University:
Kentucky State University
Course:
CHE 101 | General Chemistry I
Academic year:

2021
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135

Pages:

48
Author:

vulcanism9v4

Acid-Base Equilibria CHEM-102 1 Example 9: Predict the direction of the following reaction in aqueous solution: HNO2(aq) + CN-(aq) HCN(aq) + NO-2 (aq) Solution HNO2 is a stronger acid than HCN. Thus, CN- is a stronger base than NO-2 . The net reaction will proceed from left to right as written because HNO2 is a better proton donor than HCN (and CN- is a better proton acceptor than NO2 ). CHEM-102 2 Example 8: Calculate the pH of a : (a) 1.0 x 10-3 M HCl solution (a) H+(aq) + Cl-(aq) HCl(aq) 1.0 x 10-3 M 1.0 x 10-3 M Final (M): → H+(aq) + Cl-(aq) 1.0 x 10-3 -1.0 x 10-3 0.0 +1.0 x 10-3 0.0 +1.0 x 10-3 0.0 1.0 x 10-3 1.0 x 10-3 HCl(aq) Initial (M): Change (M): (b) 0.020 M Ba(OH)2 solution [H+] = 1.0 x 10-3 M pH = -log (1.0 x 10-3) = 3.00 CHEM-102 3 (b) Ba(OH)2 is a strong base 0.020 M 2 X 0.020 M = 0.040 M [OH-] = 0.040 M pOH = -log 0.040 = 1.40 pH = 14.00 - pOH = 14.00 - 1.40 = 12.60 CHEM-102 4 Calculate the pH of a : H3O+ (aq) + F- (aq) HF (aq) + H2O (l) HNO2 (aq) + H2O (l) F- (aq) + H2O (l) H3O+ (aq) + NO2- (aq) OH- (aq) + HF (aq) NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq) Use ICE Table CHEM-102 5 Solving weak acid ionization problems: 1. Identify the major species that can affect the pH. • In most cases, you can ignore the autoionization of water. • Ignore [OH-] because it is determined by [H+]. 2. Use ICE to express the equilibrium concentrations in terms of single unknown x. 3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4. Calculate concentrations of all species and/or pH of the solution. CHEM-102 6 Given :Kb/Ka , [ ]o Calculate: [H+] [OH−] →pH/pOH Use ICE Table Kb/Ka in terms of x Kw = [H+][OH−] KaKb = Kw Prevent Quadratic Equation Assumption → X~0 → Validation 5% Rule = X x 100% [ ]0 Given : pH/pOH , [ ]o , Ioniz. % Calculate: Kb/Ka Find: [H+] = 10−pH [OH−] = 10−pOH X x 100% Ioniz. % = [ ]0 X= Ioniz. % x [ ]0 100 % Now → we have x x=[H+] , x=[OH−] Solve for X → X=[H+] , X=[OH−] Substitute x in ICE Table ➢ pH = -log [H+] pH + pOH = 14.00 ➢ pOH = -log [OH-] X x 100% ➢ Ioniz. % = [ ]0 Find Ka / Kb Kw Ka = Kb Weak Acids (HA) and Acid Ionization Constants HA (aq) + H2O (l) HA (aq) H3O+ (aq) + A- (aq) H+ (aq) + A- (aq) [H+][A-] Ka = [HA] Ka is the acid ionization constant Ka CHEM-102 weak acid strength 8 Weak Bases and Base Ionization Constants NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+][OH-] Kb = [NH3] Kb is the base ionization constant Kb weak base strength Solve weak base problems like weak acids except solve for [OH-] instead of [H+]. Kw = [H+][OH−] CHEM-102 9 Ionization Constants of Conjugate Acid-Base Pairs HA (aq) A- (aq) + H2O (l) H2O (l) H+ (aq) + A- (aq) OH- (aq) + HA (aq) H+ (aq) + OH- (aq) Ka Kb Kw KaKb = Kw Weak Acid and Its Conjugate Base Kw Ka = Kb CHEM-102 Kw Kb = Ka 10 Ionized acid concentration at equilibrium percent ionization = x 100% Initial concentration of acid For a monoprotic acid HA, Percent ionization = [H+] [HA]0 x 100% [HA]0 = initial concentration Simply: X x 100% Percent ionization = [ ]0 CHEM-102 11 Before Solving Quadratic Equation → we need to do Approximation -b ± b 2 - 4ac x= 2a → Assume X is very small → so X~0 But, we need to test the approximation, 5% Rule = CHEM-102 X x 100% [ ]0 12 Given :Kb/Ka , [ ]o Calculate: [H+] [OH−] →pH/pOH Use ICE Table Kb/Ka in terms of x Kw = [H+][OH−] KaKb = Kw Prevent Quadratic Equation Assumption → X~0 → Validation 5% Rule = X x 100% [ ]0 Given : pH/pOH , [ ]o , Ioniz. % Calculate: Kb/Ka Find: [H+] = 10−pH [OH−] = 10−pOH X x 100% Ioniz. % = [ ]0 X= Ioniz. % x [ ]0 100 % Now → we have x x=[H+] , x=[OH−] Solve for X → X=[H+] , X=[OH−] Substitute x in ICE Table ➢ pH = -log [H+] pH + pOH = 14.00 ➢ pOH = -log [OH-] X x 100% ➢ Ioniz. % = [ ]0 Find Ka / Kb Kw Ka = Kb Example 9: Calculate the pH of a 0.036 M nitrous acid (HNO2) solution: HNO2(aq) H+(aq) + NO-2 (aq) we ignore the ionization of H2O → so, the major source of H+ ions is the acid. CHEM-102 14 Letting x be the equilibrium concentration of H+ and NO-2 ions in mol/L, we summarize: Initial (M): Change (M): Equilibrium (M): HNO2(aq) H+(aq) + (aq) 0.036 -x 0.036 - x 0.00 +x x 0.00 +x x [H + ][NO-2 ] Ka = [HNO 2 ] 2 x 4.5 × 10-4 = 0.036-x Applying the approximation → 0.036 - x ≈ 0.036, we obtain CHEM-102 15 2 2 x x 4.5 × 10-4 =  0.036-x 0.036 x 2 = 1.62  10−5 x = 4.0  10−3 M To test the approximation, X 5% Rule = x 100% [ ]0 4.0 × 10-3 M × 100% = 11% 0.036 M Because this is greater than 5%, our approximation is not valid. CHEM-102 16 we must solve the quadratic equation, as follows: x 2 + 4.5 × 10-4 x - 1.62 × 10-5 = 0 -4 -4.5 × 10 ± x= ( 4.5 × 10 -4 ) 2 -b ± b 2 - 4ac x= 2a - 4(1)(-1.62 × 10-5 ) 2(1) = 3.8 × 10-3 M or -4.3 × 10-3 M because the concentration of ions cannot be negative. Therefore, the solution is given by the positive root, x = 3.8 x 10-3 M. At equilibrium: [H+] = 3.8 x 10-3 M pH = -log (3.8 x 10-3 ) = 2.42 CHEM-102 17 Example 10: The pH of a 0.10 M solution of formic acid (HCOOH) is 2.39. What is the Ka of the acid? pH = -log [H+] 2.39 = -log [H+] Taking the antilog of both sides: [H+] = 10-2.39 = 4.1 x 10-3 M CHEM-102 18 Initial (M): Change (M): HCOOH(aq) H+(aq) + HCOO-(aq) 0.10 0.00 +4.1 x 10-3 0.00 +4.1 x 10-3 4.1 x 10-3 4.1 x 10-3 −4.1 x 10-3 Equilibrium(M) (0.10 − 4.1 x 10-3) [H + ][HCOO- ] Ka = [HCOOH] (4.1 × 10-3 )(4.1 × 10-3 ) = (0.10 - 4.1 × 10-3 ) = 1.8 × 10-4 CHEM-102 19 Diprotic and Triprotic Acids • May yield more than one hydrogen ion per molecule. • Ionize in a stepwise manner; that is, they lose one proton at a time. • An ionization constant expression can be written for each ionization stage. • Consequently, two or more equilibrium constant expressions must often be used to calculate the concentrations of species in the acid solution. CHEM-102 20 CHEM-102 21 Example 11: Oxalic acid (H2C2O4) is a poisonous substance used chiefly as a bleaching and cleansing agent. Calculate the concentrations of all the species present at equilibrium in a 0.10 M solution. H2C2O4(aq) Initial (M): Change (M): Equilibrium (M): + Ka = CHEM-102 0.10 -x 0.10 - x ][HC2O −4 ] [H [H 2C2O 4 ] 2 x 6.5 × 10-2 = 0.10 - x H+(aq) + HC2O-4 (aq) 0.00 +x x 0.00 +x x x2 x2 6.5 × 10 =  0.10 - x 0.10 x 2 = 6.5  10−3 -2 x = 8.1  10 −2 M 22 To test the approximation, 8.1 × 10-2 M × 100% = 81% 0.10 M the approximation is not valid. Therefore, we must solve the quadratic equation x2 + 6.5 x 10-2x - 6.5 x 10-3 = 0 The result is x = 0.054 M. -b ± b 2 - 4ac x= 2a [H+] = 0.054 M [ HC2O-4] = 0.054 M [H2C2O4] = (0.10 - 0.054) M = 0.046 M CHEM-102 23 Letting y be the equilibrium concentration of H+ and C2O24 ions in mol/L, we summarize: Initial (M): Change (M): Equilibrium (M): HC2O-4 (aq) H+(aq) + C2O2(aq) 4 0.054 -y 0.054 +y 0.00 +y 0.054 - y 0.054 + y y [H + ][C2O24− ] Ka = [HC2O −4 ] (0.054 + y )(y ) 6.1 × 10 = (0.054 - y ) -5 CHEM-102 24 Applying the approximation 0.054 + y ≈ 0.054 and 0.054 - y ≈ 0.054, we obtain (0.054)(y ) = y = 6.1 × 10-5 M (0.054) test the approximation→ 6.1 × 10-5 M × 100% = 0.11% 0.054 M The approximation is valid. CHEM-102 25 At equilibrium, [H2C2O4] = 0.046 M [HC2O-4 ] = (0.054 - 6.1 x 10-5) M = 0.054 M [H+] = (0.054 + 6.1 x 10-5) M = 0.054 M -5 M [ C2O2] = 6.1 x 10 4 [OH-] = 1.0 x 10-14/0.054 = 1.9 x 10-13 M CHEM-102 26 Example 12: What is the pH of a 0.40 M ammonia solution? NH3(aq) Initial (M): Change (M): Equilibrium (M): Base (aq) + OH-(aq) + H2O(l) 0.40 0.00 0.00 -x +x +x 0.40 - x x x [NH +4 ][OH - ] Kb = [NH3 ] x2 1.8 × 10 = 0.40 - x -5 Applying the approximation 0.40 - x ≈ 0.40, we obtain CHEM-102 27 x2 x2 1.8 × 10 =  0.40 - x 0.40 x 2 = 7.2  10 −6 -5 x = 2.7  10−3 M Test the approximation: 2.7 × 10-3 M × 100% = 0.68% 0.40 M It is valid At equilibrium, [OH-] = 2.7 x 10-3 M. Thus pOH = -log (2.7 x 10-3 ) = 2.57 pH = 14.00 - 2.57 = 11.43 CHEM-102 28 Example 13: Base Calculate ka and pH of a 0.40 M ammonia solution if the ionization percent is 0.68% ? Ioniz. % = X x 100% [ ]0 Ioniz. % x [ ]0 X= 100 % X= 0.68 % x 0.40 100 % X = [OH−] NH3(aq) Initial (M): Change (M): Equilibrium (M): CHEM-102 (aq) + OH-(aq) + H2O(l) 0.40 0.00 0.00 -x +x +x 0.40 - x x x 29 NH3(aq) Initial (M): 0.40 0.00 0.00 -x +x +x 0.40 - x x x Change (M): Equilibrium (M): (aq) + OH-(aq) + H2O(l) )2 ( 0.40- pOH = -log [OH-] pOH = -log ( ) Or Kw = [H+][OH−] pOH= 2.57 pH + pOH = 14.00 pH = -log [H+] pH= 11.43 CHEM-102 30 Example 14: Salt For a 0.25 M solution of potassium fluoride (KF). (a) Calculate the pH (b) What is the percent hydrolysis? KOH + HF → KF S.B W.A Basic KF(aq) Initial (M): Change (M): Final (M): 0.25 −0.25 0 → K+(aq) + 0 +0.25 0.25 F−(aq) 0 +0.25 0.25 Of these ions, only the fluoride ion will react with water F−(aq) + H2O(l) CHEM-102 HF(aq) + OH−(aq) 31 Let x be the equilibrium concentration of HF and OH− ions in mol/L, we summarize: F−(aq) Initial (M): Change (M): Equilibrium (M): + H2O (l) HF(aq) + OH−(aq) 0.25 0.00 0.00 −x +x +x 0.25 − x x x Because Kb is very small and the initial concentration of the base is large, we can apply the approximation 0.25 − x ≈ 0.25: CHEM-102 32 At equilibrium: [OH−] = 1.9 x 10-6 M pOH = -log (1.9 x 10-6 ) = 5.72 pH = 14.00 − 5.72 = 8.28 Thus the solution is basic. The percent hydrolysis is given by % hydrolysis = [F−]hydrolyzed [F−]initial −6 1.9 × 10 = × 100% 0.25 = 0.00076% CHEM-102 33 Example 15: Salt Calculate the pH of a ~~M solution of NH4Cl. NH4+(aq) + Cl−(aq) NH4Cl NH3 HCl W.B S.A Acidic NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) Use ICE Table CHEM-102 pH = -log [H+] 34 Molecular Structure and Acid Strength H X H+ + X- The stronger the bond The weaker the acid HF << HCl < HBr < HI acidity increases CHEM-102 35 CHEM-102 36 Molecular Structure and Oxoacid Strength Z dO d+ H Z O- + H+ The O-H bond will be more polar and easier to break if: • Z is very electronegative or • Z is in a high oxidation state CHEM-102 37 Molecular Structure and Oxoacid Strength 1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number. •• •• •• •• •• •• Acid strength increases with increasing electronegativity of Z •• •• O O •• •• •• •• H O Cl O H O Br O •• •• • • •• •• • • Cl is more electronegative than Br HClO3 > HBrO3 CHEM-102 acidity increases 38 Molecular Structure and Acid Strength 2. Oxoacids having the same central atom (Z) but different numbers of attached groups. Acid strength increases as the oxidation number of Z increases. HClO4 > HClO3 > HClO2 > HClO CHEM-102 39 Example 12: Predict the relative strengths of the oxoacids in each of the following groups: (a) HClO, HBrO, and HIO (b) HNO3 and HNO2 (a)electronegativity decreases from Cl to I, the Cl atom attracts the electron pair it shares with the O atom to the greatest extent. Consequently, the O−H bond is the most polar in HClO and least polar in HIO. Thus, the acid strength decreases as follows: HClO > HBrO > HIO (b) The structures of HNO3 and HNO2 are shown in Figure 16.5. Because the oxidation number of N is +5 in HNO3 and +3 in HNO2, HNO3 is a stronger acid than HNO2. Acid-Base Properties of Salts Acid Solutions: Salts derived from a strong acid and a weak base. NH4Cl (s) NH4+ (aq) H 2O NH4+ (aq) + Cl- (aq) NH3 (aq) + H+ (aq) Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid. Al(H2O)3+ 6 (aq) CHEM-102 Al(OH)(H2O)52+(aq) + H+ (aq) 41 Acid Hydrolysis of Al3+ CHEM-102 42 Acid-Base Properties of Salts Solutions in which both the cation and the anion hydrolyze: • Kb for the anion > Ka for the cation, solution will be basic • Kb for the anion < Ka for the cation, solution will be acidic • Kb for the anion  Ka for the cation, solution will be neutral CHEM-102 43 CHEM-102 44 Predict whether the following solutions will be acidic, basic, or nearly neutral: (a)NH4I The solution is acidic (b) NaNO2 The solution is basic (c) FeCl3 (d) NH4F The solution is acidic The solution is acidic (a)The cation is NH+4 , which will hydrolyze to produce NH3 and H+. The I- anion is the conjugate base of the strong acid HI. Therefore, I- will not hydrolyze and the solution is acidic. (b) The Na+ cation does not hydrolyze. The NO-2 is the conjugate base of the weak acid HNO2 and will hydrolyze to give HNO2 and OH-. The solution will be basic. (C) Fe3+ is a small metal ion with a high charge and hydrolyzes to produce H+ ions. The Cl- does not hydrolyze. Consequently, the solution will be acidic. + (d) Both the NH4 and F- ions will hydrolyze. the Ka of NH+4 (5.6 x 10-10) is greater than the Kb for F- (1.4 x 10-11). Therefore, the solution will be acidic. Oxides of the Representative Elements In Their Highest Oxidation States CHEM-102 Na2O (s) + H2O (l) 2NaOH (aq) CO2 (g) + H2O (l) H2CO3 (aq) N2O5 (g) + H2O (l) 2HNO3 (aq) 47 Acid-Base Properties of Salts Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3-). NaCl (s) H 2O Na+ (aq) + Cl- (aq) Basic Solutions: Salts derived from a strong base and a weak acid. CH3COONa (s) H2O Na+ (aq) + CH3COO- (aq) CH3COO- (aq) + H2O (l) CHEM-102 CH3COOH (aq) + OH- (aq) 48

Acid-Base Equilibria

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