Acids and Bases
Operational definitions are based on observed properties. Compounds can be
Classified as acid or base by observing these sets of properties.
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Properties of Acids
 Taste sour (acere – Latin for sour) (Lemons, vinegar)
 Cause certain organic dyes to change colour (Turns blue litmus paper to red
– BAR)
 Acid properties are destroyed by Bases (React with bases to form a salt and
water)
 Acid solutions are Electrolytes (substance in solution that conduct an
electric current – Acids can be strong or weak electrolytes)
 Acids react (corrode) with active metals (Group I and II as well as Zn and
Aluminum) (Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g))
 Acids react with carbonates (CO32-) and hydrogen carbonates (HCO31-) to
produce carbon dioxide gas {2HCl(aq) + Na2CO3(s) → 2NaCl(aq) + H2O(l) +
CO2(g)}
 Certain nonmetal oxides will dissolve to produce acid solutions. (SO3(g) + H2O →
H2SO4(aq) (SO3(g) is the acid anhydride – without water)
2
Properties of Bases
 Bases taste bitter; mustard and soap
 Bases cause weak organic acids (dyes) to change colour (red
litmus paper to blue {BB} Basic Blue
 Acids destroy base properties - react with acids to form salts
and water
 Bases are electrolytes {strong or weak}
 Feel soapy, slippery
 Bases are formed when the oxide of some metals dissolve in
water (CaO(s) + H2O → Ca(OH)2(aq) {CaO is the base anhydride}
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Acid/Base definitions
• Definition #1: Arrhenius (traditional)
– Acids are compounds with ionizable hydrogen–
produce H+ ions (or hydronium ions H3O+) in
solution
– Bases are compounds that produce OH- ions
in solution (problem: some bases don’t have
hydroxide ions!)
The reaction between an acid and a base:
H+(aq) + OH-(aq) → H2O (l)
4
Arrhenius acid is a substance that produces H+ (H3O+) in water.
The HCl molecule is ionized. (ionization)
Arrhenius base is a substance that produces OH- in water.
The ions are dissociated. (dissociation)
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Some acids have more than one ionizable
hydrogen
• H2SO4 → H+(aq) + HSO41-(aq
• HSO41- → H+(aq) + SO42-(aq)
H2SO4 is diprotic
• H3PO4(aq) → H+(aq) + H2PO41-(aq)
• H2PO41-(aq) → H+(aq) + HPO42-(aq)
• HPO42-(aq) → H+(aq) + PO43-(aq)
Phosphoric acid is a triprotic acid.
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Water self-ionization
H2O
↔ H (aq) + OH (aq)
+
-
[H+] = [OH-] = 10-7M at SATP
Keq = [H+][ OH-]
[H2O(l)]
Kw = [H+][ OH-] = 10-7 x 10-7 (at 25ºC)
Kw = 10-14 at SATP
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H2O ↔ H+(aq) + OH-(aq)
What happens to this equilibrium if HCl(g) dissolves in the water?
HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)
•
Increasing
Decreasing
• H2O ↔ H3O+(aq) + OH-(aq)
• [H+] > [OH-] = acidic
• What happens when sodium hydroxide dissolves?
NaOH(s) + H2O → Na+(aq) + OH-(aq)
•
Decreasing Increasing
• H2O ↔ H3O+(aq) + OH-(aq)
• [H+] < [OH-] = basic (alkaline solution)
If [H+] = 10-7 then [OH-] = 10-7 solution is neutral (SATP)
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pH and logs
• [H+] is important in the study of acid-base
chemistry. pH is the widely used scale to
show [H+].
• pH = -log[H+] or pH =
1 .
log[H+]
[H+] = 10 – pH (the antilog)
A logarithm is the power to which ten must
be raised to get a number.
log1000 = log(103) = 3
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pH calculations
•
•
•
•
•
For a neutral solution
pH = -log[H+]
pH = -log [10-7]
pH = - [-7]
pH = 7 at SATP
•
•
•
•
•
Example:
[H+] = 5 x 10-3
pH = -log [5 x 10-3]
pH = -log [0.005]
pH = - (-2.3) = 2.3
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pH and pOH
•
•
•
•
pOH = - log [OH-] or [OH-] = 10 - pOH
Kw = [H+] x [OH-] = 1 x 10-14 (at 25ºC)
pKw= pH + pOH
14= pH + pOH
•
•
•
•
•
•
•
•
•
Example:
If pH = (2.3) what is the [OH-]?
pH + pOH = 14
pOH = 14 – pH
pOH = 14 – 2.3
pOH = 11.7
pOH = -log [OH-]
[OH-] = inverse log -11.7 or (10 - 11.7)
[OH-] = 2.0 x 10-12
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[H3O+], [OH-] and pH
• What is the pH of the 0.0010 M NaOH
solution?
• [OH-] = 0.0010 (or 1.0 X 10-3 M)
• pOH = - log 0.0010
• pOH = 3
• pH + pOH = 14
• pH = 14 – 3 = 11
• OR Kw = [H3O+] [OH-]
• 1.0 x10-14 = [H3O+] x 1.0 X 10-3
• [H3O+] = 1.0 x 10-11 M
• pH = - log (1.0 x 10-11) = 11.00
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Problem 1: The pH of rainwater collected in a certain region of the
northeastern New Brunswick on a particular day was 4.82. What is
the H+ ion concentration of the rainwater?
[H+] = 1.51 x 10-5
Problem 2: The OH- ion concentration of a blood sample is 2.5 x 10-7M.
What is the pH of the blood?
pOH = 6.6
pH = 7.4
Problem 3: A chemist dilutes concentrated hydrochloric acid to
make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+],
pH, [OH-], and pOH of the two solutions at 25°C.
a) [H3O+] = [3.0], pH = - 0.48,
pOH = 14.48, [OH-] = 3.3 x 10-15
b) [H3O+] = [2.4x10-3], pH = 2.62,
pOH = 11.38, [OH-] = 4.2 x 10-12
Problem 4: What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67?
Is this an acid, base, or neutral?
[H3O+] = 2.14 x10-4, pOH = 10.33, [OH-] = 4.68x 10-11 It is an acid.
Problem 5: Problem #4 with pH = 8.05?
[H3O+] = 8.92 x10-9, pOH = 5.95, [OH-] = 1.12x 10-6 It is an base.
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Acid/Base Definitions
• Definition #2: Brønsted – Lowry
– Acids – proton donor
A “proton” is a hydrogen ion (the atom lost it’s electron)
– Bases – proton acceptor (accepts a hydrogen ion)
No longer needs to contain the OH- ion
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A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
NH3
Base
base
+
H2O
Acid
acid
NH4+ + OHAcid
Base
conjugate
acid
conjugate
base
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The Bronsted-Lowry concept
H
Cl H
acid
+
O
H
base
+
H
HO
+
Cl
H
conjugate acid conjugate base
conjugate acid-base pairs
• Acids and bases are identified based on
whether they donate or accept H+.
• “Conjugate” acids and bases are found on the
products side of the equation. A conjugate
base is the same as the starting acid minus H+.
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Practice problems
Identify the acid, base, conjugate acid,
conjugate base, and conjugate acid-base pairs:
CH3COOH(aq) + H2O(l)  CH3COO–(aq) +H3O+(aq)
acid
base
conjugate base conjugate acid
conjugate acid-base pairs
OH –(aq) + HCO3–(aq)  CO32–(aq) + H2O(l)
base
acid
conjugate base conjugate acid
conjugate acid-base pairs
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Base
Conjugate acid
\
\
NH3(g) + H2O(l) ↔ NH4+(aq) + OH-(aq)
/
/
Acid
Conjugate Base
HCl(aq) + H2O(l) ↔ H3O+(aq) + Cl-(aq)
Acid
Base
Conjugate Conjugate
Acid
Base
The water has acted as both an acid and a base,
depending on what it is mixed with. Substances
that can act as both an acid and a base are
amphoteric (also called amphiproteric).
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Strong acid and base :
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq)
At equilibrium the ionic form is favored
Weak acid and base :
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq)
At equilibrium the molecular form is favored
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CH3COOH(aq) + H2O(l) ↔ H+(aq) + CH3COO-(aq)
Keq= [H+] [CH3COO-] .
[CH3COOH] [H2O(l)]
[H2O] is a constant, so collect the constants
(Keq)[H2O(l)] = [H+] [CH3COO-]
[CH3COOH]
(Keq)[H2O] is represented Ka(ionization constant for an acid)
Ka = [H+]e [CH3COO-]e = 1.8 x 10-5
[CH3COOH]e
Ka < 1 weak acid
General Formula for the ionization constant of a weak acid.
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• a) What is the pH of an ethanoic acid solution with a concentration
of 0.100M?
• Ka CH3COOH = 1.82 x 10-5
• [H+] = x
• [CH3COOH]i = 0.100M
• CH3COOH(aq) ↔ H+(aq) + CH3COO -(aq)
• Ka = [H+]e [CH3COO-]e
[H+]e= [CH3COO-]e 1:1 ratio
•
[CH3COOH]e
• [H+]e = x = [CH3COOH]R
([CH3COOH]R – ionized)
• Ka =
x2 e
[CH3COOH]e = [CH3COOH]i - [H+]e = 0.100M
•
[CH3COOH]e Because it is a very weak acid [CH3COOH]R = 0
x2 = Ka x [CH3COOH]e
• x2 = 1.82 x 10-5 x 0.100
• x2 = 1.82 x 10-6
• [H+] = x = 0.00135M
• pH = -log[H+]
pH = -log[0.00135]
pH = 2.87
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• b) What is the percent ionization of
this acetic (ethanoic) acid solution?
• % ionization =
[H+]
x 100
•
[CH3COOH]
•
= 0.00135 x 100 = 1.35%
•
0.100
•
(very low degree of ionization)
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• The value of Ka for phosphoric acid, H3PO4(aq), is 7.0 x 10-3 at
25C.
a) Calculate the [H3O+] in a 0.10 M solution of H3PO4.
• H3PO4 ↔ H+(aq) + H2PO4-(aq)
• Ka= [H+] [H2PO4-]
[H+] = [H2PO4-] = x [H3PO4]e = [H3PO4]i - x
•
[H3PO4]
• K a = x2 .
•
0.10 – x
• 7.0 x 10-3 = x2
.
•
0.10 – x
• x2 = -7.0 x 10-3x + 7.0 x 10-4
• x2 + 7.0 x 10-3 x – 7.0 x 10-4 =0
• x = -b± √ b2 – 4ac
•
2a
•
• x = -7.0 x 10-3 ± √ (7.0 x 10-3)2 – 4 x 1 x – 7.0 x 10-4
•
2x1
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• x = -7.0 x 10-3 ± √ 4.9 x 10-5 – -2.8 x 10-3
•
2
• x = -7.0 x 10-3 ± √ 2.85 x 10-3
•
2
•
• x = -7.0 x 10-3 ± 5.34 x 10-2
•
2
•
• x = 4.64 x 10-2
•
2
•
• x = 2.32 x 10-2 = [H+]
• pH = - log[2.32x10-2] = 1.63
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