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Lecture

Acid-Base Equilibria 1.pdf

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Department
Chemistry
Course
CHEM102
Professor
Sai Yiu
Semester
Winter

Description
UNIVERSITY OF ALBERTA INTRODUCTORY UNIVERSITY CHEMISTRY II (CHEM 102) Acid-Base Equilibria 1 Brłnsted-Lowry definitions : An acid-base reaction is a proton transfer process HA + B HB + + A- Acid Base + + Acid: Proton (H , H O3) donor during acid-base reaction. An acid must therefore contain H in its formula. For example: + NH 4 , HCl, H2SO ,4HCN, HNO , HA3 (CH COOH) 3 + Base : Proton (H ) acceptor during acid-base reaction. Hence, a base must contain lone pair of electrons in formula so to form a bond with the hydrogen ion. For example: - - - - - - - 2- NH 3 OH , Cl , HSO , 4N , NO , Ac3(CH COO ), 3O 3 Conjugate acids and base: For the following acid base reactions: HCl (aq) + H O (l) ▯ H O (aq) + Cl (aq)- 2 3 Acid Base NH (aq) + H O (l) ▯ NH + (aq) + OH (aq) 3 2 4 Base Acid And their reverse reactions are: + - + - NH 4 (aq) + OH (aq) ▯ NH (aq)3+ H O (l) 2 NH 4 is acid and OH is base Acid Base + - Thus, NH , 4he product from NH becomes3an acid and OH , the product from H O 2 becomes a base. HCl (aq) + H O (l) ▯ H O (aq) + Cl (aq)- 2 3 Acid Base Conjugate acid Conjugate base + - NH (3q) + H O 2l) ▯ NH 4 (aq) + OH (aq) Base Acid 1 Therefore, every acid has a conjugate base and every base has a conjugate acid Acid – H = conjugate base (increase one negative charge) Base + H = conjugate acid (less one negative charge) Give the conjugate acid for the following base: - OH Give the conjugate base for the following acid: H 2O 3 Every acid-base reaction has two conjugate acid-base pairs. + - For example: NH 3aq) + H 2 (l) ▯ NH 4 (aq) + OH (aq) stronger base stronger acid weaker acid weaker base Formation of the hydronium ion (hydrated proton) + + It should be noted that proton H in water exists as a hydronium i3n H O + + H (aq) + H2O (l) ▯ H 3 (aq) + + Thus H (aq) = H 3 (aq) O H H H Hydroniumion 2 Strengths of acids and bases Strong and Weak acids The reaction of an acid, HA, with water is: + - HA (l) + H 2O (l) ▯ H 3 (aq) + A (aq) [H O ] [A ] 3 Ka = [HA] K a= acid dissociation constant (acid ionization constant) A strong acid is one that dissociates completely in water. Kc>> 1 11 For example: K aor hydroiodic acid (HI) = 1.0 x 10 A weak acid is one that dissociates slightly in water. K < 10 -2 -10 c For example: K aor hydrocyanic acid (HCN) = 4.0 x 10 Another example of weak acid: CH CO3H (acetic acid), K = 1a8 x 10 -5 Below are some common strong acids and bases that you must memorize: Strong acids Strong bases (Mostly hydroxides of group 1A and 2A) HCl LiOH HBr NaOH HI KOH HNO 3 RbOH H 2O 4 CsOH HClO 4 Mg(OH) 2 Ca(OH) 2 Sr(OH) 2 Strong and Weak bases The reaction of a base, B, with water is: + - B + H 2 (l) ▯ HB (aq) + OH (aq) + - [HB ] [OH ] Kb= [B] K = base dissociation constant (base ionization constant) b 3 For ammonia, + - NH (3q) + H O 2l) ▯ NH 4 (aq) + OH (aq) + - [NH 4 ] [OH ] K b= = 1.8 x 10-5 [NH ]3 Ammonia is a weak base . K is a special K applied to the reaction (dissociation) of an acid in water. For acids a c K a> 1 K b is a special Kcapplied to the reaction (dissociation) of a base in water. For strong bases K b> 1 Strong acid Weak base Weak acid Strong base Note: Weak acid is associated with a strong conjugate base Strong acid is associated with a weak base Fig 18.9, p797 4 Direction of acid base equilibria An acid-base reaction proceeds to a greater extent in the direction in which a stronger acid and stronger base form a weaker acid and weaker base. For example: - + H 2S (aq) + NH (a3) ▯ HS (aq) + NH 4 (aq) Stronger acid Stronger base Weaker base Weaker acid Thus, every acid-base equilibrium is always displaced towards the formation of the weaker acid/ base. Self ionization of Water: Water ionizes according to the following equilibrium: + - H 2O (l) + H 2 (l) ▯ H O3(aq) + OH (aq) Note that water acts as both an acid and a base in the above equation. + - Therefore, [H 3 ] [OH ] = K w (Ion-product constant) -14 o K w 1.0 x 10 (at 25 C) + -14 -7 Therefore in pure water, [H O3] = √(1.0 x 10 ) M = 1.0 x 10 M Hence, • In a neutral solution [H O ] = 1.0 x 10 M -7 3 + -7 • In an acidic solution [H 3 ] > 1.0 x 10 M + -7 • In a basic solution [H 3 ] < 1.0 x 10 M The pH and pOH + pH = - log [H 3 ] Similarly, pOH = - log [OH ] - o -7 Therefore pH of water at 25 C = - log 1.0 x 10 = 7.00 pH of an acidic solution < 7 pH of a basic solution >7 5 A word about the significant figures in logarithms For example: log 5.43 x 10 10= 10.735 3 significant 3 significant figure number figures The pre-decimal part accounts for the exponential 3 significant and so does NOT count figures log 5.43 x 10 10 = 10.735 3 significant The exponential figure number part does not count -4 What is the pH of a solution with hydrogen concentration = 3.2 x 10 M. + -4 pH = -log [H ] = -log (3.2 x 10 ) = - (-3.49) = 3.49 -4 What is the pH of a NaOH solution with hydroxide ion concentration = 2.9 x 10 M. Mathematical relationship for pH and pOH + - -14 o K w [H O 3 [OH ] = 1.0 x 10 at 25 C + - -14 Therefore, pK = -wlog [H O ] 3OH ] = - log 1.0 x 10 + - - log [H 3 ] + log [OH ] = 14 - log [H O ] - log [OH ] = 14 3 At 25 C pH + pOH = 14 6 Example problem 1 What is the pH of a 7.52 x 10 M CsOH solution? + - CsOH (aq) → Cs (aq) + OH (aq) + - 2 What
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