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Bronsted-Lowry, HA = H+ + A-
(conjugate base)
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Ka = [H+] [A-] / [HA]
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For Ka < 1, compound mostly associated
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For Ka > 1, compound mostly disassociated
into proton (hydronium in water) and conjugate base
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For H+ + A- = HA, K = [HA] / ([H+] [A-])
= 1/Ka
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pKa = -log (Ka), Ka = 10-pKa
(anti log of pKa)
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lower pKa, more acidic compound, less basic
conjugate base
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higher pKa, less acidic compound, more basic
conjugate base
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pKa of water = -log([H+] [-OH] / 56)
= -log((10-7) (10-7) / 56) = 7 + 7 + log(56) = 15.7
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pH = -log[H+]
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Ka = 10-pKa = (x) (x) / (I-x), compound
of initial concentration I disassociates into [H+] = [A-]
= x
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Ka (I-x) = x2
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x2 +Ka x -I Ka = 0
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[H+] = x = (-Ka +
Ö(Ka2+4 I Ka)) / 2,
negative answer does not make sense
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pH = -log(x)
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For HA + B- = A- + HB, K = Kf Kb
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Where HA = H+ + A-, Kf
= Kaf
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Where H+ + B- = HB, Kb
= 1/Kab
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K = (10-pKaf) (10--pKab) = 10(-pKaf + pKab)
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If -pKaf + pKab > 0, pKaf
< pKab, equilibrium favored towards the right, products
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If -pKaf + pKab < 0, pKaf
> pKab, equilibrium favored towards the left, reactants
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A mixture of HA and B- with initial concentrations I1
and I2, respectively, has a final concentration of A- and HB
of:
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K = 10(-pKaf + pKab)
= (x) (x) / ((I1-x) ((I2-x))
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K (I1-x) ((I2-x) = x2
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K (I1I2 - (I1+I2)
x + x2) = x2
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(K-1) x2 -K (I1+I2) x + K
I1 I2 = 0
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x = (K (I1+I2) ±
Ö((K (I1+I2))2-4
(K-1) K I 1 I2 )/(2 (K - 1)), check that the
solution is positive.
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The percentage of HA that is disassociated is x / I1
100 %
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The final ratio of [A-]/[HA] = x / (I1-x)
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Lewis Acids and Bases
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Nucleophiles and Electrophiles
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Reasons why some compounds are more acidic than others
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Electronegativity, examples: CH4, NH3,
H2O, HF
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Hybridization: sp3, sp2, sp
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Delocalization, Resonance effect, examples: carboxylic
acids and phenols
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Inductive effect, examples: halo substituted carboxylic
acids
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Sterics, examples: amines
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Hard (ionic) and Soft (covalent)
pKa values
CH4
49 |
NH3
36 |
H2O
15.7 |
HF
3.2 |
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PH3
26 |
H2S
7 |
HCl
-7 |
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H2Se |
HBr
-9 |
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HI
-10 |
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