CEE 680 Lecture #10 2/5/2020 Print version Updated: 5 February 2020 Lecture #10 Acids & Bases: Analytical Solutions with simplifying assumptions IV (Stumm & Morgan, Chapt.3 ) (Benjamin, Chapt. 4) David Reckhow CEE 680 #10 1 Bases: Sodium Acetate Example 10 -3 M in 1 L 1. List all species present Five total H + , OH ‐ , HAc, Ac ‐ , Na + 2. List all independent equations equilibria 1 K a = [H + ][Ac ‐ ]/[HAc] = 10 ‐ 4.77 K w = [H + ][OH ‐ ] = 10 ‐ 14 2 mass balances 5 3 C Na = [Na + ] = 10 -3 C = [HAc]+[Ac ‐ ] = 10 ‐ 3 proton balance: (proton rich species) = (proton poor species) Ac - H 2 O [HAc] + [H + ] = [OH ‐ ] 4 David Reckhow CEE 680 #8 2 1
CEE 680 Lecture #10 2/5/2020 5 C Na = [Na + ] = 10 -3 NaAc Example (cont.) K w = [H + ][OH - ] 2 3. Combine equations and solve for H + [OH - ] = K w /[H + ] 4 [H + ] = [OH ‐ ] ‐ [HAc] 2+4 [H + ] = K W / [H + ] ‐ [HAc] [H + ] = K W / [H + ] ‐ [H + ]C/{K a +[H + ]} 1+2+3+4 [H + ] 2 = K W ‐ C[H + ] 2 /{K a +[H + ]} 3 C = [HAc]+[Ac - ] K a [H + ] 2 + [H + ] 3 = K W K a + K w [H + ] ‐ C[H + ] 2 [Ac - ] = C-[HAc] [H + ] 3 + {C+ K a }[H + ] 2 ‐ K w [H + ] ‐ K W K a = 0 1 K a = [H + ][Ac - ]/[HAc] 4. Solve for other species K a = [H + ]{C-[HAc]} /[HAc] 1+3 K a [HAc]= [H + ]C-[H + ][HAc] [HAc] = C[H + ]/{K a +[H + ]} David Reckhow CEE 680 #8 3 Answer [OH ‐ ] = 7.67 10 ‐ 7 pOH = 6.115 pH = 7.885 David Reckhow CEE 680 #8 4 2
CEE 680 Lecture #10 2/5/2020 In ‐ class Practice 10 ‐ 4 M Sodium Acetate UMass: Alvin & Cielo, Chris UNISA: Sikelelwa 10 ‐ 3 M Sodium Cyanide UMass: Ian &JQ UNISA: For Oxalic acid, 10 ‐ 3 M Calcium Oxalate H 2 Ox HOx - Ox -2 pKa1 =1.25 UMass: pKa2 =4.27 UNISA: Alfred 10 ‐ 4 M Sodium Bicarbonate UMass: Laura, Bridgette, Isaac UNISA: David Reckhow CEE 680 #9 5 Note: Alkali metals such as Na and K do not act as acids, so they are ignored here Writing PBE’s HAc Monoprotic Acid H 2 O same as ENE [H + ] = [OH ‐ ] + [Ac ‐ ] H 2 CO 3 Diprotic Acid H 2 O same as ENE [H + ] = [OH ‐ ] + [HCO 3 ‐ ] + 2[CO 3 ‐ 2 ] HCO 3 - Diprotic: Ampholyte H 2 O Not ENE [H 2 CO 3 ] + [H + ] = [OH ‐ ] + [CO 3 ‐ 2 ] e.g., NaHCO 3 CO 3 -2 Diprotic Base H 2 O Not ENE ‐ ] + [H + ] = [OH ‐ ] 2[H 2 CO 3 ] + [HCO 3 David Reckhow CEE 680 #10 6 3
CEE 680 Lecture #10 2/5/2020 Guide to Simplified Acid/Base Solutions #1 Neutral ] [ H K w if C<10 ‐ 8 Acid Addition, C>10 ‐ 6.5 Acidic 2 K K 4 K C a a a [ H ] if K a C > 10 ‐ 13 2 ] [ H C Strong Acid 2 C C 4 K w [ H ] if K a > 10C 2 ] [ H K C Weak Acid a ] [ H K C K if C > 100K a a w David Reckhow CEE 680 #10 7 Guide to Simplified Acid/Base Solutions #2 Base Addition , C>10 ‐ 6.5 Basic K K 2 4 K C if K b C > 10 ‐ 13 b b b [ OH ] 2 Strong Base ] [ OH C 2 C C 4 K if K b > 10C w [ OH ] 2 Weak Base ] [ OH K C b ] if C > 100K b [ OH K C K b w Very Dilute Systems if 10 ‐ 8 < C < 10 ‐ 6.5 try strong acid/base or weak acid/base assumption otherwise may need to use general solution David Reckhow CEE 680 #10 8 4
CEE 680 Lecture #10 2/5/2020 Exact Solutions: Summary Monoprotic Acids: [H + ] 3 + K a [H + ] 2 ‐ {K w + K a C}[H + ] ‐ K W K a = 0 Bases: [H + ] 3 + {C+K a }[H + ] 2 ‐ K w [H + ] ‐ K W K a = 0 Diprotic Acids: [H + ] 4 + K 1 [H + ] 3 + {K 1 K 2 ‐ K w ‐ K 1 C}[H + ] 2 ‐ K 1 {2CK 2 +K w }[H + ] ‐ K W K 1 K 2 = 0 Ampholytes: [H + ] 4 + {C+K 1 }[H + ] 3 + {K 1 K 2 ‐ K w }[H + ] 2 ‐ K 1 {CK 2 +K w }[H + ] ‐ K W K 1 K 2 = 0 Bases: [H + ] 4 + {2C+K 1 }[H + ] 3 + {CK 1 +K 1 K 2 ‐ K w }[H + ] 2 ‐ K 1 K w [H + ] ‐ K W K 1 K 2 = 0 David Reckhow CEE 680 #10 9 To next lecture David Reckhow CEE 680 #10 10 5
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