Chem_2_Ch_7 c

Chem_2_Ch_7 c - Chapter 7 Acids and Bases Acids and Bases...

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Unformatted text preview: Chapter 7 Acids and Bases Acids and Bases Reviewed Arrhenius Acid: Produces H+ Arrhenius Base: Produces OH BronstedLowry Acid: H+ donor BronstedLowry Base: H+ acceptor Strong Acids and their Ionization Constants Strong acids are 100% ionized. HCl H+ + Cl a K = [H+][Cl]/[HCl] = Very large number Strong Acids HCl, Hydrochloric acid HBr, Hydrobromic acid HNO3, Nitric acid H2SO4, Sulfuric acid HClO3, Chloric acid HClO4, Perchloric acid Weak Acids and their Ionization Constants Weak acids are weakly ionized. HA H+ + A K = [H+][A]/[HA] = small number a Ionization of a Weak Acid HA H+ + A- A Strong Acid (Very Large K ) A Weak Acid a (Small K ) a Conjugate Acid Base Pair HA H+ + A Acid Conjugate Base reaction) Base Conjugate Acid (A reaction with water: hydrolysis A + H2O HA + OH Acid-Base Equilibrium, Conjugate acid/base pairs HA + B HA : H3O+ HAc H2O A : H2O Ac OH acid base acid BH + A + base B : OH + NH3 H2O Reactions of acids and bases with water Recall:H3O+ = H+ HA + H2O acid base H3O + A + acid base a K= acid base [H+][A] [HA] H2O + A 2 acid AH + OH base [AH][OH] K= [A] b Autoionization of H2O H2O + H + OH + acid base + + Kw=[H ][OH ] 14 Kw = [H ][OH ] = 1.0 x 10 at 25C [H ] = [OH ] = 1.0 x 10 M 7 The relationship between Ka, Kb and Kw is ... HA a H +A + H2O +A HA +OH K= [H+][A] [HA] a X [HA][OH] K= [A] b K = [H+][OH] b = Kw K Ka = 1x1014 /Kb o r Kb = 1x1014 /Ka Ka = Kw/Kb or b = Kw/Ka Kb refers to the reaction of a base with Water (i.e. a hydrolysis reaction). Kw = KaKb = 1 x 1014 so... A + H2O HA + OH Introduction to pH H2O + H + OH + acid base + + Kw=[H ][OH ] 14 7 Kw = [H ][OH ] = 1.0 x 10 at 25C [H ] = [OH ] = 1.0 x 10 M + Useful Equations [H+][OH] = 1 x 1014 pH = log[H+] pOH = log[OH] pH + pOH = 14 The pH Scale Basic pH = log[H+] > 7.0 1.0 M NaOH 14 7 0 Neutral pH = log[H+] = 7.0 Pure Water Acidic pH = log[H+] < 7.0 1.0 M HCl Question What is the pH of a 0.001 M HCl solution? Answer For all strong monoprotic acids, the molar concentration of the H+, [H+], is equal to the molar concentration of the parent acid; stoichiometry! pH = Log[H+] = Log(0.001) = (3) pH = 3 1 Question If 1.00 mL of 10 M HCl is diluted with water to a final volume of 100.0 mL, what is the pH of the final solution? HCl(aq) H+(aq) + Cl(aq) 100 % ionized Answer Using: M1V1 = M2V2 Here, M2 = M1(V1/V2) = (10 M)(1.00/100.0) = 0.10 M pH = log(0.10) = log(101) = 1.0 Question If 10.00 mL of 1.0 M NaOH is diluted with water to a final volume of 100.0 mL, what is the pH of the final solution? a) 1 b) 13 c) 7 Answer NaOH is a strong base: NaOH Na+ + OH 100 % dissociation Diluted [OH]: M1V1 = M2V2 = (1.0 M)(10.00 mL/100.0 mL) = 0.10 M pOH = log[OH] = log(0.10) = log(101) = 1.0 We know that pH + pOH = 14; so pH = 14 pOH = 14 1 = 13 Question Which of the following aqueous solutions has the highest pH? 12 1) 103M NaOH 2) 10 M HCl 3) 10 M HCl Answer 103M NaOH (NaOH is a strong base) pOH = 3, so the pH = 11 [OH] = 103M 10 M HCl 12 6 [H ] = 10 + + 6 12 + pH = 6 7 10 M HCl [H ] = 10 ? but water has [H ] = 10 12 pH = 7 The pH of Weak Acids HA H+ + A Weak acids are weakly ionized. So, [H+] = [HA] Here, we need to use acid dissociation constants, Ka. We will also need to use the IRE table. Question What is the pH of a 0.15 M solution of a weak acid, HA, with a Ka of 1.8 x 105? HA a) 1.8 b) 2.8 H+ + A c) 3.5 Answer I R E HA 0.15 x 0.15 x x H+ 0 +x + 0 +x A x Ka = [H+][A]/[HA] = 1.8 x 105 a) 1.8 b) 2.8 c) 3.5 Answer ( a closer look), Cont. Ka = [H+][A]/[HA] = 1.8 x 105 Using the quadratic equation, we solve for x; (x)(x)/(0.15 x) = 1.8 x 105 x = [H+] = 0.00165 pH = log[H+] = log(0.00165) = 2.8 Question Now, using the 5% rule, find the pH of a 0.15 M solution of a weak acid, HA, with a Ka of 1.8 x 105? HA a) 1.8 b) 2.8 H+ + A c) 3.5 Answer I R E HA 0.15 x 0.15 x x H+ 0 +x + 0 +x A x Ka = [H+][A]/[HA] = 1.8 x 105 (x)(x)/(0.15 x) = 1.8 x 105 Answer ( 5 % rule), Cont. Using the 5% rule, we assume x<< 0.15; so now we have, (x)(x)/(0.15 x) = 1.8 x 105 x2/0.15 = 1.8 x 105 ; x = 0.00164 Check: (0.00164/0.15) x 100 = 1.1 % 1.1 % < 5 %, so assumption is valid. pH = log[H+] = log(0.00164) = 2.8 % Dissociation HAc H+ + Ac % Dissociation = [amount dissociated] x 100 [initial con.] Question What is the pH of a solution of 0.15 M HAc, which is only 1.1 % ionized? a) 0.82 b) 2.8 c) 2.3 Answer HAc H+ + Ac % Dissociated = [amount dissociated] x 100 [initial con.] (x/0.15) x 100 = 1.1% X = [HAc dissociated], which is [H+] = 0.00165 M pH = log[H+] = log(0.00165) = 2.8 Question What (if any) is the difference between the strength of an acid (or base) and the concentration of an acid (or base)? Answer Strength = extent of ionization HA H+ + A. Stronger acid = larger Ka Concentration = how much HA is present higher molarity = more concentrated Question Consider solutions of a weak acid HA. If the initial concentration of HA is increased, what happens to the percent dissociation of the acid at constant temperature? A) increase B) decrease C) stay the same What happens to the pH of the solution? A) increase B) decrease C) stay the same Answer HA H+ + A K = [H+][A]/[HA] (weak acid) Let initial conc. = [HA]0 Let fraction dissociated = x Equilibrium concs: [HA] = [HA]0(1x) [HA]0 [H+] = [A] = [HA]0x Ka = [H+][A]/[HA] At equ, Ka = ([HA]0x)2/[HA]0 = [HA]0x2 So, if we increase [HA]0, x must decrease % dissociation decreases when initial HA concentration increases Question What happens to the pH of the solution if the initial [HA] is increased? A) increase B) decrease C) stay the same Answer HA H+ + A K = [H+][A]/[HA] Increasing [HA] will increase [H+] (and [A]) so as to keep K constant at a given temperature. pH decreases when initial HA concentration is increased Weak Acid/Base Pairs (HAc/Ac-) HAc + H2O acid base acid H3O + Ac + base a acid H2O + Ac base acid HAc + OH base K = 1.8x105 pK = 4.7 K = 5.6x1010 pK = 9.3 b a Weak Base/Acid Pairs (NH3/NH4+) NH3 + H2O base acid base OH + NH + 4 acid K = 1.8x105 pK = logK = 4.7 b H2O + NH base acid + 4 NH3 + H3O + base acid b a b a K = 5.6x1010 a Conjugate Acid/Base Pair "A Buffers Solution" HAc+H2O a H3O +Ac + H2O +Ac HAc +OH K= [H3O+][Ac] [HAc] a X [HAc][OH] K= [Ac] b K = [H3O+][OH] b = Kw K Conjugate Base/Acid Pair "A Buffers Solution" NH3+H2O NH +OH H2O +NH4+ + 4 NH3 +H3O + [NH4+][OH] K= [NH3] b X a K= [H3O+][NH3] [NH4+] K a = [H3O+][OH] = Kw K b Question Which of the following 0.1 M solutions has the highest pH ? 1) NaAc 2) NH3 3) NH4Cl Answer Basic NH + H O H2O + Ac Acidic H2O + NH4+ 3 2 OH + NH HAc + OH 4+ K = 1.8x10 Kb = 5.6x1010 b 5 NH3 + H3O+ Ka = 5.6x1010 Larger Kb means Larger [OH ] Produced 1) NaAc 2) NH3 3) NH4Cl Question Does the hydroysis of Na2CO3 produce an acidic, neutral or basic solution? Na2CO3 + H2O NaHCO3 + NaOH Hint: Ka = 5.6 x 1011 for NaHCO3 , so... Kb = 1 x 1014/5.6 x 1011 = 1.8 x 104 Molecular equation: Na2CO3 + H2O NaHCO3 + NaOH Answer Total ionic equation: 2Na+ + CO32 + H2O 2Na+ + HCO3 + OH Net ionic equation: CO32 + H2O HCO3 + OH Question HX and HY are two different weak acids. HX has a larger Ka than HY. You have aqueous solutions of the two sodium salts, NaX and NaY, at equal concentrations. Which solution has the higher pH? Answer X + H2O HX + OH Y + H2O HY + OH K < K since K K Question Cl is the conjugate base of the acid HCl. Are solutions containing chloride ion (such as NaCl) basic? If not, why? Answer Molecular Eq.: Cl + H2O HCl + OH Total ionic Eq.: Cl + H2O H+ + Cl + OH Net ionic Eq.: H2O H+ + OH Kw = [H+][OH] = 1x1014. [H+] = 1x107 ; so, pH = log[H+] = log(1x107) = 7 NaCl dissolves in water to give a pH = 7.0 Question What is the pH of a 0.10 M solution of NH4Cl? Kb for NH3 is 1.8 x 105. Hint NH4+ is the conjugate acid of NH3. The reaction involved is: NH4+(aq) NH3(aq) + H+(aq) for which Ka = Kw/Kb Answer Ka = Kw/Kb = 1 x 1014/1.8 x 105 = 5.6 x 1010 NH4+ NH3 + H+ I R E 0.10 x 0.10 x 0 0 + x + x + x + x Ka = [NH3][H+]/[NH4+] = (x)(x)/(0.10x) x2/(0.10 x) = 5.6 x 1010 Answer, Cont. x2/(0.10 x) = 5.6 x 1010 Assuming x << 0.10, we have: x2/0.10 = 5.6 x 1010 x2 = 5.6 x 1011 x = 7.5 x 106; Check assumption with 5% rule: (7.5 x 106/ 0.10) x 100 = 0.0075% 0.0075% < 5%, assumption is valid. Answer, Cont. x = [H+] = 7.5 x 106 Therefore, the pH is: pH = log[H+] = log(7.5 x 106) pH = 5.1 Polyprotic acids Have more than one acidic proton Example: phosphoric acid H3PO4(aq) H+(aq) + H PO (aq) Ka1 = 7.5 x 103 2 4 H PO (aq) H+(aq) + HPO (aq) 2 4 K = 6.2 x 108 42 Polyprotic Acids Another example: carbonic acid, H2CO3 (formed by reaction of CO2 with H2O) H2CO3 H+ + HCO pKa1 = 6.1 3 HCO H+ + CO32 pKa2 = 10.2 trend for all weak acids: pKa1 < pKa2 < pKa3 Ka1 > Ka2 > Ka3 3 Question What is the pH of a 0.010 M H2SO4 solution? (Hint: K = Very Large while K = 0.012) a) 1.8 b) 1.7 c) 2.4 Answer The Ka1 for H2SO4 is VERY large H2SO4 H+ + HSO4 I 0.01 0 R 0.01 +0.01 Epseudo 0 0.01 0 +0.01 0.01 Answer, Cont. HSO4 Ka2 = 0.012 HSO4 H+ + SO42 I 0.01 0.01 0 R x +x +x E 0.01x 0.01+x x Ka2 = [H+][SO42]/[HSO4] Ka2 = (0.01+x)(x)/(0.01x) = 0.012 Note: 0.01 comes from 1 ionization Answer, Cont. Ka2 = (0.01+x)(x)/(0.01x) = 0.012 The Ka2 > 105, 5% rule won't work; using the quadratic eq. to solve for x, we get x = 0.0045; [H+] = (0.010+0.0045) = 0.0145 pH = log[H+] = log(0.0145) = 1.8 Question For phosphoric acid, H3PO4 H+ + H2PO4 Ka1 = 7.5 x 103 H2PO4 H+ + HPO42 Ka2 = 6.2 x 108 HPO42 H+ + PO43 Ka3 = 4.8 x 1013 If you add 0.1 mole of H3PO4 to 1.0 L of water, what will be the major phosphoruscontaining species at equilibrium? Answer H3PO4 H+ + H2PO4 Ka1 = 7.5 x 103 Ka1 is much less than 1, so the first equilibrium will lie on the side of the reactant (H3PO4). The other two reactions have even smaller Ka's and will be negligible. The dominant species is H3PO4 (and the solution is acidic). Question Which carbonate species is present in the highest concentration at pH 7? H2CO3 H+ + HCO3 pKa1 = 6.1 HCO3 H+ + CO32 pKa2 = 10.2 Ka1 = 7.9x107 1) H2CO3 Ka2 = 6.3x1011 2 2) HCO3 3) CO3 Answer H2CO3 H+ + HCO3 pKa1 = 6.1 [HCO3][H ]/[H2CO3] = Ka1 = 7.9 x 107 + at pH 7, [H ] = 10 , so [HCO3]/[H2CO3] = 7.9 + 7 HCO3 H+ + CO32 pKa2 = 10.2 [CO32][H ]/[HCO3] = Ka2 = 6.3 x 1011 + The Relationships between [H+], pH, Ka and pKa [H ] + Ka+ pH = log[H ] pK = logK Question Find the pH of an aqueous solution containing 0.010 M each of HCl, H2SO4, and HCN H2SO4: Ka1 = large, Ka2 = 1.2 x 102 HCN: Ka = 6.2 x 1010 Answer 0.010M H2SO4 0.010M H+ + 0.010M HSO4 [H ] = 0.020 M 0.010 M HCl 0.010 M H+ HCN Ka = 6.2 x 1010; so, [H+] is negligible. HSO4 ionization: HSO4 H+ + SO42 I 0.01 0.02 0 R x +x +x E 0.01 x 0.02 +x x Ka2 = [H+][SO42]/[HSO4] = 1.2 x 102 Answer, cont. [H+][SO42]/[HSO4] = Ka2 (0.020 + x)x/(0.010 x) = 0.012 0.020x + x2 = 0.00012 0.012x x2 + 0.032x 0.00012 = 0 x = 0.0034 (must use the quadratic eq.) so total [H ] = 0.020 + x = 0.0234 M + pH = 1.63 Question Citric acid is a triprotic acid: Ka1 = 8.4 x 104 Ka2 = 1.8 x 105 Ka3 = 4.0 x 106 What is the pH of a 0.15 M solution of citric acid? Answer We start with all H3A, 1st reaction is Ka1 = 8.4 x 10 H3A I R E 0.15 x 0.15x 4 H2A 0 +x + H+ 0 +x x x x2/(0.15 x) = 8.4 x 10 Answer, Cont. Second acid dissociation, equil. concs.: H2A HA2 + H+ I 0.011 0 0.011 R x +x +x E 0.011x x 0.011 + x Ka2 = x(0.011+x)/(0.011x) = 1.8 x105 x<< 0.011; x = 1.8 x 10 ; passes 5% rule. [HA2] is much smaller than [H+]. We can neglect the third acid 5 Question Which of the following solution(s) will be neutral (pH = 7)? A) 0.2 M NH4Cl B) 0.2 M HCl C) 0.2 M NaOH D) 0.2 M NH3 E) 0.2 M NaNO3 Answer Which of the following solution(s) will be neutral (pH = 7)? 4+ A) 0.2 M NH4Cl (NH is a weak acid) B) 0.2 M HCl (strong acid) C) 0.2 M NaOH (basic) D) 0.2 M NH3 (basic) E) 0.2 M NaNO (neutral) ...
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