This is a draft cheat sheet. It is a work in progress and is not finished yet.
ACID
PROTON DONOR |
H+ |
Arrhenius Concept |
increases H+ in water (ex. H3O+) |
Bronsted-Lowry Concept |
can donate a proton (H+) |
Lewis Concept |
electron pair acceptor (covalent bond created) |
strong acids |
dissociate fully (100% ionized) |
weak acids |
dissociate partially (<100% ionized) |
ACID/BASE REACTIONS
can be ions or substances |
not limited to (aq) solutions |
some react as acid or base depending on other reactant |
salt |
ionic compound, product of acid base reaction, does NOT contain OH-, not metal oxide |
ex. HCl (aq) + KOH (aq) ----> KCl (aq) + H2O
AUTOIONIZATION OF WATER
small % of water undergoes ionization to produce ions |
Kw = 1 × 10^-14 |
ex. H2O (l) + H2O (l) --> H3O+ (aq) + OH- (aq) |
in pure water.... [H3O+] × [OH-] = 10-14 and [H3O+] = [OH-] = 10-7 |
the product between the molar concentrations of hydronium ion and hydroxide ion is a constant
pH SCALE
ACID |
BASE |
pH = -log[H3O+] OR [H3O+] = 10^-pH |
pOH = -log[OH-] OR [OH-] =10^-pH |
pH = -log[H+]
-pH = log[H3O+]
10 10
[H+] = 10^-pH
ACID IONIIZATION CONSTANT, Ka
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WEAK ACID IONIZATION |
equilibria of weak acid |
HA (aq) + H2O⇄H3O+(aq) + A- |
Ka = [H+][A-]/[HA] |
pKa = -logKa |
pH = -log [H3O+] |
[H3O+] = 10-pH |
pOH = -log[OH-] |
[OH-] = 10-pH |
p = -log |
Degree of Ionization, α |
neutral molecule splits into charged ions when exposed in a solution |
α of weak acid/base in water = fraction of total concentration that has formed ions |
ex. HA(aq) + H2O⇄H3O+(aq) + A- (aq) |
% ionization (weak acid) = |
[A-] eq/[HA] orginally × 100% |
the larger the Ka, the stronger the acid
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BASE
PROTON RECEPTOR |
OH- |
Arrhenius Concept |
increase [OH-] in water |
Bronsted-Lowry Concept |
can accept a proton (H+) |
Lewis Concept |
electron pair donor (covalent bond created) |
strong bases |
dissociate fully (100% ionized) |
weak bases |
dissociate partially (>100%) |
bitter taste, slippery feeling
BRONSTED ACID/BASE STRENGHT
acid 1 + base 2 <--> (conjugate) base 1 + (conjugate) acid 2
rxn direction favors weaker acid and base
AUTOIONIZATION OF WATER PROBLEMS
1. hydrogen (hydronium) ion concentration when [OH-] = 2× 10-3M |
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Kw = [H+][OH-] = 1 × 10-14 |
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[H+][ 2×10-3] = 1 × 10-14 |
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[H+] = 5.0 × 10-12 M |
2. hydroxide ion concentration in 0.002M HCl solution |
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HCl (aq) → H+ (aq) + Cl- (aq) |
HCl is a strong acid, will dissociate 100% |
Kw = [H+][OH-] = 1 × 10-14 |
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[0.002M][OH-] = 1 × 10-14 |
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[OH-] = 5.0 × 10-12 |
CALCULATING pH OF STRONG ACID & BASE
1. pH of 0.10M Ba(OH)2 solution |
Ba(OH)2 + H2O ---> Ba2+ + 2OH- |
I |
0.10M \\\ 0 0 |
C |
-0.10 \\\ +0.10 +2(0.10) |
E |
0M \\\ 0.10 0.20 |
pOH = -log[OH-] = -log[0.20M] = 0.6990 |
pH = 14.00-0.6990 = 13.30 |
ACID EQUILLIBRIUM CALCULATIONS (pH from Ka)
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calculate pH of 1.0M Acetic acid soln, using approximations. Ka=1.8×10-5
HC2H3O2 + H2O ⇄ H3O+ + C2H3O2- I 1.0 M \\ 0M 0M
C -x \\\\ +x +x
E 1-x \\\\ x x
Ka = [H3O+][C2H3O2-]/[HC2H3O2]
Ka = x2/1.0 - x = 1.8 × 10-5
√x2 = √1.8 × 10-5
x = 4.24 × 10-3M
%ionization = [H3O+]eq/[HC2H3O2]org × 100%
=4.24×10-3M/1.0M ×100% = 0.424% <5% can neglect!
pH = -log[4.24×10-3] =2.30
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calculate pH of 1.0×10-5M Acetic acid soln (diluted concentration)
HC2H3O2 + H2O ⇄ H3O+ + C2H3O2- I 1.0×10-5M \\\\ 0M 0M
C 1.0×10-5M-x \\\ +x +x
E 1.0×10-5M-x \\\\ x x
Ka =x2/ 1.0×10-5M - x= 1.8×10-5M
√x2 = √1.8 × 10-10
x = 1.34× 10-5
%ionization = [H3O+]/[HC2H3O2] × 100%
=13% > 5% can't neglect!
x2/ 1.0×10-5M = 1.8×10-5M
x2 + 1.8×10-5x - 1.8×10-10=0
1.8×10-5±√(1.8e-5)2 - 4(1)(-1.8×10-10)/ 2(1)
= 7.15×10-6M = [H3O+]
pH = -log[7.15×10-6M]= 5.14
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