Autoionization of Water

Water itself is a weak acid and weak base: Two water molecules react to form H 3 O + and OH - (but only slightly) H 2 O H 2 O H 3 O + OH - Autoionization of Water H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) [H 3 O + ][OH - ] K = [H 2 O] 2 The ion-product for water, K w : = [H + ][OH - ] = K w Set = 1.0 since nearly pure liquid water even when salts, acids, or bases present. K = K w = [H 3 O + ][OH - ] = 1.0 x 10-14 (at 25 o C) For pure water the concentration of hydroxyl and hydronium ions must be equal: [H 3 O + ] = [OH - ] = 1.0 x 10-14 = 1.0 x 10-7 M (at 25 o C) 1

The Meaning of K b, the Base Dissociation Constant For the generalized reaction between a base, B(aq), and water: B (aq) + H 2 O (l) BH + (aq) + OH - (aq) Equilibrium constant: K = K b = [BH + ][OH - ] [B] The stronger the base, the higher the [OH - ] at equilibrium and the larger the K b. The Relation Between K a and K b for a Conjugate Acid-Base Pair Acid HA + H 2 O H 3 O + + A - Base A - + H 2 O HA + OH - 2 H 2 O H 3 O + + OH - [H 3 O + ] [A - ] [HA] x [HA] [OH - ] = [H 3 O + ] [OH - ] [A - ] K a x K b = K w For HNO 2 : K a = 4.5 x 10-4 and for NO - 2 : K b = 2.2 x 10-11 K a x K b = (4.5 x 10-4 )(2.2 x 10-11 ) = 9.9 x 10-15 or ~ 10 x 10-15 = 1 x 10-14 = K w 2

ph of a Strong Acid Calculate ph of 1.0 M HClO 4 major species : H +, ClO 4 - and H 2 O perchloric acid is a strong acid: FULLY dissociated Therefore, 1.0 M HClO 4 1.0 M H + ph = - log [H + ] = - log [1.0] = 0.00 Calculate ph of 0.01 M HClO 4 ph = - log [H + ] = - log [0.01] = 2.0 Calculate ph of 0.02 M HClO 4 ph = - log [H + ] = - log [0.02] = 1.7 Two New Definitions Remember: K w = [H+][OH-] so that [OH-] = K w / [H+] 1) poh = -log [OH-] = - log (K w /[H + ]) = - log K w (- log [H + ]) = pk w ph = - log (1.0 x 10-14 ) ph poh = 14.00 ph 2) It is convenient to express equilibrium constants for acid dissociation in the form pk a = - log K a 3

ph of a Strong Base Calculate ph of 1.0 M NaOH major species : Na +, OH - and H 2 O sodium hydroxide is a strong base: FULLY dissociated Therefore, 1.0 M NaOH 1.0 M [OH - ] ph = - log [H + ]; poh= - log [OH - ]; pk w = ph + poh = 14 ph = 14 poh ph = 14 (-log 1.0) = 14.0 Calculate ph of 0.01 M NaOH ph= 14 (-log 0.01) = 14 2 = 12.0 Calculate ph of 0.02 M NaOH ph= 14 (-log 0.02) = 14-1.70 = 12.3 Calculating [H 3 O + ], ph, [OH - ], and poh Problem: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H 3 O + ], ph, [OH - ], and poh of the two solutions at 25 o C. Solution: (a) [H + ] = 3.0 M ph = -log[h + ] = -log(3.0) = -0.48 [OH - ] = K w = 1 x 10-14 = 3.3 x 10-15 M [H + ] 3.0 poh = - log(3.3 x 10-15 ) = 14.52 (b) [H + ] = 0.0024 M ph = -log[h + ] = -log(0.0024) = 2.62 [OH - ] = K w = 1 x 10-14 = 4.2 x 10-12 M [H + ] 0.0024 poh = -log(4.2 x 10-12 ) = 11.38 4

ph= -log[h + ] poh= - log [OH - ] pk w = 14 = ph + poh K w = [H + ][OH - ] Remember The Acid-Dissociation Constant (K a ) Acids dissociate into ions in water: HA (aq) + H 2 O (l) HA (aq) H 3 O + (aq) + A - (aq) H + (aq) + A - (aq) K a = [H 3 O + ][A - ] [HA] = [H + ][A - ] [HA] In a dilute solution of a WEAK acid, most of HA remains undissociated and therefore [HA] init = [HA] eq K a = [H + ][A - ]/[HA] = [H + ] 2 / [HA] init 5

Calculate the ph of a 1.00 M HNO 2 Solution Problem: Calculate the ph of a 1.00 M solution of nitrous acid HNO 2. Solution: HNO 2 (aq) H + (aq) + NO 2 - (aq) K a = 4.0 x 10-4 Initial concentrations: [H + ] = 0, [NO 2- ] = 0, [HNO 2 ] = 1.00 M Final concentrations: [H + ] = x, [NO 2- ] = x, [HNO 2 ] = 1.00 M - x [H K a = + ] [NO 2- ] = 4.0 x 10-4 = [HNO 2 ] (x) (x) 1.00 - x Assume 1.00 x 1.00 to simplify the problem. x 2 4.0 x 10-4 or x 2 4.0 x 10-4 1.00 x = 2.0 x 10-2 = 0.02 M = [H + ] = [NO 2- ] ph = - log[h + ] = - log(2.0 x 10-2 ) = 1.70 Determining Concentrations from K a and Initial [HA] Problem: Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H + ] and ph of a 0.125 M HClO solution? K a = 3.5 x 10-8 Plan: We need to find [H + ]. First we write the balanced equation and the expression for K a and solve for the hydronium ion concentration. Solution: HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) [H K a = + ] [ClO - ] = 3.5 x 10-8 [HClO] Concentration (M) HClO H 2 O H + ClO - Initial 0.125 ---- 0 0 Change -x ---- +x +x Equilibrium 0.125 - x ---- x x (x)(x) K a = = 3.5 x 10-8 0.125-x Assume (0.125 x) 0.125 x 2 = 4.375 x 10-9 x = 6.61 x 10-5 ph = -log(h + ) = -log(6.61 x 10-5 ) = 4.18 6

Finding K a for a Weak Acid from the ph I Problem: The weak acid hypochlorous acid is formed in bleach solutions. If the ph of a 0.12 M solution of HClO is 4.19, what is the value of the K a of this weak acid? Plan: We are given [HClO] initial and the ph, which will allow us to find [H 3 O + ] and then the hypochlorite anion concentration, so we can write the reaction and expression for K a and solve directly. Solution: Calculating [H 3 O + ] : [H 3 O + ] = 10 -ph = 10-4.19 = 6.46 x 10-5 M Concentration (M) HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) Initial 0.12 ---- 1 x 10-7 0 Change -x ---- +x +x Equilibrium 0.12 -x ---- 1 x 10-7 + x x Assumptions: [H 3 O + ] = [H 3 O + ] HClO since HClO is a weak acid, we assume (0.12 M x) 0.12 M Finding the K a for a Weak Acid from the ph II HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) x = [H 3 O + ] = [ClO - ] = 6.46 x 10-5 M [H 3 O + ] [ClO - ] K a = (6.46 x 10-5 M) (6.46 x 10-5 M) = = 3.48 x 10-8 [HClO] 0.12 M K a = 3.48 x 10-8 In text books it is found to be: 3.5 x 10-8 Checking: 1. For [H 3 O + 1 x 10 ] from water : -7 M x 100 = 0.155% 6.46 x 10-5 M assumptions are OK 2. For [HClO] dissoc : x 100 = 0.0538 % 6.46 x 10-5 M 0.12 M 7

Overview: Solving Weak Acid Equilibrium Problems List the major species in the solution. Choose the species that can produce H + and write balanced equations for the reactions producing H +. Comparing the values of the equilibrium constants for the reactions you have written, decide which reaction will dominate in the production of H +. Write the equilibrium expression for the dominant reaction. List the initial concentrations of the species participating in the dominate reaction. Define the change needed to achieve equilibrium; that is, define x. Write the equilibrium concentrations in terms of x. Substitute the equilibrium concentrations into the equilibrium expression. Solve for x the easy way assume that [HA] 0 x [HA] 0 Verify the approximation is valid (5% rule). Calculate [H + ] and ph. Mixtures of Several Acids - I Calculate the ph of a solution that contains 1.00 M HF (K a = 7.2 x 10-4 ) and 5.00 M HOCl (K a = 3.5 x 10-8 ). Also calculate the concentrations of the F- and OCl- ions at equilibrium. Three components produce H + : HF (aq) H + (aq) + F - (aq) K a = 7.2 x 10-4 HOCl (aq) H + (aq) + OCl - (aq) K a = 3.5 x 10-8 H 2 O (aq) H + (aq) + OH - (aq) K a = 1.0 x 10-14 Even though HF is a weak acid, it has by far the greatest K a, so it will be the dominant producer of H +. Solve for its I.C.E. table first, then use those concentrations in the second most important equilibrium. Move down the list, in each case setting K a = [H + ] [A - ] [HA] = specific value 8

Mixtures of Several Acids - II Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [HF] 0 = 1.00 [HF] = 1.00 x [F - ] = 0 x mol HF [F - ] = x [H + ] = ~ 0 dissociates [H + ] = x [H K a = + ] [F - ] (x) (x) = 7.2 x 10-4 = [HF] 1.00-x x 2 1.00 Therefore, x = 2.7 x 10-2 [F - ] = [H + ] = x = 2.7 x 10-2 and ph = 1.57 Mixtures of Several Acids - III The concentration of H + comes from the first part of this problem: Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [HOCl] 0 = 5.00 [HOCl] = 5.00 x [OCl - ] = 0 x mol HOCl [OCl - ] = x [H + ] = 2.7 x 10-2 dissociates [H + ] = 2.7 x 10-2 + x [H K a = + ] [OCl - ] (2.7 x 10 = 3.5 x 10-8 = -2 + x) (x) [HOCl] 5.00-x (2.7 x 10-2 ) x 5.00 x = 6.5 x 10-6 M = [OCl - ] Therefore, ph = 1.56 [F - ] = 2.7 x 10-2 M [OCl - ] = 6.5 x 10-6 M 9

PERCENT DISSOCIATION amount dissociated (mol/l) % dissociation = x 100 initial concentration (mol/l) x % dissociation = x 100 [HA] o for a given acid, % dissociation increases as the acid is diluted for a weak acid, [H+] decreases with increasing dilution as [HA] does, but % dissociation increases as dilution increases Figure 7.5: Effect of dilution on the percent dissociation and [H+] K a = [H+ ] [A - ] [HA] 10

Problem: Calculate the percent dissociation of a 1.00 M hydrocyanic acid solution, K a = 6.20 x 10-10 HCN (aq) + H 2 O (l) H 3 O + (aq) + CN - (aq) HCN H 3 O + CN - [H 3 O + ][CN - ] Initial 1.00 M 0 0 K a = [HCN] Change -x +x +x Final 1.00 x x x (x)(x) K a = = 6.20 x 10 (1.00-x) -10 Assume 1.00-x 1.00 K a = = 6.2 x 10 1.00-10 x = 2.49 x 10-5 2.49 x 10-5 % dissociation = x 100 = 0.00249% 1.00 x 2.. N Weak Bases Many compounds with an electron-rich nitrogen are weak bases. The common structural feature is an N atom that has a lone electron pair in its Lewis structure. EXAMPLES: 1) Ammonia (:NH 3 ) 2) Amines (general formula RNH 2, R 2 NH, R 3 N) H 3 C H CH 3 Dimethylamine H 3 C.. N CH 3 CH 3 Trimethylamine H 3 C.. N.. N H H Methylamine H H C 2 H 5 Ethylamine 11

General reaction (R is a hydrocarbon group or a H): NR 3 (aq) + H 2 O(l) HNR 3+ (aq) + OH - (aq) K b = [HNR 3+ ] [OH - ] [NR 3 ] Determining ph from K b and Initial [B] I Problem: Ammonia is commonly used cleaning agent and is a weak base, with a K b of 1.8 x 10-5. What is the ph of a 1.5 M NH 3 solution? Plan: Ammonia reacts with water to form [OH - ]. Calculate [H 3 O + ] and the ph. The balanced equation and K b expression are: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) K b = [NH 4 + ] [OH - ] [NH 3 ] Concentration (M) NH 3 H 2 O NH 4 + OH - Initial 1.5 ---- 0 0 Change -x ---- +x +x Equilibrium 1.5 - x ---- x x making the assumption: since K b is small, (1.5 M x) 1.5 M 12

Determining ph from K b and Initial [B] II Substituting into the K b expression and solving for x: [NH 4+ ] [OH - ] (x)(x) K b = = = 1.8 x 10-5 [NH 3 ] 1.5 x 2 = 2.7 x 10-5 x = 5.20 x 10-3 = [OH - ] = [NH 4+ ] Calculating ph: K [H 3 O + w 1.0 x 10-14 ] = = = 1.92 x 10-12 [OH - ] 5.20 x 10-3 ph = -log[h 3 O + ] = - log (1.92 x 10-12 ) ph = 11.72 Polyprotic Acids 13

Calculate the ph of 3.00 x 10-3 M Sulfuric Acid Polyprotic acid, more than one proton to lose!!! Multiple K a s H 2 SO 4 (aq) HSO 4- (aq) + H + (aq) K a1 = LARGE HSO 4- (aq) SO 4- (aq) + H + (aq) K a2 = 1.2 x 10-2 Because first dissociation has a very large K a it can be treated as a strong acid (~100% dissociated). Calculate the ph of 3.00 x 10-3 M Sulfuric Acid HSO 4- (aq) SO 4- (aq) + H + (aq) K a = 1.2x10-2 Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [HSO 4- ] 0 = 0.00300 x mol/l HSO - 4 [HSO 4- ] = 0.00300 x [SO 2-4 ] 0 = 0 dissociates [SO 2-4 ] = x [H + ] 0 = 0.00300 to reach [H + ] = 0.00300 + x equilibrium From dissociation of H 2 SO 4 [H + ][SO 2-4 ] K a2 = 1.2 x 10-2 (0.00300 + x)(x) = = [HSO 4- ] (0.00300 x) No obvious approximation looks valid, so we must solve with the quadratic formula. 0 = x 2 + 0.015x 3.6 x 10-5 -b + a = 1 x = - b 2 4ac b = 0.015 x = 2.10 x 10-3 2a c = -3.6 x 10-5 [H + ] = 0.00300 + x = 0.00510 ph = 2.29 Note: Change in [H+] does not violate huge K a for 1 st dissociation: [H 2 SO 4 ] = 0. 14

SALTS and their ph sodium acetate (CH 3 COONa) solution is basic CH 3 COO - + H 2 O CH 3 COOH + OH - ammonium chloride (NH 4 Cl) solution is acidic NH 4 + + H 2 O NH 3 + H 3 O + sodium chloride solution is neutral NaCl + H 2 O Na + + Cl - + H 2 O Effects of Salts on ph and Acidity Salts that consist of cations of strong bases and the anions of strong acids have no effect on the [H + ] when dissolved in water. Examples: NaCl, KNO 3, Na 2 SO 4, NaClO 4, KBr, etc. A salt whose cation alone has neutral properties (such as Na + or K + ) and whose anion is the conjugate base of a weak acid, produces a basic solution in water, since anion captures some H +. Examples: NaF, KCN, NaC 2 H 3 O 2, Na 3 PO 4, Na 2 CO 3, K 2 S, Na 2 C 2 O 4, etc. A salt whose anion alone has neutral properties and whose cation is the conjugate acid of a weak base OR a small and highly charged metal ion will produce an acidic solution in water. Examples: NH 4 Cl, AlCl 3, Fe(NO 3 ) 3, etc. 15

See Table 7.6 Like Example 7.12 - I Calculate the ph of a 0.45 M NaCN solution. The K a value for HCN is 6.2 x 10-10. Solution: Since HCN is a weak acid, the cyanide ion must have significant affinity for protons. CN - (aq) + H 2 O (l) HCN (aq) + OH - (aq) K b = [HCN] [OH- ] [CN - ] The value of K b can be calculated from K w and the K a value for HCN. K = K w 1.0 x 10-14 K a (for HCN) = b 6.2 x 10 = 1.61 x 10-5 -10 Initial Concentration (mol/l) [CN - ] 0 = 0.45 [HCN] 0 = 0 [OH - ] 0 = 0 X mol/l CN - reacts with H 2 O to reach equilibrium Equilibrium Concentration (mol/l) [CN - ] = 0.45 x [HCN] = x [OH - ] = x 16

Like Example 7.12 - II Thus: [HCN] [OH - ] K b = 1.61 x 10-5 = [CN - = ] (x)(x) 0.45 - x The approximation is not valid by the 5% rule, so you have to use the quadratic equation. x = 2.68 x 10-3 x = [OH - ] = 2.68 x 10-3 M poh = -log [OH - ] = 2.57 ph = 14.00 poh = 14.00-2.57 = 11.43 Small and highly charged cations produce acidic solutions in water. Examples: Al 3+, Fe 3+, etc.. 17

Predicting the Relative Acidity of Salt Solutions Problem: Determine whether an aqueous solution of iron(iii) nitrite, Fe(NO 2 ) 3, is acidic, basic, or neutral. Plan: The formula consists of the small, highly charged, and therefore weakly acidic, Fe 3+ cation and the weakly basic NO - 2 (anion of the weak acid HNO 2 ). To determine the relative acidity of the solution, we write equations that show the reactions of the ions with water, and then find K a and K b of the ions to see which ion reacts to form H + or OH - to a greater extent. Solution: Writing the reactions with water: Fe(H 2 O) 3+ 6 (aq) + H 2 O (l) Fe(H 2 O) 5 OH 2+ (aq) + H 3 O + (aq) NO 2 - (aq) + H 2 O (l) HNO 2(aq) + OH - (aq) Obtaining K a and K b of the ions: for Fe 3+ (aq) K a = 6 x 10-3 for NO 2 - (aq), K b must be determined: K b of NO - K 2 = w 1.0 x 10 = -14 = 2.5 x 10-11 K a of HNO 2 4.1 x 10-4 Since K a of Fe 3+ > K b of NO 2-, the solution is acidic. Summary: Solving Acid-Base Equilibria Problems List the major species in solution. Look for reactions that can be assumed to go to completion, such as a strong acids/bases dissociating or H + reacting with OH -. For a reaction that can be assumed to go to completion: a) Determine the concentrations of the products. b) Write down the major species in solution after the reaction. Look at each major component of the solution and decide whether it is an acid or a base. Pick the equilibrium that will control the ph. Use known values of the dissociation constants for the various species to determine the dominant equilibrium. a) Write the equation for the reaction and the equilibrium expression. b) Set up the I.C.E. table and define x. c) Compute the equilibrium concentrations in terms of x. d) Substitute the [ ] eq into the equilibrium expression and solve for x. e) Check the validity of the approximation. f) Calculate the ph and other concentrations as required. 18

Summary: General Strategies for Solving Acid-Base Problems Think Chemistry. Focus on the solution components and their reactions. It will almost always be possible to choose one reaction that is the most important. Be systematic. Write down all the things you know, including things like the equilibrium expression, list what you can calculate, and list what s needed. Be flexible. Although all acid-base problems are similar in many ways, important differences do occur. Treat each problem as a separate entity. Do not try to force a given problem to match any you have solved before. Look for both the similarities and the differences. Be patient. The complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into its workable steps. Be confident. Look within the problem for the solution, and let the problem guide you. Assume that you can think it out. Do not rely on memorizing solutions to problems. In fact, memorizing solutions is usually detrimental because you tend to try to force a new problem to be the same as one you have seen before. Understand and think - don t just memorize. 19