NAME: ___________________________________ UNIT #11: Acids and Bases pH and pOH Neutralization Reactions Oxidation and Reduction 1. SELF-IONIZATION OF WATER a) Water molecules collide, causing a very small number to ionize in a reversible reaction: H 2 O(l) + H 2 O(l) ↔ H 3 O + (aq) + OH - (aq) water molecules hydronium ion hydroxide ion b) The hydronium ion (H 3 O + ) consists of a water molecule attached to a hydrogen ion (H + ) by a covalent bond; thus, H 3 O + and H + can be used interchangeably in a chemical equation to represent a hydrogen ion in aqueous solution. Simplified equation for self-ionization of water: H 2 O(l) ↔ H + (aq) + OH - (aq) c) Water is considered neutral since it produces equal numbers of H + and OH - ions. Hydrogen ion concentration, shown as [H + ], for water = 1.0 x 10 -7 M Hydroxide ion concentration, shown as [OH - ], for water = 1.0 x 10 -7 M Self-ionization produces a tiny number of ions but explains how pure water can behave as a very weak electrolyte. d) Water is the usual solvent for acids and bases; the dissociation of acids or bases in an aqueous solution increases either the [H + ] or [OH - ], resulting in an aqueous solution that is no longer neutral. 2. DEFINITION OF ACIDS AND BASES a) Arrhenius Model of Acids and Bases 1. Acid: A substance that contains hydrogen and ionizes to produce hydrogen ions in aqueous solutions. Ex. HCl(g) → H + (aq) + Cl - (aq) 2. Base: A substance that contains a hydroxide group and dissociates to produce hydroxide ions in aqueous solution. Ex. NaOH(s) → Na + (aq) + OH - (aq) 3. The Arrhenius model is limited because it does not describe all bases such as NH 3 (ammonia) and NaHCO 3 (baking soda/sodium bicarbonate). These substances dissociate to form OH - in an aqueous solution but do not have the hydroxide group in their formula and therefore, would not meet the definition of an Arrhenius base. NH 3 (aq) + H 2 O(l) ↔ NH 4 + + OH - NaHCO 3 (s) + H 2 O(l) ↔ H 2 CO 3 (aq) + Na + (aq) + OH - (aq) sodium bicarbonate carbonic acid
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NAME: ___________________________________
UNIT #11: Acids and Bases
pH and pOH
Neutralization Reactions
Oxidation and Reduction
1. SELF-IONIZATION OF WATER
a) Water molecules collide, causing a very small number to ionize in a reversible reaction:
H2O(l) + H2O(l) ↔ H3O+(aq) + OH
-(aq)
water molecules hydronium ion hydroxide ion
b) The hydronium ion (H3O+) consists of a water molecule attached to a hydrogen ion
(H+) by a covalent bond; thus, H3O
+ and H
+ can be used interchangeably in a chemical
equation to represent a hydrogen ion in aqueous solution.
Simplified equation for self-ionization of water:
H2O(l) ↔ H+(aq) + OH
-(aq)
c) Water is considered neutral since it produces equal numbers of H+ and OH
- ions.
Hydrogen ion concentration, shown as [H+], for water = 1.0 x 10
-7M
Hydroxide ion concentration, shown as [OH-], for water = 1.0 x 10
-7M
Self-ionization produces a tiny number of ions but explains how pure water can behave
as a very weak electrolyte.
d) Water is the usual solvent for acids and bases; the dissociation of acids or bases in an
aqueous solution increases either the [H+] or [OH
-], resulting in an aqueous solution
that is no longer neutral.
2. DEFINITION OF ACIDS AND BASES
a) Arrhenius Model of Acids and Bases
1. Acid: A substance that contains hydrogen and ionizes to produce hydrogen ions in
aqueous solutions.
Ex. HCl(g) → H+(aq) + Cl
-(aq)
2. Base: A substance that contains a hydroxide group and dissociates to produce
hydroxide ions in aqueous solution.
Ex. NaOH(s) → Na+(aq) + OH
-(aq)
3. The Arrhenius model is limited because it does not describe all bases such as NH3
(ammonia) and NaHCO3 (baking soda/sodium bicarbonate). These substances
dissociate to form OH- in an aqueous solution but do not have the hydroxide
group in their formula and therefore, would not meet the definition of an
Arrhenius base.
NH3(aq) + H2O(l) ↔ NH4+ + OH
-
NaHCO3(s) + H2O(l) ↔ H2CO3 (aq) + Na+(aq) + OH
-(aq)
sodium bicarbonate carbonic acid
b) Bronsted-Lowry Model of Acids and Bases
1. Acid: A substance that is a hydrogen ion (H+) donor; also referred to as a
proton donor since H+ consists of 1 proton only.
2. Base: A substance that is a hydrogen ion (H+) or proton acceptor.
3. Using the example cited above:
NH3(aq) + H2O(l) ↔ NH4+ + OH
-
Bronsted-Lowry acid: H2O since it donates a H+ to NH3
Bronsted-Lowry base: NH3 since it accepts a H+ from H2O
NaHCO3(s) + H2O(l) ↔ H2CO3 (aq) + Na+(aq) + OH
-(aq)
Bronsted-Lowry acid: H2O since it donates a H+ to NaHCO3, with Na
+ released
Bronsted-Lowry base: NaHCO3 since it accepts a H+ from H2O, with Na
+ released
3. CHARACTERISTICS OF ACIDS AND BASES
a) Acids
1. pH between 0 and 6.9; the lower the pH value, the stronger the acid.
2. React with certain metals (aluminum, magnesium and zinc) to produce H2 gas.
3. Acids are corrosive.
4. Acids taste sour.
Ex. Carbonic and phosphoric acids give carbonated beverages their sharp taste.
Citric and ascorbic acids give lemons and grapefruit their tart taste.
Acetic acid makes vinegar taste sour.
5. Causes blue litmus paper to turn pink.
6. Conducts electricity.
b) Bases
1. pH between 7.1 and 14; the higher the pH value, the stronger the base.
2. Taste bitter.
Ex. Soap has a bitter taste.
3. Feel slippery
Ex. Soap has a slippery texture.
4. Causes red litmus paper to turn blue.
5. Conducts electricity.
c) Water: has a neutral pH of 7.0 but can act as a very weak acid or base due to the
tiny degree of self-ionization.
4. CALCULATING pH and pOH FROM SIMPLE CONCENTRATIONS
a) pH and pOH : Since [H+] and [OH
-] concentrations are usually very small numbers
expressed in scientific notation, scientists have adopted a shorthand method to
express these concentrations in an aqueous solution. This is known as the pH scale
and ranges from 0 to 14.
b) pH = -log[H+] and pOH = -log[OH
-]
Additionally, pH + pOH = 14
c) For concentrations of H+ and OH
- that are 1.0 x 10
n, calculating the pH or pOH of the
solution is relatively easy. The log (n) is the number to which 10 is raised.
1. [H+] = 1.0 x 10
-4 the log is -4
pH = -log[1.0 x 10-4
] or –log[10-4
]
pH = -(-4) = 4 this represents an acidic solution with a pH of 4
Since pH + pOH = 14, then the pOH is 10
2. [OH-] = 1.0 x 10
-2 the log is -2
pOH = -log[1.0 x 10-2
] or –log[10-2
]
pOH = -(-2) = 2
Since pH + pOH = 14, then the pH is 12 this represents a basic solution with a
pH of 12
3. For concentrations with the coefficient 1.0, the pH and pOH will be a simple whole
number between 0 and 14.
d) Whether a solution is acidic, basic or neutral is based only on pH, a measurement of
the number of H+ ions in solution.
1. Acidic solutions have more H+ ions than OH
- ions and thus, have pH’s below 7
2. Basic solutions have more OH- ions than H
+ ions and thus, have pH’s above 7.
3. Neutral solutions have equal numbers of H+ ions and OH
- ions and thus, have a pH
of 7.
5. CALCULATING pH and pOH FROM MORE COMPLEX CONCENTRATIONS (HONORS ONLY)
a) Calculating pH and pOH from concentrations in which the coefficient of the [H+] and
[OH-] is not 1.0 is more difficult and requires a calculator with a LOG function since the
calculation must include both the coefficient and the power to which 10 is raised.
b) [H+] = 3.8 x 10
-10
In your calculator, enter the -log of 3.8 x 10-10
to calculate the pH.
In a TI-83 or equivalent, punch in (-), followed by the LOG button and a left parenthesis
“(“ will appear; enter 3.8, followed by the 2nd
function button, followed by the button
beneath EE, followed by (-), followed by 10, followed by the right parenthesis “)” and
press ENTER.
ANSWER: pH = 9.4 and pOH = 4.6 (Remember, pH + pOH = 14) A basic solution
c) [OH] = 7.9 x 10-14
In your calculator, enter the -log of 7.9 x 10-14
to calculate the pOH as instructed
above.
ANSWER: pOH = 13.1 and pH = 0.9 An acidic solution
d) For concentrations with a coefficient other than 1.0, the pH and pOH will be a more
precise number usually carried out to 1 decimal place.
6. CALCULATING CONCENTRATIONS FROM pH and pOH
a) The [H+] and [OH
-] values can be calculated from the pH or pOH using the reverse
process. This is relatively simple for pH or pOH values that are whole numbers, such as
1, 2, 3, etc.
Ex. pH = 6.0
Since pH = -log[H+], then the [H
+] must equal 1.0 x 10
-6
If pH = 6.0, then pOH = 8; therefore, [OH-] must equal 1.0 x 10
-8
b) HONORS ONLY
1. For pH and pOH values that are not whole numbers, the calculation is more
difficult. This requires a calculator with a LOG and ANTILOG function.
2. If pH = -log[H+], then multiplying both sides of the equation by -1 gives:
-pH = log [H+]
To calculate [H+] using this equation, you must take the antilog of both sides of the