Chapter 8 Monoprotic AcidBase Equilibria 8 1 Strong

Chapter 8 Monoprotic Acid-Base Equilibria 8 -1 Strong Acids and Bases HBr is a strong acid, so the reaction goes to completion. HBr + H 2 O H 3 O 3+ + Brp. H = -log[H+] = -log(0. 10) = 1. 00

Box 8 -1 Concentrated HNO 3 Is Only Slightly Dissociated 2 The figure shows a Raman spectrum of solutions of nitric acid of increasing concentration.

KOH is a strong base (completely dissociated), so [OH-] = 0. 10 M. p. H = -log[H+] = 13. 00 Relation between p. H and p. OH: p. H + p. OH = -log. KW = 14. 00 at 250 C (9 -1)

The Dilemma [H+] = KW/(1. 0 X 10 -8) = 1. 0 X 10 -6 M p. H = 6. 00 The Cure KW Step 1 Pertinent reactions. The only one is H 2 O = H+ + OH-. Step 2 Charge balance. The species in solution are K+, OH-, and H+. So, [K+] + [H+] = [OH-] (9 -2) Step 3 Mass balance. All K+ comes from the KOH, so [K+] = 1. 0 X 10 -8 M. Step 4 Equilibrium constant expression. KW = [H+][OH-] = 1. 0 X 10 -14. Step 5 Count. There are three equations and three unknowns ([H+], [OH-], [K+]), so we have enough information to solve the problem.

Step 6 Solve. Because we are seeking the p. H, let’s set [H+] = x. Writing [K+] = 1. 0 X 10 -8 M in Equation 9 -2, we get [OH-] = [K+] + [H+] = 1. 0 X 10 -8 + x [H+] = 9. 6 X 10 -8 M p. H = -log[H+] = 7. 02

1. When the concentration is “high” (≥ 10 -6 M), p. H is calculated by just considering the added H+ or OH-. That is, the p. H of 10 -5. 00 M KOH is 9. 00. 2. When the concentration is “low” (≤ 10 -8 M), the p. H is 7. 00. We have not added enough acid or base to change the p. H of the water itself. 3. At intermediate concentrations of 10 -6 to 10 -8 M, the effects of water ionization and the added acid or base are comparable. Only in this region is a systematic equilibrium calculation necessary.

8 -2 Weak Acids and Bases Let’s review the meaning of the acid dissociation constant, Ka, for the acid HA: Weak-acid equilibrium: A weak acid is one that is not completely dissociated. That is, Reaction 9 -3 does not go to completion. For a base, B, the base hydrolysis constant, Kb, is defined by the reaction Weak-base equilibrium: A weak base is one for which Reaction 9 -4 does not go to completion. p. K is the negative logarithm of an equilibrium constant: p. KW = -log KW p. Ka = -log Ka p. Kb = -log Kb

The acid HA and its corresponding base, A-, are said to be a conjugate acid -base pair, because they are related by the gain or loss of a proton. Relation between Ka and Kb for conjugate pair: Ka · Kb = K w (9 -5) Weak Is Conjugate to Weak The conjugate base of a weak acid is a weak base. The conjugate acid of a weak base is a weak acid.

Using Appendix G lists acid dissociation constants. Each compound is shown in its fully protonated form.

Names refer to neutral molecules.

8 -3 Weak-Acid Equilibria

A Typical Weak-Acid Problem Reactions: Charge balance: Mass balance: Equilibrium expressions: x 2 + (1. 07 X 10 -3) x – 5. 35 X 10 -5 = 0 x = 6. 80 X 10 -3 (negative root rejected) [H+] = [A-] = x = 6. 80 X 10 -3 M [HA] = F – x = 0. 0432 M p. H = -logx = 2. 17

[A-] (from HA dissociation) = 6. 8 X 10 -3 M [H+] from HA dissociation = 6. 8 X 10 -3 M [OH-] (from H 2 O dissociation) = 1. 5 X 10 -12 M [H+] from H 2 O dissociation = 1. 5 X 10 -12 M

Fraction of Dissociation The fraction of dissociation, α, is defined as the fraction of the acid in the form A-: Fraction of dissociation of an acid: Weak electrolytes (compounds that are only partially dissociated) dissociate more as they are diluted. The Essence of a Weak-Acid Problem Equation for weak acids:

Box 8 -2 Dyeing Fabrics and the Fraction of Dissociation

Dyes are colored molecules that can form covalent bonds to fabric. For example, Procion Brilliant Blue M-R is a dye with a blue chromophore (the colored part) attached to a reactive dichlorotriazine ring:

CelluloseㅡCH 2 OH Ka≈10 -15 = ROH cellulose ㅡ CH 2 O- + H+ RO- Fraction of dissociation fraction of dissociation

A handy tip: Equation 9 -11 can always be solved with the quadratic formula. However, an easier method worth trying first is to neglect x in the denominator.

8 -4 Weak-Base Equilibria Kb B + H 2 O = BH+ + OH[B] = F – [BH+] = F-x Equation for weak base:

A Typical Weak-Base Problem

B + H 2 O = BH+ + OH- 0. 0372 -x x x [H+] = KW/[OH-] = 1. 0 X 10 -14/3. 1 X 10 -4 = 3. 2 X 10 -11 p. H = -log[H+] = 10. 49 What fraction of cocaine has reacted with water? We can formulate α for a base, called the fraction of association: Fraction of association of a base:

Conjugate Acids and Bases-Revisited Earlier, we noted that the conjugate base of a weak acid is a weak base, and the conjugate acid of a weak base is a weak acid. Isomer of hydroxyl benzoic acid ortho para Ka Kb(= Kw/Ka) 1. 07 X 10 -3 2. 9 X 10 -5 9. 3 X 10 -12 3. 5 X 10 -10 p. H of 0. 050 0 M o-hydroxybenzoate = 7. 83 p. H of 0. 050 0 M p-hydroxybenzoate = 8. 62

8 -5 Buffers A buffered solution resists changes in p. H when acids or bases are added or when dilution occurs. The buffer is a mixture of an acid and its conjugate base. Mixing a Weak Acid and Its Conjugate Base If you mix A moles of a weak acid with B moles of its conjugate base, the moles of acid remain close to A and the moles of base remain close to B. HA = H+ + A 0. 10 -x x x Fraction of dissociation p. Ka = 4. 00

A- + H 2 O = HA + OH 0. 10 -x x x p. Kb = 10. 00 Fraction of dissociation HA dissociates very little, and adding extra A- to the solution will make the HA dissociate even less. Similarly, A- does not react very much with water, and adding extra HA makes A- react even less.

Henderson-Hasselbalch Equation The central equation for buffers is the Henderson-Hasselbalch equation, which is merely a rearranged form of the Ka equilibrium expression. Henderson-Hasselbalch equation for an acid: Henderson-Hasselbalch equation for a base:

Challenge Show that, when activities are included, the Henderson-Hasselbalch equation is Properties of the Henderson-Hasselbalch Equation Regardless of how complex a solution may be, whenever p. H = p. Ka, [A‑] must equal [HA]. This relation is true because all equilibria must be satisfied simultaneously in any solution at equilibrium. If there are 10 different acids and bases in the solution, the 10 forms of Equation 9 -16 must all give the same p. H, because there can be only one concentration of H+ in a solution.

A Buffer in Action

Notice that the volume of solution is irrelevant, because volume cancels in the numerator and denominator of the log term: moles of B/L of solution moles of BH+/L of solution moles of BH+

Box 8 -3 Strong Plus Weak Reacts Completely B + H+ = BH+ Weak Strong base acid OH- + HA = A- + H 2 O Strong base Weak acid Ka(for HA) H+ + OH- = H 2 O Strong acid Strong base Ka(for BH+) Kb(for A-)

Demonstration 9 -2 How Buffers Work

Percentage of reaction completed Calculated p. H

Procedure: All solutions should be fresh. Prepare a solution of formaldehyde by diluting 9 m. L of 37 wt% formaldehyde to 100 m. L. Dissolve 1. 5 g of Na. HSO 310 and 0. 18 g of Na 2 SO 3 in 400 m. L of water, and add ~1 m. L of phenolphthalein solution (Table 11 -4). Add 23 m. L of formaldehyde solution to the well-stirred buffer solution to initiate the clock reaction. The time of reaction can be adjusted by changing the temperature, concentrations, or volume.

We see that the p. H of a buffer does not change very much when a limited amount of strong acid or base is added. But why does a buffer resist changes in p. H? It does so because the strong acid or base is consumed by B or BH+.

Preparing a Buffer in Real Life! 1. Weigh out 0. 100 mol of tris hydrochloride and dissolve it in a beaker containing about 800 m. L of water. 2. Place a calibrated p. H electrode in the solution and monitor the p. H. 3. Add Na. OH solution until the p. H is exactly 7. 60. 4. Transfer the solution to a volumetric flask and wash the beaker a few times. Add the washing to the volumetric flask. 5. Dilute to the mark and mix.

Buffer Capacity 14 Buffer capacity, β, is a measure of how well a solution resists changes in p. H when strong acid or base is added. Buffer capacity is defined as Buffer capacity: Buffer capacity reaches a maximum when p. H = p. Ka. That is, a buffer is most effective in resisting changes in p. H when p. H = p. Ka (that is, when [HA] = [A-]). In choosing a buffer, seek one whose p. Ka is as close as possible to the desired p. H. The useful p. H range of a buffer is usually considered to be p. Ka ± 1 p. H unit. Buffer p. H depends on Ionic Strength and Temperature The correct Henderson-Hasselbalch equation, 9 -18, includes activity coefficients.

When What You Mix Is Not What You Get Mass balance: Charge balance: FHA + FA- = [HA] + [A-] [Na+] + [H+] = [OH-] + [A-] [HA] = FHA – [H+] + [OH-] [A-] = FA- + [H+] – [OH-] (9 -20) (9 -21)

The Henderson-Hasselbalch equation (with activity coefficients) is always true, because it is just a rearrangement of the Ka equilibrium expression.