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pH and Water

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    1343
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    pH

    The pH of a solution ranges from 1-14, 1-6 are acidic, 7 is neutral, and 8-14 are basic. It is a measure of the amount of hydrogen ion concentration in a solution and any change greater than 0.5 can cause loss of function or death to any organism. Maintaining a constant pH is vital to existence, because pH also deals with the conformation and formation of hydrogen bonding. The change in pH can cause bonds to weaken, break, or not form at all.

    Water and pH

     
    Water is essential for life; and it has unusual physical properties related to the weak 
    forces that connect atoms
    • covalent: strong (142 –O:O- to 946 N:::N, -C:C- 343 kJ/mol; 0.15 nm)
    • hydrogen: weak (12-30 kJ/mol ; 0.3 nm; combine two electronegative atoms, O, N, e.g. O...H-O: water, note 10-ps “flickering clusters”)
    • ionic: moderate (20 kJ/mol ; 0.25 nm in crystal; but moderated by H2O: F =Q1Q2/εr2 and ε [dielectric constant] = 1 in vacuum, 80 in H2O)
    • Vanderwaal’s: weak (0.4-4.0 kJ/mol ; 0.1 nm)
    • hydrophobic: moderate (<40 kJ/mol; depend on hydrocarbons disrupting the H-bonds “flickering clusters” of H2O)
    Dissociation of H-O bonds in water
    H2O + H2O ↔ H3O+ + OH-
    [H3O+] = [OH-] = 10-7 in pure H2O
    as with hydrogen bonds, note “flickering” as H is traded around H2O molecules
    Note the convention:
    H2O ↔ H+ + OH- and [H3O+] = [H+]
    Keq = [H+][OH-]/[H2O]
    Because [H2O] is approx. constant, define Kw = Keq *[H2O]
    Kw = [H+][OH-] = (10-7)( 10-7) = 10-14 M2
    Define the p function: p[X] = - log10[X]
    pKw = pH + pOH = 7 + 7 = 14  (because –log(10-14) = 14)
     

    Strong Acids and Bases

    pH of 0.1 M HCl = 1, etc.
    pOH of 0.1 M NaOH = 1; pH of 0.1 M NaOH = 13; etc
     
    What is the pH of 1 nM HCl? ([H+] = 10-7+ 10-9M)
    Can you made a solution of pH 0? (pH 0: - log10(X) = 0, X = exp10(-0) = 1 M)
    **Note that a base can be either an OH- donor or an H+ acceptor
     

    Weak Acids

    HA↔ H+ + A- HA is “conjugate acid”; Ais conjugate base
    Ka = [H+][ A-]/[HA] = [H+]*[ A-]/[HA]
    pKa = pH – log ([ A-]/[HA])
     
    Example: CH3COOH ↔ H+ + CH3COO- Ka = 1.74 x 10-5; pKa = 4.76
    What is the pH of a 0.1 M CH3COOH solution? What is [H+]?
    CH3COOH ↔ H+ + CH3COO-
    0.1 – X         X         X
     
    Ka = [H+]*[CH3COO-]/[CH3COOH] = 1.74 x 10-5
    X*X = (0.1-X)* 1.74 x 10-5
    Assume (0.1-X) = 0.1 if error < 5%
    X = sqrt (1.74 x 10-6) = 1.32 x 10-3 = [H+]
    pH = - log10[H+] = 2.88
     

    Weak Bases

    A- + H2O ↔ HA + OH- HA is “conjugate acid”; A is conjugate base
    Keq = [HA][OH-]/[ A-] [H2O]
    Kb = Keq* [H2O] = [HA][OH-]/[ A-] Show that Kw = Ka * Kb
    pKb = pOH – log ([HA]/[ A-]) pKw = pKa + pKb = 14  Useful!
     
    Example: imidazole C3N2H4 + H2O ↔ C3N2H5+ + OH-
    Kb = 0.98 x10-7; pKb = 7.01
     
    What is the pH of a 0.1M imidazole solution in water? What is [H+]?
    CH3COOH ↔ H+ + CH3COO-
    0.1 – X         X         X
     
    Kb = [C3N2H5+][OH-]/[C3N2H4] = 0.98 x 10-7
    X*X = (0.1-X)* 0.98 x 10-7 Assume (0.1-X) = 0.1 
    X = sqrt (0.98 x 10-8) = 1.0 x 10-4 = [OH-]
    pOH = 4; pH = 14 – 4 = 10; [H+] = 10-10 M