Weak AcidBase Equilibrium A molecule with negligible acidity contains hydrogen but does not demonstrate any acidic behavior in water. Its conjugate base is extremely strong. Possibly explained by Lewis theory. No measurable H +. 1. A strong acid completely transfers its protons to water leaving no undissociated (ionized) molecules. Its conjugate base is extremely weak. Easily explained by Arrhenius Concentration of H + is defined stoichometrically by initial acid concentration 2. A weak acid only partially dissociates in aqueous solution and therefore exists in the solution as a mixture of the acid and its conjugate base. Its conjugate base is fairly strong. Explained by Arrhenius but better explained by BL. Concentration of H + must be determined applying equilibrium calculations. Dissociation Constants For a generalized acid dissociation, HA(aq) + H 2 O(l) the equilibrium expression would be K c = [H 3 O + ] [A ] [HA] A (aq) + H 3 O + (aq) This equilibrium constant is called the aciddissociation constant, K a. 1
Weak AcidBase Equilibrium For the weak acid acetic acid, the equilibrium exists as: CH 3 CO 2 H + H 2 O CH 3 CO 2 + H 3 O + Therefore, we can define K as K = [CH 3 CO 2 ] [H 3 O + ] / [CH 3 CO 2 H] = K a Where: K a = acid dissociation constant for Acetic Acid The aciddissociation constant (K a ) describes the ratio of H + and acid anion to acid molecules for weak acid dissociation using the Law of Mass Action. Using what we now know about equilibrium, we can calculate the concentrations of dissociated ions and undissociated acid molecules at equilibrium, thereby allowing us to find K a for any acid. (AP equation Sheet) Calculating K a from the ph The ph of a 0.10 M solution of formic acid, HCOOH, at 25 C is 2.87. Calculate K a for formic acid at this temperature. We know that K a = [H 3 O + ] [HCOO ] [HCOOH] Two things to note: A. We neglected the [H + ] due to the autoionization of water in aqueous solutions only one source of H + is dominant. Even weak acids have a much higher H + concentration than water B. Given the number of significant figures, 1. Niacin, one of the B vitamins, has the following molecular structure: [RCO 2 H] o [RCO 2 H] @equ Weak acids usually have very little dissociation (K<<1). We will learn to make this assumption as long as we do not affect the H + concentration by more than 5% A 0.020 M solution of niacin has a ph of 3.26, What is the K a for Niacin? 2
Dissociation Constants The greater the value of K a, the stronger the acid. Percent Ionization Another measure of acid strength, besides Ka, is the acid percent ionization (dissociation) The percent dissociation of an acid,or base, is the percent of initial concentration that is dissociated in a weak acidbase solution. As the weak acid/base concentration increases, the percent ionization decreases Calculating Percent Ionization [HCOOH], M [H 3O + ], M [HCOO ], M Initially 0.10 0 0 Weak Bases Change 4.2 10 3 +4.2 10 3 +4.2 10 3 At equilibrium 0.10 4.2 10 3 = 0.0958 = 0.10 4.2 10 3 4.2 10 3 [H 3 O + ] eq Percent ionization = [HA] 100 initial From our earlier example, [H 3 O + ] eq = 4.2 10 3 M [HCOOH] initial = 0.10 M 4.2 10 Percent ionization = 3 100 0.10 = 4.2% Bases react with water to produce hydroxide ion. 2012 Pearson Education, Inc. Weak Bases The equilibrium constant expression for this reaction is K b = [HB] [OH ] [B ] 3. For base reactions, K b can be determined following the same equilibrium guidelines. Calculate the K b for a 0.050 M CO 3 2 solution having a ph of 11.48. where K b is the basedissociation constant. 2012 Pearson Education, Inc. (AP equation Sheet) 3
The relationship between K a and K b conj. is described by the ionproduct constant of water (K w ) 4. Calculate K b for the fluoride ion, using Appendix D. K b for F is not given, but its conjugate acid is. K a x K b conj. = K w The product of the aciddissociation constant and basedissociation constant for the conjugate acidbase pair equals the ionproduct constant of water Recall the autoionizatoin of water: Where: K w = [H 3 O + ][OH ] = K w = 1.0 x 10 14 (@ 25.0 o C) This only describes the concentration of hydronium and hydroxide ions in water, not the acid or base strength of water. You may also recall that the concentration of a solid or pure liquid, not a solution, is equal to its density(g/cm 3 ) divided by its molecular weight (g/mol); for water this is 55.556 g/mol. From: K a = [H 3 O + ][OH ] / [H 2 O] = K w = 1.0 x 10 14 We get: K a H20 = K b H20 = (1.0 e 7 )(1.0 e 7 ) / (55.556) = 1.8 e 16 And from: K ah2o K boh = K w We get: K boh = K ah+ = K w / K ah2o = 1.0 e14 / 1.8 e 16 = 55.556 Looking at the relative acid base strengths, this should make since: K a H20 = K b H20 = 1.8 e 16 K b OH = K a H+ = 55.556 The magnitude of K a describes the weak acid strength, just as K eq described the magnitude of a reaction. The larger the value of K a, the greater the dissociation, the further to the right the equilibrium is shifted, meaning the greater the acid strength. Examine the Acetic acid equilibrium again. 4
CH 3 CO 2 H + H 2 O CH 3 CO 2 + H 3 O + A 1 B 1 B 2 A 2 Look Up K a values for both acids We know that K<<1; so, the equilibrium favors the reactants. But what does that say about the rest of the species in equilibrium? Because H 3 O + is a stronger BL acid than acetic the reaction lies to the left. This brings us to an important observation: In every acidbase reaction, equilibrium favors the transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base. 2. Write the dissociation (ionization) reaction for each of the following and label as acid/base or conjugate acid/base. Then determine if the equilibrium will lie to the right or the left using what you know about acid/base strengths. a. NH 3 + H 2 O b.hso 4 + CO 3 2 c. PO 4 3 + H 2 O 5