cm4106 review of lesson 2
DESCRIPTION
Acid-BaseTRANSCRIPT
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CM4106 Chemical Equilibria & Thermodynamics
Lesson 2Acid-Base EquilibriaA Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/
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Fundamentals:
Acid Base
1. proton H+ donor2. electron acceptor
(vacant orbital)
1. proton H+ acceptor2. electron donor
(lone pair)
1. Identify acids/ bases
Acid Conjugate Base– H+
+ H+
Base Conjugate Acid+ H+
– H+
2. Identify conjugate acids/ bases
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Fundamentals: 3. Identify stronger acids/ bases
(A) In any acid-base reaction, the equilibrium will favor the reaction where the stronger acid reacts with the stronger base. i.e. (In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.)
HCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq)
CH3CO2H (aq) + H2O (l) ⇌ H3O+ (aq) + CH3CO2- (aq)
(B) The stronger an acid, the weaker its conjugate base.
The stronger a base, the weaker its conjugate acid.
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p = -log10
At 25ºC:
pH + pOH = pKw = 14
pKa + pKb = 14
[H+][OH–] = 10-14
Ka x Kb = Kw = 10-14
Important relations
Fundamentals:
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(I) Calculations
pH, pKa, [H+], Ka pOH, pKb, [OH–], Kb
Step 1: Determine what is present in the solution.
(A)Acid : Strong Acid vs Weak Acid Monoprotic Acid / Diprotic Acid / Triprotic Acid
(B) Base: Strong Base vs Weak Base
Step 2: Use the appropriate equations for the respective species.
Take note of concentration of acid / bases (For dilute solutions, we need to take into consideration of [H+] / [OH-] from auto-ionization from water)
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pH of Acid/ Base
Strong acids dissociate completely into ions in aqueous solution.
[H+] = [HA]
Strong bases dissociate completely into ions in aqueous solution.
[OH-] = [B]
HA(aq) ⇌ H+(aq) + A-
(aq)
I [HA] 0 0
C - x + x + x
E [HA] - x + x + x
B + H2O ⇌ BH+ + OH-
I [B] - 0 0
C - x - + x + x
E [B] - x - x x
Strong Acid
Strong Base
Weak Acid
Weak Base
Ka =x2
([HA] – x)
Kb =x2
([B] – x)
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pH of Acid/ Base
1. Determine if acid/ base is strong or weak (more common)
2. For weak acids, if asked to determine Ka, pH or [H+],
you can save time by using the formula:
[H+] = Ka × c [OH–] = Kb × cExample:Calculate the pH of 0.50M HF solution at 25ºC where the Ka = 7.1 x 10-4
Assumption: x is negligible
[H+][F-][HF]
Ka =x2
(0.50 – x)= = 7.1 x 10-4
Assumption: For weak acids, x must be very small 0.50 – x ≈ 0.5
x2
0.50= 7.1 x 10-4 x = [H+] = 0.0188 M
pH = 1.73 (to 2 d.p.)
Assumption is valid; x < 5% of [HF]initial
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Calculate the pH of the following solutions at 298K
0.10 mol dm−3 CH3COOH (pKa = 4.75) Weak acid solution (monoprotic acid)
CH3COOH(aq) + H2O (l) ⇌ CH3COO(aq) + H3O+
(aq)
Initial (M) 0.10 - 0.00 0.00
Change (M) - x - + x + x
Eqm (M) 0.10 - x - + x + x
Assumption: For weak acids, x must be very small 0.10 – x ≈ 0.10
x = [H3O+] = 1.338 x 10-3 M
pH = 2.87 ( 2 d.p.)
[CH3COO][H3O+]
[CH3COOH]Ka =
x2
(0.10 – x)= 10 4.75= = 1.79 x 10-5
x2
0.10 = 1.79 x 10-5 Assumption is justified, x < 5% of [CH3COOH]initial
Concentration: 2 s.f.pH: 2 d.p.
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Calculate the pH of the following solutions at 298K
0.30 mol dm3 ethylamine, CH3CH2NH2 (pKb= 3.27)
pH = 14.00 – 1.896 pH = 12.10 (2 d.p.)pOH = - log10[OH]
= - log10[0.01269 ]= 1.896
CH3CH2NH2 (aq) + H2O (l) ⇌ CH3CH2NH3+(aq) +
OH-(aq)
Initial (M) 0.30 - 0.00 0.00
Change (M) -x - + x + x
Eqm (M) 0.30 - x - + x + x
[CH3CH2NH3+][OH-]
[CH3CH2NH2]Kb =
x2
(0.30 – x)= 10 3.27= = 5.37 x 104
Assumption: For weak bases, x must be very small 0.30 – x ≈ 0.30
x2
0.30 = 5.37 x 10-4
x = [OH] = 0.01269 M
Assumption is justified, x < 5% of [CH3CH2NH2]initial
Concentration: 2 s.f.pH: 2 d.p.
Weak base solution (monoprotic base)
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pH of salt solutions
Weak Acid Salt
Weak Basic Salt
CH3COONa (aq) → CH3COO- (aq) + Na+ (aq)
conjugate base
CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
NH4Cl (aq) → NH4+ (aq) + Cl- (aq)
conjugate acid
Salt solutions can be (i) neutral (ii) weak acids or (iii) weak bases
pH > 7
pH < 7