Chem_2_Ch_7

Chem_2_Ch_7 - Chapter 7 Acids and Base s Acids and Bases...

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Unformatted text preview: Chapter 7 Acids and Base s Acids and Bases Reviewed Arrhe Acid: Produce H+ nius s Arrhe Base Produce OHnius : s Bronste d-Lowry Acid: H+ donor Bronste d-Lowry Base H+ acce : ptor Strong Acids and their Ionization Constants S trong acids are100%ionize d. HC H+ + Cll a K = [H+][Cl-]/[HC = Ve largenum r l] ry be Strong Acids HC Hydrochloric acid l, HBr, Hydrobrom acid ic HNO3, Nitric acid H2S 4, S O ulfuric acid HC 3, C lO hloric acid HC 4, Pe lO rchloric acid Weak Acids and their Ionization Constants We acids arewe ak akly ionize d. HA H+ + AK = [H+][A-]/[HA] = sm num r all be a Ionization of a Weak Acid HA H+ + A- AS trong Acid (Ve LargeK ) ry A We Acid a ak (S all K ) m a Conjugate Acid Base Pair HA H+ + A - Acid C onjugateBase re action) Base C onjugateAcid A-(A reaction water: hydrolysis+ H2O with HA + OH Acid-Base Equilibrium, Conjugate acid/base pairs acid base acid + base - HA + B BH + A HA : H3O+ HAc H2O A : H2O Ac OH - B : OH + - NH3 H2O Reactions of acids and bases with water acid base acid + Re call:H3O+ = H+ base a HA + H2O H3O + A [H+][A-] [HA] K= acid base - acid base - H2O + A 2 AH + OH [AH][OH-] K= [A-] b Autoionization of H2O acid base - H2O + H + OH + + Kw=[H ][OH ] -14 Kw = [H ][OH ] = 1.0 x 10 at 25C - + [H ] = [OH ]- = 1.0 x 10 M -7 There lationship be e Ka, Kb and twe n Kw is ... HA a H +A [H+][A-] [HA] + - H2O +A- HA +OH - X [HA][OH-] K= [A-] b K= a K = [H+][OH-] b = Kw K Kw = KaKb = 1 x 10-14 so... Ka = 1x10-14 /Kb or Kbb = Kw/Ka Ka = Kw/Kb or K = 1x10-14 /Ka Kb re rs to there fe action of a basewith Wate (i.e a hydrolysis re r . action). A- + H2O HA + OH- Introduction to pH acid base - H2O + H + OH + + Kw=[H ][OH ] -14 Kw = [H ][OH ] = 1.0 x 10 at -25C [H ] = [OH ]- = 1.0 x 10 M -7 + + Useful Equations [H+][OH-] = 1 x 10-14 pH = - log[H+] pOH = -log[OH-] pH + pOH = 14 The pH Scale Basic pH = -log[H+] > 7.0 1.0 M NaOH 14 Ne utral pH = -log[H+] = 7.0 PureWate r 7 Acidic pH = -log[H+] < 7.0 1.0 M HC l 0 Question What is thepH of a 0.001 M HCl solution? Answer For all strong m onoprotic acids, the m conce olar ntration of theH+, [H+], is e qual to them conce olar ntration of thepare acid; stoichiom try! nt e pH = -Log[H+] = -Log(0.001) = -(-3) pH = 3 1 Question I f 1.00 m of 10 M HCl is dilute with wate to a L d r final volum of 100.0 m what is thepH of the e L, final solution? HC l(aq) H+(aq) + Cl-(aq) 100 %ionize d Answer Using: M1V1 = M2V2 He , M2 = M1(V1/V2) = (10 M)(1.00/100.0) re = 0.10 M pH = -log(0.10) = -log(10-1) = 1.0 Question I f 10.00 m of 1.0 M NaOH is dilute with wate L d r to a final volum of 100.0 m what is thepH of the e L, final solution? a) 1 b) 13 c) 7 Answer NaOH is a strong base : NaOH Na+ + OH- 100 %dissociation Dilute [OH-]: M1V1 = M2V2 d = (1.0 M)(10.00 m L/100.0 m = 0.10 M L) pOH = -log[OH-] = -log(0.10) = -log(10-1) = 1.0 Weknow that pH + pOH = 14; so pH = 14 pOH = 14 1 = 13 Question Which of thefollowing aque solutions ous has thehighe pH? st -12 1) 10-3M NaOH 2) 10 M HCl 3) 10 M HCl Answer 10-3M NaOH (NaOH is a strong base ) pOH = 3, so thepH = 11 [OH-] = 10-3M 10 M HC l -12 -6 [H ] = 10 + + -6 -12 + pH = 6 -7 10 M HC l [H ] = 10 ? but wate has [H ] = 10 r -12 pH = 7 The pH of Weak Acids HA H+ + AWe acids arewe ak akly ionize d. S [H+] = [HA] o, He , wene d to useacid dissociation re e constants, Ka. Wewill also ne d to usetheI RE table e . Question What is the pH of a 0.15 M solution of a we acid, HA, with a Ka of 1.8 x 10-5? ak HA a) 1.8 b) 2.8 H+ + Ac) 3.5 Answer HA I 0.15 R -x E 0.15 x x H+ 0 +x x + 0 +x A- Ka = [H+][A-]/[HA] = 1.8 x 10-5 a)1.8 b) 2.8 c) 3.5 Answer ( a closer look), Cont. Ka = [H+][A-]/[HA] = 1.8 x 10-5 (x)(x)/(0.15 x) = 1.8 x 10-5 Using thequadratic e quation, wesolvefor x; x = [H+] = 0.00165 pH = -log[H+] = -log(0.00165) = 2.8 Question Now, using the5%rule find the pH of a , 0.15 M solution of a we acid, HA, with a ak Ka of 1.8 x 10-5? HA a) 1.8 b) 2.8 H+ + Ac) 3.5 Answer HA I 0.15 R -x E 0.15 x x H+ 0 +x x + 0 +x A- Ka = [H+][A-]/[HA] = 1.8 x 10-5 (x)(x)/(0.15 x) = 1.8 x 10-5 Answer ( 5 % rule), Cont. (x)(x)/(0.15 x) = 1.8 x 10-5 Using the5%rule weassum x<< 0.15; so now we , e have , x2/0.15 = 1.8 x 10-5 ; x = 0.00164 Che (0.00164/0.15) x 100 = 1.1 % ck: 1.1 %< 5 % so assum , ption is valid. pH = -log[H+] = -log(0.00164) = 2.8 % Dissociation HAc H+ + Ac%Dissociation = [am ount dissociate x 100 d] [initial con.] Question What is thepH of a solution of 0.15 M HAc, which is only 1.1 % ionize d? a)0.82 b) 2.8 c) 2.3 Answer HAc H+ + Ac%Dissociate = [am d ount dissociate x 100 d] [initial con.] (x/0.15) x 100 = 1.1% X = [HAc dissociate which is [H+] = 0.00165 M d], pH = -log[H+] = -log(0.00165) = 2.8 Question What (if any) is thediffe ncebe e thestre re twe n ngth of an acid (or base and theconce ) ntration of an acid (or base )? Answer S ngth = e nt of ionization tre xte HA H+ + A-. S tronge acid = large Ka r r C ntration = how m HA is pre nt once uch se highe m r olarity = m conce ore ntrate d Question Conside solutions of a we acid HA. If theinitial r ak conce ntration of HA is incre d, what happe to the ase ns pe nt dissociation of theacid at constant te pe rce m rature ? A) incre ase B) de ase C) stay thesam cre e What happe to thepH of thesolution? ns A) incre ase B) de ase C) stay thesam cre e Answer HA H+ + AK = [H+][A-]/[HA] (we acid) ak Le initial conc. = [HA] 0 t Le fraction dissociate = x t d Equilibriumconcs: [HA] = [HA] 0(1-x) [HA] 0 [H+] = [A-] = [HA] 0x Ka = [H+][A-]/[HA] At e Ka = ([HA] 0x)2/[HA] 0 = [HA] 0x2 qu, S if weincre [HA] 0, x m de ase o, ase ust cre %dissociation de ase whe initial HA conce cre s n ntration incre s ase Question What happe to thepH of the ns solution if theinitial [HA] is incre d? ase A) incre ase B) de ase C) stay thesam cre e Answer HA H+ + AK = [H+][A-]/[HA] I ncre asing [HA] will incre [H+] (and [A-]) so as to ase ke p K constant at a give te pe e n m rature . pH de ase whe initial HA conce cre s n ntration is incre d ase Weak Acid/Base Pairs (HAc/Ac-) acid base acid + base a a HAc + H2O H3O + Ac acid base acid base - K = 1.8x10-5 pK = 4.7 K = 5.6x10-10 pK = 9.3 b H2O + Ac- HAc + OH Weak Base/Acid Pairs (NH3/NH4+) base acid base - acid + 4 NH3 + H2O OH + NH K = 1.8x10-5 pK = -logK = 4.7 b base acid + 4 base acid + b a b H2O + NH NH3 + H3O a K = 5.6x10-10 a Conjugate Acid/Base Pair "A Buffers Solution" HAc+H2O a H3O +Ac + - H2O +Ac- HAc +OH - [H3O+][Ac-] [HAc] X [HAc][OH-] K= [Ac-] b K= a K = [H3O+][OH-] b = Kw K Conjugate Base/Acid Pair "A Buffers Solution" NH3+H2O NH +OH + 4 - H2O +NH4+ NH3 +H3O + [NH4+][OH-] K= [NH3] b X a [H3O+][NH3] [NH4+] K= K a = [H3O+][OH-] = Kw K b Question Which of thefollowing 0.1 M solutions has thehighe pH ? st 1) NaAc 2) NH3 3) NH4C l Answer Basic NH + H O H2O + AcAcidic H2O + NH4+ 3 2 OH + NH HAc + OH4+ K = 1.8x10 Kb = 5.6x10-10 b -5 NH3 + H3O+ - Ka = 5.6x10-10 Large Kb m ans Large [OH ] Produce r e r d 1) NaAc 2) NH3 3) NH4C l Question Doe thehydroysis of Na2C 3 produce s O an acidic, ne utral or basic solution? Na2C 3 + H2O NaHC 3 + NaOH O O Hint: Ka = 5.6 x 10-11 for NaHC 3 , so... O Kb = 1 x 10-14/5.6 x 10-11 = 1.8 x 10-4 Answer Mole cular e quation: Na2C 3 + H2O NaHC 3 + NaOH O O Total ionic e quation: 2Na+ + C 3-2 + H2O 2Na+ + HC 3- + OHO O Ne ionic e t quation: C 3-2 + H2O HC 3- + OH O O - Question HX and HY aretwo diffe nt we acids. re ak HX has a large Ka than HY. r You haveaque solutions of thetwo sodiumsalts, ous NaX and NaY, at e qual conce ntrations. Which solution has thehighe pH? r Answer X- + H2O HX + OHY- + H2O HY + OHK < K sinceK K Question C - is theconjugatebaseof theacid HCl. Are l solutions containing chlorideion (such as NaCl) basic? If not, why? Answer Mole cular Eq.: Cl- + H2O HCl + OHTotal ionic Eq.: C - + H2O H+ + Cl- + OHl Ne ionic Eq.: t H2O H+ + OHKw = [H+][OH-] = 1x10-14. [H+] = 1x10-7 ; so, pH = -log[H+] = -log(1x10-7) = 7 NaC dissolve in wate to givea pH = 7.0 l s r Question What is thepH of a 0.10 M solution of NH4Cl? Kb for NH3 is 1.8 x 10-5. Hint NH4+ is theconjugateacid of NH3. There action involve is: d NH4+(aq) NH3(aq) + H+(aq) for which Ka = Kw/Kb Answer Ka = Kw/Kb = 1 x 10-14/1.8 x 10-5 = 5.6 x 10-10 NH4+ I R E 0.10 -x 0.10 x NH3 + H+ 0 0 +x +x +x +x Ka = [NH3][H+]/[NH4+] = (x)(x)/(0.10-x) x2/(0.10 x) = 5.6 x 10-10 Answer, Cont. x2/(0.10 x) = 5.6 x 10-10 Assum x << 0.10, wehave ing : x2/0.10 = 5.6 x 10-10 x2 = 5.6 x 10-11 x = 7.5 x 10-6; C ck assum he ption with 5%rule : (7.5 x 10-6/ 0.10) x 100 = 0.0075% 0.0075%< 5% assum , ption is valid. Answer, Cont. x = [H+] = 7.5 x 10-6 The fore thepH is: re , pH = -log[H+] = -log(7.5 x 10-6) pH = 5.1 Polyprotic acids Havem than oneacidic proton ore Exam : phosphoric acid ple H3PO4(aq) H+(aq) + H PO (aq) 2 4- Ka1 = 7.5 x 10-3 H PO (aq) H+(aq) + HPO (aq) 2 442- K = 6.2 x 10-8 Polyprotic Acids Anothe e ple carbonic acid, H2CO3 r xam : (form d by re e action of CO2 with H2O) H2C 3 H+ + HCO O 3- pKa1 = 6.1 HC H+ + CO32O pKa2 = 10.2 tre for all we acids: pKa1 < pKa2 < pKa3 nd ak 3- Ka1 > Ka2 > Ka3 Question What is thepH of a 0.010 M H2S 4 O solution? (Hint: K = Ve LargewhileK = 0.012) ry a)1.8 b) 1.7 c) 2.4 Answer The Ka1 for H2S 4 is VERY large O H2S 4 H+ + HS 4O O I 0.01 R -0.01 Epseudo 0 0 +0.01 0.01 0 +0.01 0.01 Answer, Cont. HS 4- Ka2 = 0.012 O HS 4- O H+ + S 4-2 O I 0.01 0.01 0 R -x +x +x E 0.01-x 0.01+x x Ka2 = [H+][S 4-2]/[HS 4-] O O Ka2 = (0.01+x)(x)/(0.01-x) = 0.012 Note 0.01 com s from1 ionization : e Answer, Cont. Ka2 = (0.01+x)(x)/(0.01-x) = 0.012 TheKa2 > 10-5, 5%rulewon't work; using thequadratic e to solvefor x, wege q. t x = 0.0045; [H+] = (0.010+0.0045) = 0.0145 pH = -log[H+] = -log(0.0145) = 1.8 Question For phosphoric acid, H3PO4 H+ + H2PO4H2PO4- H+ + HPO42HPO42- H+ + PO43Ka1 = 7.5 x 10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13 I f you add 0.1 m of H3PO4 to 1.0 L of wate what ole r, will bethem phosphorus-containing spe s at ajor cie e quilibrium ? Answer H3PO4 H+ + H2PO4Ka1 = 7.5 x 10-3 Ka1 is m le than 1, so thefirst e uch ss quilibriumwill lieon thesideof there actant (H3PO4). Theothe two re r actions havee n sm r Ka's and will bene ve alle gligible The . dom inant spe s is H3PO4 (and thesolution is acidic). cie Question Which carbonatespe s is pre nt in thehighe cie se st conce ntration at pH 7? H2C 3 H+ + HC 3O O HC 3- H+ + C 32O O pKa1 = 6.1 Ka1 = 7.9x10-7 pKa2 = 10.2 1) H2CO3 Ka2 = 6.3x10-11 -2 2) HC 3 O 3) C 3 O Answer H2C 3 H+ + HC 3O O pKa1 = 6.1 [HC 3-] [H ] /[H2C 3] = Ka1 = 7.9 x 10-7 O O + at pH 7, [H ] = 10 , so [HCO3-]/[H2CO3] = 7.9 + -7 HCO3- H+ + C 32O pKa2 = 10.2 [CO32-] [H ] /[HC 3-] = Ka2 = 6.3 x 10-11 O + The Relationships between [H+], pH, Ka and pKa [H ] + Ka + pH = -log[H ] pK = -logK Question Find thepH of an aque solution containing 0.010 ous M e of HC H2S 4, and HC ach l, O N H2S 4: Ka1 = large Ka2 = 1.2 x 10-2 O , HC Ka = 6.2 x 10-10 N: Answer 0.010M H2S 4 0.010M H+ + 0.010M HS 4O O [H ] = 0.020 M 0.010 M HC 0.010 M H+ l HC Ka = 6.2 x 10-10; so, [H+] is ne N gligible . HS 4- ionization: O HS 4O I 0.01 R x E 0.01 x H+ + S 42O 0.02 0 +x +x 0.02 +x x Ka2 = [H+][S 4-2]/[HS 4-] = 1.2 x 10-2 O O Answer, cont. [H+][S 42-]/[HS 4-] = Ka2 O O (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 (m usethequadratic e ust q.) so total [H ] = 0.020 + x = 0.0234 M + pH = 1.63 Question C acid is a triprotic acid: itric Ka1 = 8.4 x 10-4 Ka2 = 1.8 x 10-5 Ka3 = 4.0 x 10-6 What is thepH of a 0.15 M solution of citric acid? Answer Westart with all H3A, 1st re action is Ka1 = 8.4 x 10 -4 H3A I R E 0.15 -x 0.15-x H2A0 +x x + H+ 0 +x x x2/(0.15 - x) = 8.4 x 10 Answer, Cont. S cond acid dissociation, e e quil. concs.: H2A- HA-2 + H+ I 0.011 0 0.011 R -x +x +x E 0.011-x x 0.011 + x Ka2 = x(0.011+x)/(0.011-x) = 1.8 x10-5 x<< 0.011; x = 1.8 x 10 ; passe 5%rule s . -5 [HA2-] is m sm r than [H+]. Wecan ne ct thethird acid uch alle gle Question Which of thefollowing solution(s) will bene utral (pH = 7)? A) 0.2 M NH4C l B) 0.2 M HC l C 0.2 M NaOH ) D) 0.2 M NH3 E) 0.2 M NaNO3 Answer Which of thefollowing solution(s) will bene utral (pH = 7)? 4+ A) 0.2 M NH4C (NH is a we acid) l ak B) 0.2 M HC (strong acid) l C 0.2 M NaOH (basic) ) D) 0.2 M NH3 (basic) E) 0.2 M NaNO (ne utral) ...
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