V total
    Note that Vtotal = Vwater + Vindicator + VHCl added = 30.00 + 0.25 + V NaOH added
    3) Calculate [OH-] and the corresponding pH:
    NaOH is a strong acid and therefore it dissociates completely in water and therefore:
    [NaOH] = [OH-]
    pOH = - log [OH-]
    pH = 14 - pOH
    Part D: Addition of NaOH to Acetate Buffer (Weak acid/Conjugate base Combination)
    1) VNaOH = 0 mL (before addn of NaOH)
    Change of Concentrations Due to Dilution (M1V1 = M2V2)
    [OAc-] = 0.100 M
    50 mL
    [OAc-] = (0.100)(50)/95 = 0.0526 M
    [HOAc] = 0.100 M
    45 mL
    [HOAc] = (0.100)(45)/95 = 0.0474 M
    Vtotal = 95 mL
    You took 30.00 mL of the buffer:
    [OAc-] = 0.0526M
    30 mL
    [OAc-] =
    (0.0526)(30.00)
    =
    0.0522 M
    30.25 + 0
    [HOAc] = 0.0474 M
    30 mL
    [HOAc] =
    (0.0474)(30.00)
    =
    0.0470 M
    30.25 + 0
    Plug these values in the ICE Table:
    HOAc + H2O < = > H3O+ + OAc-]
    Initial 0.0470 M 10-7 0.0522 M
    Change -x +x +x
    Equilibrium 0.0470 - x x 0.0522 + x
    Then into the equilibrium expression:
    Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0.0522 + x)
    [HOAc] 0.0470 - x
    5% of 0.0474 = 2.37 x 10-3
    assume x is less than this value
    1.76 x 10-5 = x (0.0522 )
    0.0470
    Solving for x = [H3O+]
    [H3O+] = 1.58 x 10-5
    pH = - log [H3O+] = 4.80 M (Same as in Part B)
    2) Addn of 0.100 M NaOH
    Calculate these concentrations for each volume of base added corresponding to the increments added in Table 12.4 in the manual:
    [OAc-] = 0.0526M
    30 mL
    [OAc-] =
    (0.0526)(30.00)
    30.25 + VNaOH added
    [HOAc] = 0.0474 M
    30 mL
    [HOAc] =
    (0.0474)(30.00)
    30.25 + VNaOH added
    Remember that the volumes are additive (i.e., the volume of NaOH added after the second addition of the base is 0.50 + 1.00 = 1.50 mL, third time is 1.50 + 1.50 = 3.0 mL and so on ).
    You are adding an acid to the buffer, calculate [NaOH] for each volume of acid added:
    [NaOH]
    =
    0.100 M
    >x
    V NaOH added
    30.25 + VNaOH added
    You will be subtracting this value from [HOAc] and adding it to [OAc-] in the ICE table:
    Example: After addition of 0.50 mL of NaOH, [NaOH] = 0.00163
    HOAc + H2O < = > H3O+ + OAc-]
    Initial 0. 0462 - 0.00163 10-7 0.0513 + 0.00163
    Change -x +x +x
    Equilibrium 0.0446 - x x 0.0529 + x
    Then into the equilibrium expression:
    Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0. 0529 + x)
    [HOAc] 0.0446 - x
    5% of 0.0446 = 2.23 x 10-3
    assume x is less than this value
    1.76 x 10-5 = x (0. 0529)
    0.0446
    Solving for x = [H3O+]
    [H3O+] = 1.48 x 10-5 < 2.23 x 10-3
    pH = - log [H3O+] = 4.83 M
  4. mudplayerx

    mudplayerx New Member

    Thanks for the extensive help reefreak!
    After rereading my work and looking at your work I have come to the conclusion that the professor wanted me to use a simpler approach to "theoretical ph."
    I think I'm supposed to use pH= -LogPka + [A-]/[HA] (Henderson Hasselbach formula) after finding the equilibrium concentrations from
    CH3COOH <----> CH3COO- + H+
    Does that sound right?
    ps- I am to find the theoretical pH of both solutions individually, not of both solutions mixed together. Sorry for the confusion.
  5. scotts

    scotts New Member

    The sad part is that MANY years ago I could have helped you out. Now when I open my college books I think, WOW I used to know this stuff.
  6. reefreak29

    reefreak29 New Member

    pH = pKa + log [A-]/[HA] :thinking:
    this is correct
  7. mudplayerx

    mudplayerx New Member

    Man I hate buffer solutions and acids/bases. This chapter takes soooo much pencilwork!
  8. reefreak29

    reefreak29 New Member

    Originally Posted by mudplayerx
    Man I hate buffer solutions and acids/bases. This chapter takes soooo much pencilwork!
    tep it sucks and i have to work with it every day

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