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Home , Acid strength, Base (chemistry), Conjugate acid, Weak base

AP NOTES 9-1 / THEORY

DEFINITIONS:

*Arrhenius Concept

+ Acid – substance which produces (or ions – H3O ) in

Base – substance which produces ions in aqueous solution

*Bronsted – Lowry Concept

Acid – substance which acts as a (H+) donor

Base – substance which acts as a proton (H+) acceptor

*Lewis Concept

Lewis Acid – substance which acts as an pair acceptor (must have a “hole”)

Lewis Base – substance which acts as an donor (must have a “” of )

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Conjugate Acid – a base after it has gained a proton

Conjugate Base – an acid after it has lost a proton

+ - HA + H2O → H3O + A acid base conjugate base

Conjugate Acid-Base Pair – two substances related to each other by the donating and accepting of a single proton

In the above equation, the conjugate acid-pairs are:

EXAMPLE: For the following reaction, determine the acid, base, conjugate acid, conjugate base and conjugate acid-base pair(s):

+ 2- H2S + NH3 → NH4 + S

Amphoteric Substances – substances which can act as either an acid or a base

Water is a common example:

- + HNO3 + H2O → NO3 + H3O

+ - NH3 + H2O → NH4 + OH

- EXAMPLE: Show how HCO3 could act as both an acid and a base.

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STRONG AND BASES

Strong Acids – acids which are mostly dissociated (whose equilibrium lies far to the right)

HCl HBr HI H2SO4 HNO3 HClO4 HMnO4 HCrO4

Strong Bases – bases which are mostly dissociated (whose equilibrium lies far to the right)

Strong bases include the of Groups IA and IIA (except for Mg) *Note: The hydroxides of group IIA are not very soluble

CALCULATING THE STRENGTH OF STRONG ACIDS AND BASES

The auto-ionization of :

+ - 2H2O(l) ↔ H3O (aq) + OH (aq)

or

+ - H2O(g) ↔ H (aq) + OH (aq)

The (or -product constant) for water:

+ - + - Kw = [H3O ] [OH ] or Kw = [H ] [OH ]

o Kw varies with temperature, but at 25 C:

[H+] = [OH-] = 1.0 x 10-7 M

so

-14 2 . -2 Kw = 1.0 x 10 mol L

o -14 No matter what the solution contains, at 25 C Kw = 1.0 x 10

EXAMPLE: If the [H+] in an aqueous solution is 3.4 x 10-5 at 25oC, what is the [OH-]?

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o -13 EXAMPLE: At 60 C the value of Kw is 1.0 x 10 . Using Le Chatelier’s Principle, predict whether the reaction + - H2O(g) ↔ H (aq) + OH (aq)

is endothermic or exothermic.

Calculate [H+] and [OH-] of at neutral solution at this temperature.

The pH Scale is a convenient method of determining the acidity of a solution.

Calculating pH:

The pH of a solution can be determined using the following equation:

pH = - log [H+]

For example, if [H+] = 1.00 x 10-7 M then pH = 7.000

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*Note: Since pH is a logarithmic function

1. the number of places past the decimal in the pH value = the number of total sig figs in the concentration value

2. a change in pH of 1 will change the concentration by a factor of 10

In addition:

pOH = - log [OH-]

pH + pOH = 14

pK = - log K

[H+] = 10-pH [OH-] = 10-pOH

EXAMPLE: Calculate the pH of a solution in which [H+] = 3.52 x 10-5 M.

EXAMPLE: Calculate the [H+] of a solution which has a pH of 8.937.

-4 + - EXAMPLE: For a 7.88 x 10 M solution of HNO3, calculate [H ], [OH ], pH, and pOH.

EXAMPLE: For a 0.00821 M solution of NaOH, calculate [H+], [OH-], pH, and pOH.

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AP CHEMISTRY NOTES 9-2 ACID / BASE EQUILIBRIA

Weak Acids – acids which are mostly dissociated (their equilibrium lies far to the left)

Common organic acids: CH3COOH - C6H5COOH - HCOOH -

The acidic hydrogen on organic acids is the hydrogen in the carboxyl group (-COOH)

Within a series, increases with increasing numbers of - the electronegative oxygen atoms draw electrons away from the O-H bond, making it weaker.

Weaker bond to H = stronger acid (since the H dissociates more easily)

For example:

HClO4 > HClO3 > HClO2 > HClO and H2SO4 > H2SO3

Also:

HOCl > HOBr > HOI

QUESTION: Why is HF the only weak hydrohalic acid?

Weak Bases – bases which are mostly dissociated (their equilibrium lies far to the left)

Common organic bases: CH3NH2 - C2H5NH2 -

These bases are “dirty nasty hydrogen stealers” just like NH3.

EXAMPLE: Show the ionization of methylamine in water.

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CALCULATING THE STRENGTH OF WEAK ACIDS AND BASES

Calculation of the pH of weak acid and base solution involves the use of constants (a form of equilibrium constant) – either Ka or Kb – in an equilibrium problem.

-4 o EXAMPLE: Calculate the pH of a 1.00 x 10 M solution of acetic acid. The Ka of acetic acid (at 25 C) is 1.8 x 10-5.

Calculate the percent dissociation of acetic acid in this example if

- EXAMPLE: Calculate the [OH ] and the pH for a 15.0 M solution of . The Kb for ammonia is 1.8 x 10-5.

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EXAMPLE: 0.15 M solution of cyanic acid (also known as hydrocyanic acid) has a pH of 2.67. What is the concentration?

What is the ionization constant, Ka, for cyanic acid at this temperature?

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AP CHEMISTRY NOTES 9-3 EQUILIBRIA

Neutral – salts that are formed from the cation of a strong base and the anion of a strong acid; they form neutral solutions when dissolved in water (ie. NaCl)

Acidic Salts – salts that are formed from the cation of a and the anion of a strong acid; they form acidic solutions when dissolved in water (ie. NH4Cl – a salt of a weak base)

+ - NH4Cl → NH4 + Cl

+ The “weak” part of the salt (NH4 ) will react with water (hydrolyze) to form an equilibrium system:

+ + NH4 + H2O ↔ NH3 + H3O

+ The Ka of NH4 cannot be found in a table. However, the Kb of NH3 (the conjugate base) is known to be 1.8 x 10-5.

In addition, for a conjugate acid-base pair:

-14 Ka of the acid x Kb of the base = 1 x 10

+ EXAMPLE: Determine the Ka of NH4 .

Basic Salts – salts that are formed from the cation of a strong base and the anion of a weak acid: they form basic solutions when dissolved in water (ie. NaF – a salt of a weak acid)

NaF → Na+ + F-

The “weak” part of the salt (F-) will hydrolyze to form an equilibrium system:

- -4 EXAMPLE: Determine the Kb of F if the Ka of HF is 7.2 x 10 .

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EXAMPLE: Calculate the pH of a 0.15 M solution of acetate. (The Ka of acetic acid is 1.8 x 10-5.)

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