Question What is the pH of a 10 3 M solution of HCl? A. 7 B. 3 C. 0 - PDF document

CEE 680 Lecture #9 2/3/2020 Print version Updated: 3 February 2020 Lecture #8 Acids & Bases: Analytical Solutions with simplifying assumptions II (Stumm & Morgan, Chapt.3 ) (Benjamin, Chapt. 3) David Reckhow CEE 680 #8 1 Question  What is the pH of a 10 ‐ 3 M solution of HCl? A. 7 B. 3 C. 0 D. 9 E. Impossible to tell None of the above F. David Reckhow CEE 680 #8 2 1

CEE 680 Lecture #9 2/3/2020 Hydrochloric Acid Example: 10 ‐ 3 M  1. List all species present Four total  H + , OH ‐ , HCl, Cl ‐  2. List all independent equations  equilibria 1  K a = [H + ][Cl ‐ ]/[HCl] = 10 +3 2  K w = [H + ][OH ‐ ] = 10 ‐ 14  mass balances 3  C = [HCl]+[Cl ‐ ] = 10 ‐ 3  proton balance:  (proton rich species) =  (proton poor species) HCl H 2 O  [H + ] = [OH ‐ ] + [Cl ‐ ] 4 David Reckhow CEE 680 #8 3 HCl: Exact Solution [H + ] 3 + K a [H + ] 2 - { K w + K a C} [H + ] - K W K a = 0  Exact solution: pH = 3.0000004  [H + ] = 1.00 x 10 ‐ 3 [OH - ] = K w /[H + ]  [OH ‐ ] = 1.00 x 10 ‐ 11 [Cl - ]=K a C/{K a +[H + ]}  [Cl ‐ ] = 1.00 x 10 ‐ 3 [HCl] = C-[Cl - ]  [HCl] = 1.00 x 10 ‐ 11 David Reckhow CEE 680 #8 4 2

CEE 680 Lecture #9 2/3/2020 HCl example (cont.)  Can we simplify? H + + ] ] 3 3 H + + ] ] 2 2 - H + + ] H + + ] [H + K K a [H - K K w [H ] - -K K a C[ [H ] - - K K W K a a = 0 0 [ + a [ w [ a C W K = 1.000E-9 0.001000 1.000E-17 0.001000 1.000E-11 0  What about the PBE?  And the MBE too? [H + ] = [OH ‐ ] + [Cl ‐ ]  C = [HCl] + [Cl ‐ ]  ~0 ~0 David Reckhow CEE 680 #8 5 Simplified HCl Example K w = [H + ][OH - ] 2  3. Use simplified PBE & MBE [OH - ] = K w /[H + ] 4  [H + ] = [OH ‐ ] + [Cl ‐ ]  [H + ]  [Cl ‐ ] Assumes [H + ]>>[OH - ] C  [HCl]+[Cl - ] 3 C  [Cl - ] [Cl - ]  C  [H + ] = C 3+4 Assumes [HCl]<<[Cl - ]  [H + ] = C 1 K a = [H + ][Cl - ]/[HCl]  4. Solve for other species K a = [H + ] C / [HCl] 1+3 [HCl] = [H + ] C /K a David Reckhow CEE 680 #8 6 3

CEE 680 Lecture #9 2/3/2020 Assumptions  Use both & Compare answers  Exact: pH = 3.0000004  Simplified: pH = 3.0000000  Use simplified equation, and check assumptions!  [OH ‐ ] << [H + ]  1.00 x 10 ‐ 11 << 1.00 x 10 ‐ 3 yes!  [Cl ‐ ] >> [HCl]  1.00 x 10 ‐ 3 >> 1.00 x 10 ‐ 11 yes! David Reckhow CEE 680 #8 7 Simplified HCl Example for low C K w = [H + ][OH - ] 2  3. Use simplified PBE & MBE [OH - ] = K w /[H + ] 4  [H + ] = [OH ‐ ] + [Cl ‐ ] C= [HCl]+[Cl - ] 3 C  [Cl - ]  [H + ] = K W / [H + ] + [Cl ‐ ] [Cl - ]  C 2+3+4  [H + ] = K W / [H + ] + C  [H + ] 2 ‐ C[H + ] ‐ K w = 0 Assumes [HCl]<<[Cl - ]  [H + ] = {C  (C 2 + 4K w ) 0.5 }/2 1 K a = [H + ][Cl - ]/[HCl]  4. Solve for other species K a = [H + ] C / [HCl] 1+3 [HCl] = [H + ] C /K a David Reckhow CEE 680 #8 8 4

CEE 680 Lecture #9 2/3/2020 Calculation for 2nd HCl example  For a 10 ‐ 7 solution of HCl  pH = 6.79  2  C C 4 K   w [ H ] 2    7  14  14  Check Assumptions 10 10 4 x 10   [Cl ‐ ]  C = 10 ‐ 7 2  [HCl] = [H + ] C /K a  1 5   7 10  =10 ‐ 6.79 10 ‐ 7 /10 +3 =10 ‐ 16.79 2  [Cl ‐ ]>>[HCl], yes!   7 1 . 62 x 10 David Reckhow CEE 680 #8 9 Hypochlorous Acid Example  1. List all species present Four total  H + , OH ‐ , HOCl, OCl ‐  2. List all independent equations  equilibria 1  K a = [H + ][OCl ‐ ]/[HOCl] = 10 ‐ 7.6  K w = [H + ][OH ‐ ] = 10 ‐ 14 2  mass balances 3  C = [HOCl]+[OCl ‐ ] = 10 ‐ 6  proton balance:  (proton rich species) =  (proton poor species) HOCl H 2 O  [H + ] = [OH ‐ ] + [OCl ‐ ] 4 David Reckhow CEE 680 #8 10 5

CEE 680 Lecture #9 2/3/2020 HOCl Example for low C K w = [H + ][OH - ] 2  3. Combine equations and solve for H + [OH - ] = K w /[H + ] 4  [H + ] = [OH ‐ ] + [OCl ‐ ] 2+4  [H + ] = K W / [H + ] + [OCl ‐ ]  [H + ] = K W / [H + ] + K a C/[H + ] 1+2+3+4  [H + ] 2 = K W + K a C 3 C = [HOCl]+[OCl - ] [HOCl]  C  [H + ] = (K a C + K W ) 0.5 Assumes [HOCl]>>[OCl - ] 1 K a = [H + ][OCl - ]/[HOCl] K a = [H + ][OCl - ]/ C  4. Solve for other species 1+3 [OCl - ]=K a C/[H + ] David Reckhow CEE 680 #8 11 Calculation for HOCl example  For a 10 ‐ 6 solution of HOCl    [ H ] K C K  pH = 6.73 a w     7 . 6 6  14 10 10 10   14 3 . 51 x 10  Check Assumptions   7  [HOCl]  C = 10 ‐ 6 1 . 87 x 10  [OCl ‐ ]=K a C/[H + ]  =10 ‐ 7.6 10 ‐ 6 /10 ‐ 6.73 =10 ‐ 6.87  [HOCl]>>[OCl ‐ ], OK David Reckhow CEE 680 #8 12 6

CEE 680 Lecture #9 2/3/2020 In ‐ class Practice  10 ‐ 4 M Hydrofluoric Acid  JQ & Ian  Godfrey  10 ‐ 2 M Phenol  Cielo, Alvin, Chris  Hezron  10 ‐ 3 M Carbonic Acid  Laura, Isaac, Bridgette  Naeldi  10 ‐ 4 M Sulfuric Acid  Niall David Reckhow CEE 680 #9 13 NAME EQUILIBRIA pKa HClO4 = H+ + ClO4- Perchloric acid -7 STRONG HCl = H+ + Cl- Hydrochloric acid -3 H2SO4= H+ + HSO4- Sulfuric acid -3 (&2) ACIDS HNO3 = H+ + NO3- Nitric acid -0 H3O+ = H+ + H2O Hydronium ion 0 CCl3COOH = H+ + CCl3COO- 0.70 Trichloroacetic acid HIO3 = H+ + IO3- Iodic acid 0.8 CHCl2COOH = H+ + CHCl2COO- 1.48 Dichloroacetic acid HSO4- = H+ + SO4-2 2 Bisulfate ion H3PO4 = H+ + H2PO4- Phosphoric acid 2.15 (&7.2,12.3) Fe(H2O)6+ 3 = H+ + Fe(OH)(H2O)5+ 2 2.2 (&4.6) Ferric ion CH2ClCOOH = H+ + CH2ClCOO- 2.85 Chloroacetic acid C6H4(COOH)2 = H+ + C6H4(COOH)COO- 2.89 (&5.51) o-Phthalic acid C3H5O(COOH)3= H+ + C3H5O(COOH)2COO- 3.14 (&4.77,6.4) Citric acid HF = H+ + F- 3.2 Hydrofluoric acid HCOOH = H + + HCOO- 3.75 Formic Acid C2H6N(COOH)2= H+ + C2H6N(COOH)COO- 3.86 (&9.82) Aspartic acid C6H4(OH)COOH = H+ + C6H4(OH)COO- 4.06 (&9.92) m-Hydroxybenzoic acid C2H4(COOH)2 = H+ + C2H4(COOH)COO- 4.16 (&5.61) Succinic acid C6H4(OH)COOH = H+ + C6H4(OH)COO- 4.48 (&9.32) p-Hydroxybenzoic acid HNO2 = H+ + NO2- Nitrous acid 4.5 FeOH(H2O)5+ 2 + H+ + Fe(OH)2(H2O)4+ Ferric Monohydroxide 4.6 CH3COOH = H+ + CH3COO- 4.75 Acetic acid Al(H2O)6+ 3 = H+ + Al(OH)(H2O)5+ 2 4.8 Aluminum ion David Reckhow CEE 680 #9 14 7

CEE 680 Lecture #9 2/3/2020 NAME FORMULA pKa C2H5COOH = H+ + C2H5COO- 4.87 Propionic acid H2CO3 = H+ + HCO3- Carbonic acid 6.35 (&10.33) H2S = H+ + HS- Hydrogen sulfide 7.02 (&13.9) H2PO4- = H+ + HPO4-2 7.2 Dihydrogen phosphate HOCl = H+ + OCl- 7.5 Hypochlorous acid Cu(H2O)6+ 2 = H+ + CuOH(H2O)5+ Copper ion 8.0 Zn(H2O)6+ 2 = H+ + ZnOH(H2O)5+ Zinc ion 8.96 B(OH)3 + H2O = H+ + B(OH)4- Boric acid 9.2 (&12.7,13.8) NH4+ = H+ + NH3 Ammonium ion 9.24 HCN = H+ + CN- 9.3 Hydrocyanic acid C6H4(OH)COO- = H+ + C6H4(O)COO-2 9.32 p-Hydroxybenzoic acid H4SiO4 = H+ + H3SiO4- Orthosilicic acid 9.86 (&13.1) C6H5OH = H+ + C6H5O- 9.9 Phenol C6H4(OH)COO- = H+ + C6H4(O)COO-2 9.92 m-Hydroxybenzoic acid Cd(H2O)6+ 2 = H+ + CdOH(H2O)5+ Cadmium ion 10.2 HCO3- = H+ + CO3-2 Bicarbonate ion 10.33 Mg(H2O)6+ 2 = H+ + MgOH(H2O)5+ Magnesium ion 11.4 HPO4-2 = H+ + PO4-3 Monohydrogen phosphate 12.3 Ca(H2O)6+ 2 = H+ + CaOH(H2O)5+ Calcium ion 12.5 H3SiO4- = H+ + H2SiO4-2 Trihydrogen silicate 12.6 HS- = H+ + S-2 Bisulfide ion 13.9 H2O = H+ + OH- Water 14.00 NH3 = H+ + NH2- Ammonia 23 OH- = H+ + O-2 Hydroxide 24 CH4 = H+ + CH3- Methane 34 David Reckhow CEE 680 #9 15  To next lecture DAR David Reckhow CEE 680 #8 16 8