Acid-Base Reactions (1)
Chapter 9
Acids and Bases Chemistry https://www.youtube.com/watch?v=ZNo6gfCAgWE
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Content
Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction Reactions
o
Combustion Reactions
o
Synthesis Reactions
o
Decomposition Reactions
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Arrhenius acids and bases
Sevante Arrhenius proposed that substances exists as ions in solution in his
dissertation, which was awarded a fourth class (D) in 1884. He was unable to
find a job in his native Sweden.
He was awarded the Nobel Prize in 1903 for his electrolytic dissociation
theory!!!
The fundamental concept:
Acid - any substance which delivers hydrogen ion (H+) to the solution.
HA  H+ + A¯
Base - any substance which delivers hydroxide ion (OH¯) to the solution.
BOH  X+ + OH¯
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Arrhenius (or Classical) Acid-Base Definition
An acid is a substance that contains hydrogen and dissociates
in water to yield a hydronium ion : H3O+
A base is a substance that contains the hydroxyl group and
dissociates in water to yield : OH -
Neutralization is the reaction of an H+ (H3O+) ion from the
acid and the OH - ion from the base to form water, H2O.
The neutralization reaction is exothermic and releases approximately 56 kJ per
mole of acid and base.
H+(aq) + OH-(aq)
H2O(l)
H0rxn = -55.9 kJ
Problems with Arrhenius theory
Question the established theory.
Acidity did not show in other solvent. What is the solvent role?
Some salts produce acidic or basic solutions, not neutral. Why?
Which one is the base, NH3 or NH4OH? Is OH¯ really the only
base?
How can H+ be stable? Are proton donated?
17 ACID AND BASE
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In 1923 physical chemists Johannes Nicolaus Brønsted in Denmark
and Thomas Martin Lowry in England both independently proposed
the theory that carries their names. In the Brønsted-Lowry theory
acids and bases are defined by the way they react with each other,
which allows for greater generality. The definition is expressed in
terms of an equilibrium expression.
acid + base ⇌ conjugate base + conjugate acid
With an acid, HA, the equation can be written symbolically as:
HA + B ⇌ A− + HB+
The equilibrium sign, ⇌, is used because the reaction can occur in
both forward and backward directions. The acid, HA, can lose a
proton to become its conjugate base, A−. The base, B, can accept a
proton to become its conjugate acid, HB+. Most acid-base reactions
are fast so that the components of the reaction are usually in dynamic
equilibrium with each other.
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Relative Strength of Acids and Their Conjugate Bases
Acids
Conjugate Bases
Very Strong
Very Weak
Strong
Weak
Weak
Strong
Very Weak
Very Strong
______________________________
Strong acids lose protons very readily  weak conjugate bases;
Weak acids do not lose protons very readily  strong conjugate
bases.
Acid-Base Reactions
In the reaction of an acid with a base, the H+ from the acid
combines with the OH- from the base to make water; the
cation from the base combines with the anion from the acid to
make the salt
acid + base salt + water
H2SO4(aq) + Ca(OH)2(aq)  CaSO4(aq) + 2 H2O(l)
The net ionic equation for an Acid-Base reaction is always
H+ (aq) + OH- (aq)  H2O(l)
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Table 1. Strong and Weak Acid and Base
Strong acids lose protons very readily
Weak acids do not lose protons very readily
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Problem: Write balanced molecular, total ionic, and net
ionic equations for each of the following acid-base reactions
and identify the spectator ions:
(a) Hydrochloric acid(aq) + potassium hydroxide(aq)
(b) Strontium hydroxide(aq) + perchloric acid(aq)
(c) Barium hydroxide(aq) + sulfuric acid(aq)
Plan All are strong acids and bases (see Table 1), so the actual reaction is between H+
and OH-. The products are H2O and a salt solution of spectator ions. In (c), we note that
the salt (BaSO4) is insoluble (see Chapter 8), so there are no spectator ions.
Solution (a) Writing the molecular equation:
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Problem: Write balanced molecular, total ionic, and net
ionic equations for each of the following acid-base reactions
and identify the spectator ions:
(b) Writing the molecular equation:
(c) Writing the molecular equation:
This is a neutralization and a precipitation reaction, so the net ionic equation is the same
as the total ionic. There are no spectator ions.
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Problem
Specialized cells in the stomach release HCl to aid digestion. If they release too
much, the excess can be neutralized with a base in the form of an antacid.
Magnesium hydroxide is a common active ingredient in antacids. As a
government chemist testing commercial antacids, you use 0.10 M HCl to
simulate the acid concentration in the stomach. How many liters of this “stomach
acid” will react with a tablet containing 0.10 g of magnesium hydroxide?
Plan We are given the mass (0.10 g) of magnesium hydroxide,
Mg(OH)2, that reacts with the acid. We also know the acid
concentration (0.10 M) and must find the acid volume. After
writing the balanced equation, we convert the mass (g) of
Mg(OH)2 to amount (mol) and use the molar ratio to find the
amount (mol) of HCl that reacts with it. Then, we use the
molarity of HCl to fi nd the volume (L) that contains this amount
(see the road map).
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Solution Writing the balanced equation
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Molarity (M)
When working quantitatively with any solution, it is essential to know the
concentration, the quantity of solute dissolved in a given quantity of solution
(or of solvent).
o Concentration is an intensive property (like density or temperature) and thus is
independent of the solution volume:
o a 50-L tank of a solution has the same concentration (solute quantity/solution
quantity) as a 50-mL beaker of the solution.
o Molarity (M) is the most common unit of concentration. It expresses the
concentration in units of moles of solute per liter of solution:
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Problem Glycine (C H NO ) has the simplest structure of the 20
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amino acids that make up proteins. What is the molarity of a solution
that contains 53.7 g of glycine dissolved in 495 mL of solution?
Plan The molarity is the number of moles of solute in each liter of solution. We
convert the mass (g) of glycine (75.1 g) to amount (mol) by dividing by its molar
mass. We divide that number of moles by the volume (495 mL) and convert the
volume to liters to find the molarity.
Solution Finding the amount (mol) of glycine:
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