Chem_2_Ch_7 co

Chem_2_Ch_7 co - Chapter 7 Aci ds and Bases Acids and Bases...

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Unformatted text preview: Chapter 7 Aci ds and Bases Acids and Bases Reviewed Ar r heni us Aci d: Pr oduces H + Ar r heni us Base: Pr oduces OH Br onsted-L owr y Aci d: H + donor Br onsted-L owr y Base: H + acceptor Strong Acids and their Ionization Constants Str ong aci ds ar e 100% i oni zed. H Cl H + + Cl a K = [H +][Cl -]/ [H Cl ] = Ver y l ar ge Strong Acids H Cl , H ydr ochl or i c aci d H Br , H ydr obr omi c aci d H NO3, Ni tr i c aci d H 2SO4, Sul fur i c aci d H Cl O3, Chl or i c aci d H Cl O4, Per chl or i c aci d Weak Acids and their Ionization Constants Weak aci ds ar e weakl y i oni zed. H A H + + AK = [H +][A -]/ [H A] = smal l number a Ionization of a Weak Acid HA H+ + A- A Str ong Aci d (Ver y L ar ge K ) A Weak Aci d (Smal l K ) a a Conjugate Acid Base Pair HA H+ + A - Aci d Conjugate Base (A r H 2O water + A- +eacti on wi th H A : hydr ol ysi s OH r eacti on) Base Conjugate Aci d Acid-Base Equilibrium, Conjugate acid/base pairs aci d base aci d + base - HA + B BH + A H A : H 3O+ H Ac H 2O A : H 2O Ac OH - B : OH + - NH 3 H 2O Reactions of acids and bases with water aci d base aci d + Recal l :H 3O+ = H + a base - H A + H 2O H 3O + A [H +][A -] [H A] K = aci d base - aci d base - H 2O + A 2 AH + OH [AH ][OH -] K= [A -] b Autoionization of H2O aci d base - H 2O + H + OH + + K w =[ H ][ OH ] -14 K = [ H ][ OH ] = 1.0 x 10 [ H ] = [ OH ] =- 1.0 x 10 M -7 + w at 25C - The r el ati onshi p between K a, K b and K w i s ... HA a H +A + - H 2O +A - H A +OH - [H +][A -] [H A] X [H A][OH -] K = [A -] b K = a K = [H +][OH -] = K w b K K w = K aK b = 1 x 10-14 so... K a =K a = -14 /w /bKor orb = 1x10K wK K a 1x10 K K b K K b = -14 / / a K b r efer s to the r eacti on of a base wi th Water (i .e. a hydr ol ysi s r eacti on). A - + H 2O H A + OH - Introduction to pH aci d base - H 2O + H + OH + + K w =[ H ][ OH ] -14 K = [ H ][ OH ] = 1.0 x 10 [ H ] = [ OH ] =- 1.0 x 10 M -7 + w + at 25C Useful Equations [H +][OH -] = 1 x 10-14 pH = - l og[H +] pOH = -l og[OH -] pH + pOH = 14 The pH Scale Basi c pH = -l og[H +] > 7.0 1.0 M NaOH 14 Neutr al pH = -l og[H +] = 7.0 Pur e Water 7 Aci di c pH = -l og[H +] < 7.0 1.0 M H Cl 0 Question What i s the pH of a 0.001 M H Cl sol uti on? Answer For al l str ong monopr oti c aci ds, the mol ar concentr ati on of the H +, [H +], i s equal to the mol ar concentr ati on of the par ent aci d; stoi chi ometr y! pH = -L og[H +] = -L og(0.001) = -(-3) pH = 3 1 Question I f 1.00 mL of 10 M H Cl i s di l uted wi th water to a fi nal vol ume of 100.0 mL , what i s the pH of the fi nal sol uti on? H Cl (aq) H +(aq) + Cl -(aq) 100 % i oni zed Answer Usi ng: M 1V 1 = M 2V 2 H er e, M 2 = M 1(V 1/ V 2) = (10 M )(1.00/ 100.0) = 0.10 M pH = -l og(0.10) = -l og(10-1) = 1.0 Question I f 10.00 mL of 1.0 M NaOH i s di l uted wi th water to a fi nal vol ume of 100.0 mL , what i s the pH of the fi nal sol uti on? a) 1 b) 13 c) 7 Answer NaOH i s a str ong base: NaOH Na+ + OH - 100 % di ssoci ati on Di l uted [OH -]: M 1V 1 = M 2V 2 = (1.0 M )(10.00 mL / 100.0 mL ) = 0.10 M pOH = -l og[OH -] = -l og(0.10) = -l og(10-1) = 1.0 We know that pH + pOH = 14; so pH = 14 pOH = 14 1 = 13 Question Whi ch of the fol l owi ng aqueous sol uti ons has the hi ghest pH ? -12 1) 10-3M NaOH 2) 10 M H Cl 3) 10 M H Cl Answer 10-3M NaOH (NaOH i s a str ong base) pOH = 3, so the pH = 11 [OH -] = 10-3M 10 M H Cl -12 -6 [H ] = 10 + + -6 -12 + pH = 6 -7 10 M H Cl [H ] = 10 but water has [H ? ] = 10 -12 pH = 7 The pH of Weak Acids HA H+ + AWeak aci ds ar e weakl y i oni zed. So, [H +] = [H A] H er e, we need to use aci d di ssoci ati on constants, K a. We wi l l al so need to use the I RE tabl e. Question What i s the pH of a 0.15 M sol uti on of a weak aci d, H A, wi th a K a of 1.8 x 10-5? HA a) 1.8 b) 2.8 H+ + c) A3.5 Answer I R E HA 0.15 -x 0.15 x H+ 0 +x x + 0 +x x A- K a = [H +][A -]/ [H A] = 1.8 x 10-5 a) 1.8 b) 2.8 c) 3.5 Answer ( a closer look), Cont. Ka = [H +][A-]/ [H A] = 1.8 x 10-5 (x)(x)/ (0.15 x) = 1.8 x 10-5 Usi ng the quadr ati c equati on, we sol ve for x; x = [H +] = 0.00165 pH = -l og[H +] = -l og(0.00165) = 2.8 Question Now, usi ng the 5% r ul e, fi nd the pH of a 0.15 M sol uti on of a weak aci d, H A, wi th a K a of 1.8 x 10-5? HA a) 1.8 b) 2.8 H+ + c) A3.5 Answer I R E HA 0.15 -x 0.15 x H+ 0 +x x + 0 +x x A- K a = [H +][A -]/ [H A] = 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 Usi ng the 5% r ul e, we assume x<< 0.15; so now we have, x2/ 0.15 = 1.8 x 10-5 ; x = 0.00164 Check: (0.00164/ 0.15) x 100 = 1.1 % 1.1 % < 5 %, so assumpti on i s val i d. pH = -l og[H +] = -l og(0.00164) = 2.8 % Dissociation H Ac H + + Ac% Di ssoci ati on = [ amount di ssoci ated] x 100 [i ni ti al con.] Question What i s the pH of a sol uti on of 0.15 M H Ac, whi ch i s onl y 1.1 % i oni zed? a) 0.82 b) 2.8 c) 2.3 Answer H Ac H + + Ac% Di ssoci ated = [ amount di ssoci ated] x 100 [i ni ti al con.] (x/ 0.15) x 100 = 1.1% X = [H Ac di ssoci ated], whi ch i s [H +] = 0.00165 M pH = -l og[H +] = -l og(0.00165) = 2.8 Question What (i f any) i s the di ffer ence between the str ength of an aci d (or base) and the concentr ati on of an aci d (or base)? Answer Str ength = extent of i oni zati on H A H + + A -. Str onger aci d = l ar ger Ka Concentr ati on = how much H A i s pr esent hi gher mol ar i ty = mor e concentr ated Question Consi der sol uti ons of a weak aci d H A. I f the i ni ti al concentr ati on of H A i s i ncr eased, what happens to the per cent di ssoci ati on of the aci d at constant temper atur e? A) i ncr ease B) decr ease C) stay the same What happens to the pH of the sol uti on? A) i ncr ease B) decr ease C) stay the same Answer H A H + + AK = [H +][A -]/ [H A] (weak aci d) L et i ni ti al conc. = [H A] 0 L et fr acti on di ssoci ated = x Equi l i br i um concs: [H A] = [H A] 0(1-x) [H A] 0 [H +] = [A -] = [H A] 0x K a = [H +][A -]/ [H A] At equ, K a = ([H A] 0x)2/ [H A] 0 = [H A] 0x2 So, i f we i ncr ease [H A] 0, x must decr ease % di ssoci ati on decr eases when i ni ti al H A concentr ati on i ncr eases Question What happens to the pH of the sol uti on i f the i ni ti al [H A] i s i ncr eased? A) i ncr ease B) decr ease C) stay the same Answer H A H + + AK = [H +][A -]/ [H A] I ncr easi ng [H A] wi l l i ncr ease [H +] (and [A -]) so as to keep K constant at a gi ven temper atur e. pH decr eases when i ni ti al H A concentr ati on i s i ncr eased Weak Acid/Base Pairs (HAc/Ac-) aci d base aci d + base a a H Ac + H 2O H 3O + Ac aci d base aci d base - K = 1.8x10-5 pK = 4.7 H 2O + Ac- H Ac + OH K = 5.6x10-10 pK = 9.3 b Weak Base/Acid Pairs (NH3/NH4+) base aci d base - aci d + 4 NH 3 + H 2O OH + NH K = 1.8x10-5 pK = -l ogK = 4.7 b base aci d + 4 base aci d b+ a b H 2O + NH NH 3 + H 3O a K = 5.6x10-10 a Conjugate Acid/Base Pair "A Buffers Solution" H Ac+H 2O a H 3O +Ac + - H 2O +Ac- H Ac +OH - [H 3O+][Ac-] [H Ac] X [H Ac][OH -] K = [Ac-] b K = a K = [H 3O+][OH -] = K w b K Conjugate Base/Acid Pair "A Buffers Solution" NH 3+H 2O + 4 - H 2O +NH 4+ NH 3 +H 3O+ NH +OH [NH 4+][OH -] K = [NH 3] b X a [H 3O+][NH 3] [NH 4+] K = K K b a = [H 3O+][OH -] = K w Question Whi ch of the fol l owi ng 0.1 M sol uti ons has the hi ghest pH ? 1) NaAc 2) NH 3 3) NH 4Cl Answer Basi c NH + H O H 2O + AcAci di c H 2O + NH 4+ 3 2 OH + NH K = 1.8x10 H Ac + OH - K b = 5.6x10-10 4+ b -5 NH 3 + H 3O+ K a = 5.6x10-10 - L ar ger K b means L ar ger [OH ] Pr oduced 1) NaAc 2) NH 3 3) NH 4Cl Question Does the hydr oysi s of Na2CO3 pr oduce an aci di c, neutr al or basi c sol uti on? Na2CO3 + H 2O NaH CO3 + NaOH H i nt: K a = 5.6 x 10-11 for NaH CO3 , so... K b = 1 x 10-14/ 5.6 x 10-11 = 1.8 x 10-4 Answer M ol ecul ar equati on: Na2CO3 + H 2O NaH CO3 + NaOH Total i oni c equati on: 2Na+ + CO3-2 + H 2O 2Na+ + H CO3- + OH Net i oni c equati on: CO3-2 + H 2O H CO3- + OH - Question H X and H Y ar e two di ffer ent weak aci ds. H X has a l ar ger K a than H Y. You have aqueous sol uti ons of the two sodi um sal ts, NaX and NaY, at equal concentr ati ons. Whi ch sol uti on has the hi gher pH ? Answer X - + H 2O H X + OH Y- + H 2O H Y + OH K < K si nce K K Question Cl - i s the conjugate base of the aci d H Cl . Ar e sol uti ons contai ni ng chl or i de i on (such as NaCl ) basi c? I f not, why? Answer M ol ecul ar Eq.: Cl - + H 2O H Cl + OH Total i oni c Eq.: Cl - + H 2O H + + Cl - + OH Net i oni c Eq.: H 2O H + + OH K w = [H +][OH -] = 1x10-14. [H +] = 1x10-7 ; so, pH = -l og[H +] = -l og(1x10-7) = 7 NaCl di ssol ves i n water to gi ve a pH = 7.0 Question What i s the pH of a 0.10 M sol uti on of NH 4Cl ? K b for NH 3 i s 1.8 x 10-5. Hint NH 4+ i s the conjugate aci d of NH 3. The r eacti on i nvol ved i s: NH 4+(aq) NH 3(aq) + H +(aq) for whi ch K a = K w / K b Answer K a = K w / K b = 1 x 10-14/ 1.8 x 10-5 = 5.6 x 10-10 NH 4+ I R E 0.10 -x 0.10 x NH 3 + H + 0 +x +x 0 +x +x K a = [NH 3][H +]/ [NH 4+] = (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 Assumi ng x << 0.10, we have: x2/ 0.10 = 5.6 x 10-10 x2 = 5.6 x 10-11 x = 7.5 x 10-6; Check assumpti on wi th 5% r ul e: (7.5 x 10-6/ 0.10) x 100 = 0.0075% 0.0075% < 5%, assumpti on i s val i d. Answer, Cont. x = [H +] = 7.5 x 10-6 Ther efor e, the pH i s: pH = -l og[H +] = -l og(7.5 x 10-6) pH = 5.1 Polyprotic acids H ave mor e than one aci di c pr oton Exampl e: phosphor i c aci d H 3PO4(aq) H +(aq) + H PO (aq) 2 4- K a1 = 7.5 x 10-3 H PO (aq) H +(aq) + H PO (aq) 2 442- K = 6.2 x 10-8 Polyprotic Acids Another exampl e: car boni c aci d, H 2CO3 (for med by r eacti on of CO2 wi th H 2O) H 2CO3 H + + H CO 3- pK a1 = 6.1 H CO H + + CO32pK a2 = 10.2 tr end for al l weak aci ds: pK a1 < pK a2 < pK a3 3- K a1 > K a2 > K a3 Question What i s the pH of a 0.010 M H 2SO4 sol uti on? (H i nt: K = Ver y L ar ge whi l e K = 0.012) a) 1.8 b) 1.7 c) 2.4 Answer The K a1 for H 2SO4 i s VERY l ar ge H 2SO4 H + + H SO4I 0.01 R -0.01 E pseudo 0 0 +0.01 0.01 0 +0.01 0.01 Answer, Cont. H SO4- K a2 = 0.012 H SO4- H + + SO4-2 I 0.01 0.01 0 R -x +x +x E 0.01-x 0.01+x x K a2 = [H +][SO4-2]/ [H SO4-] K a2 = (0.01+x)(x)/ (0.01-x) = 0.012 Note: 0.01 comes fr om 1 i oni zati on Answer, Cont. K a2 = (0.01+x)(x)/ (0.01-x) = 0.012 The K a2 > 10-5, 5% r ul e won't wor k ; usi ng the quadr ati c eq. to sol ve for x, we get x = 0.0045; [H +] = (0.010+0.0045) = 0.0145 pH = -l og[H +] = -l og(0.0145) = 1.8 Question For phosphor i c aci d, H 3PO4 H + + H 2PO4H 2PO4- H + + H PO42H PO42- H + + PO43K a1 = 7.5 x 10-3 K a2 = 6.2 x 10-8 K a3 = 4.8 x 10-13 I f you add 0.1 mol e of H 3PO4 to 1.0 L of water , what wi l l be the major phosphor uscontai ni ng speci es at equi l i br i um? Answer H 3PO4 H + + H 2PO4K a1 = 7.5 x 10-3 K a1 i s much l ess than 1, so the fi r st equi l i br i um wi l l l i e on the si de of the r eactant (H 3PO4). The other two r eacti ons have even smal l er K a's and wi l l be negl i gi bl e. The domi nant speci es i s H 3PO4 (and the sol uti on i s aci di c). Question Whi ch car bonate speci es i s pr esent i n the hi ghest concentr ati on at pH 7? H 2CO3 H + + H CO3H CO3- H + + CO32pK a1 = 6.1 K a1 = 7.9x10-7 pK a2 = 10.2 1) H 2CO3 K a2 = 6.3x10-11 -2 2) H CO3 3) CO3 Answer H 2CO3 H + + H CO3pK a1 = 6.1 [H CO3-] [H ] / [H 2CO3] = K a1 = 7.9 x 10-7 + at pH 7, [H ] = 10 , so [H CO3-]/ [H 2CO3] = 7.9 + -7 H CO3- H + + CO32- pK a2 = 10.2 [CO32-] [H ] / [H CO3-] = K a2 = 6.3 x 10-11 + The Relationships between [H+], pH, Ka and pKa [H ] + K a+ pH = -l og[H ] pK = -l ogK Question Fi nd the pH of an aqueous sol uti on contai ni ng 0.010 M each of H Cl , H 2SO4, and H CN H 2SO4: K a1 = l ar ge, K a2 = 1.2 x 10-2 H CN: K a = 6.2 x 10-10 Answer 0.010M H 2SO4 0.010M H + + 0.010M H SO40.010 M H Cl 0.010 M H + [H ] = 0.020 M H CN K a = 6.2 x 10-10; so, [H +] i s negl i gi bl e. H SO4- i oni zati on: H SO4 H+ + SO42I 0.01 0.02 0 R x +x +x E 0.01 x 0.02 +x x K a2 = [H +][SO4-2]/ [H SO4-] = 1.2 x 10-2 Answer, cont. [H +][SO42-]/ [H SO4-] = K a2 (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 quadr ati c eq.) so total [H ] = 0.020 + x = 0.0234 M + pH = 1.63 Question Ci tr i c aci d i s a tr i pr oti c aci d: K a1 = 8.4 x 10-4 K a2 = 1.8 x 10-5 K a3 = 4.0 x 10-6 What i s the pH of a 0.15 M sol uti on of ci tr i c aci d? Answer We star t wi th al l H 3A, 1st r eacti on i s K a1 = 8.4 x 10 -4 H 3A I R E 0.15 -x 0.15-x H 2A 0 +x x + H+ 0 +x x x2/ (0.15 - x) = 8.4 x 10 Answer, Cont. Second aci d di ssoci ati on, equi l . concs.: H 2A - H A -2 + H+ I 0.011 0 0.011 R -x +x +x E 0.011-x x 0.011 + x K a2 = x(0.011+x)/ (0.011-x) = 1.8 x10-5 x<< 0.011; x = 1.8 x 10 ; passes 5% r ul e. -5 [H A 2-] i s much smal l er than [H +]. We can negl ect Question Whi ch of the fol l owi ng sol uti on(s) wi l l be neutr al (pH = 7)? A) B) C) D) E) 0.2 M NH 4Cl 0.2 M H Cl 0.2 M NaOH 0.2 M NH 3 0.2 M NaNO3 Answer Whi ch of the fol l owi ng sol uti on(s) wi l l be neutr al (pH = 7)? 4+ A) B) C) D) E) 0.2 M NH 4Cl (NH i s a weak aci d) 0.2 M H Cl (str ong aci d) 0.2 M NaOH (basi c) 0.2 M NH 3 (basi c) 0.2 M NaNO (neutr al ) ...
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This note was uploaded on 04/27/2009 for the course CHEM chem1A taught by Professor Hooker during the Spring '09 term at UCSB.

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