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Ap Chemistry Notes 9-1 Acid / Base Theory

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Ap Chemistry Notes 9-1 Acid / Base Theory AP CHEMISTRY NOTES 9-1 ACID / BASE THEORY DEFINITIONS: *Arrhenius Concept + Acid – substance which produces hydrogen ions (or hydronium ions – H3O ) in aqueous solution Base – substance which produces hydroxide ions in aqueous solution *Bronsted – Lowry Concept Acid – substance which acts as a proton (H+) donor Base – substance which acts as a proton (H+) acceptor *Lewis Concept Lewis Acid – substance which acts as an electron pair acceptor (must have a “hole”) Lewis Base – substance which acts as an electron pair donor (must have a “lone pair” of electrons) 1 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 acid 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. 2 STRONG ACIDS 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 hydroxides 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 water: + - 2H2O(l) ↔ H3O (aq) + OH (aq) or + - H2O(g) ↔ H (aq) + OH (aq) The equilibrium constant (or ion-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-]? 3 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 4 *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. 5 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 - acetic acid C6H5COOH - benzoic acid HCOOH - formic acid The acidic hydrogen on organic acids is the hydrogen in the carboxyl group (-COOH) Within a series, acid strength increases with increasing numbers of oxygen atoms - 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 - methylamine C2H5NH2 - ethylamine These bases are “dirty nasty hydrogen stealers” just like NH3. EXAMPLE: Show the ionization of methylamine in water. 6 CALCULATING THE STRENGTH OF WEAK ACIDS AND BASES Calculation of the pH of weak acid and base solution involves the use of dissociation 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 -5 1.8 x 10 . 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 ammonia. The Kb for ammonia is 1.8 x 10-5. 7 EXAMPLE: 0.15 M solution of cyanic acid (also known as hydrocyanic acid) has a pH of 2.67. What is the hydrogen ion concentration? What is the ionization constant, Ka, for cyanic acid at this temperature? 8 AP CHEMISTRY NOTES 9-3 SALT EQUILIBRIA Neutral Salts – 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 weak base 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 -5 be 1.8 x 10 . 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 . 9 EXAMPLE: Calculate the pH of a 0.15 M solution of sodium acetate. (The Ka of acetic acid is 1.8 x 10-5.) 10 .
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