Ch 13 Acids and Bases

Transcription

1 Ch 13 Acids and Bases 13-1 BrФnsted-Lowry acid-base model 13- The ion product of water [H ][OH ] = K w 13-3 ph and poh ph =log[h ] 13-4 Weak acid and their equilibrium constants 13-5 Weak base and their equilibrium constants 13-6 Acid-base properties of salt solutions 13-1 BrФnsted-Lowry acid-base model Acid: a proton (H ion) donor Base: a proton acceptor conjugated acid-base (aq) (aq) (aq) (aq) HB A HA B acid 1 base acid base 1 conjugated acid-base E 13-1: a) HNO 之 conjugated base? F 之 conjugated acid? b) HCO 3 之 conjugated base 及之 conjugated acid? a) NO ; HF b) CO 3 ; H CO 3 conjugated base conjugated acid 13- The ion product of water K = [H ][OH ] w 5 o C K w = o C K w = 13 0 o C K w = 1 at 5 o C [H ] > acidic [H ] < basic - 1 -

2 13-3 ph and poh 3 ph= log[h ] =log[h O] poh =log[oh ] 5 o C, ph < 7.0 acidic ph = 7.0 neutral ph > 7.0 basic 5 o C, ph poh =14 E 13-: 5 o C, a) [H ] and ph of a tap-water sample [OH ] =.0 7 b) [H ] and [OH ] of a human body ph = 7.4 c) poh of a solution which [H ] = 5.0[OH ] a) 5 o C b) c) 14 [H ][OH ] = [H ].0 = [H ] = 5.0 ph= log[h ] = 7.3 ph for strong acids and strong bases Strong acids: HCl, HBr, HI completely ionized HClO 4, HNO 3, H SO 4 Strong bases: LiOH, NaOH, KOH completely ionized Ca(OH), Sr(OH), Ba(OH) 8 ph = 7.4 [H ] = [OH ] = = [H ] = 5.0[OH ] [OH ][OH ] = 1.0 [OH ] = poh = log[oh ] =

3 MM = 171.3g/mole E 13-3: ph value of 1.00g Ba(OH) dissolving in water to form 1L solution. Ba(OH) HO Ba OH 13-4 Weak acids and their equilibrium constants weak acid HB OH (g) () l (aq) (aq) 1 nba(oh) = = MBa(OH) = = 5.84 M 1 3 M = 5.84 = 1.17 M poh= log[oh ] = 1.93 ph = = 1.07 (aq) () l 3 (aq) (aq) HB HO H O B 少於 1% 解離弱酸多為下列二類之一 1. Molecules containing an ionizable hydrogen atom. e: HNO (aq) H O H 3 O (aq) NO (aq). 陽離子 cations e: NH 4 NH 4 (aq) H O H 3 O (aq) NH 3 0.1M NH 4 Cl (aq) ph = 5 The equilibrium constant for a weak acid 簡化 (aq) 3 (aq) HB HO H O(aq) B (aq) (aq) HB H (aq) B [H ][B ] K a= Ka:acid equilibrium constant [HB] HNO K a = pk a = 3. Ka 酸性 HCN K a = 5.8 pk a = 9.4 pk a = logka - 3 -

4 E 13-4: Acetic acid HC H 3 O (CH 3 COOH) and hydrated zinc cation Zn(H O) 4 四水合鋅離子 a) 解離方程式? b) Which is the stronger acid? c) What is the pka of Zn(H O) 4? a)hc HO HO H O CHO 3 (aq) 3 (aq) 3 (aq) 4(aq) 3 (aq) 3 (aq) Zn(H O) HO Zn(H O) (OH) H O 5 b)hcho 3 Ka = 1.8 Table 13- Zn(HO) 4 Ka = 3.3 HC 較大 HO 3 is the stronger acid Q Ka c)pka =logka = log3.3 = 9.48 E 13-5: Aspirin: HC 9 H 7 O 4 ph =.60. Calculate K a? ; 3.60g dissolve in water per liter MHC 9HO = =.0 M ph =.60 [H ] =.5 M 9 7 4(aq) (aq) 9 7 4(aq) HC HO H CHO starting 0.000M equilibrium = M 3 [H ][C9HO 7 4] (.5 ) a [HC 9HO 7 4] 1.75 [H ] eq K = = = 3.6 percent ionization = 0 [HB] o 3.5 E 3-5: % ionization = 0= 13% 4

5 E 13-6: What is the percent ionization of the acid? % ionization = 1/5 0 = 0% Calculate of [H ] in a water solution of a weak acid. 由 K a 求 [H ] E 13-7: Nicotinic acid, HC 6 H 4 O N (HNic)(Ka = ); 0.1mole of nicotinic acid dissolve in water to foam 1.0L solution, 求 [H ] =? HNic H Nic starting 0.1 (aq) (aq) (aq) equilibrium 0.1 [H ][Nic ] Ka = = 1.4 [HNic] = Ka a 0.05 (5%) a a a 才成立 E 13-8: Calculate [H ] in a 0.0M HNO, Ka = = = = 1.4 = (aq) (aq) (aq) HNO H NO starting 0.0 equilibrium 0.0 [H ][NO ] Ka = = 6.0 [HNO ] 0.0 =

6 若 0.0 ; 0 5 = 6.0 = 不妥 >5% 7.7 a = 0.1 = = 7.75 > 0.05 a 0.1 解法一 : 二次方程式法 : = 二次方程式公式 : b± b 4ac = a 4 5 (6.0 ) 6.0 = ± (6.0 ) 46.0 = 3 3 = 7.4 M or 8.0 M or 解法二 : The method of successive approimations 連續近似值法 : = [HNO ] = = = 0.09 = 5.5 = Polyprotic weak acids 多質子弱酸 3 3 4(aq) (aq) (aq) a1 = HPO H HPO K (aq) (aq) (aq) a = HPO H HPO K (aq) (aq) (aq) a3 = HPO H PO K 4.5 K > K > K a1 a a3-6 -

7 E 13-9: 0.00M H CO 3(aq) 之 ph? Table (aq) (aq) 3(aq) a1 = HCO H HCO K (aq) (aq) 3(aq) a = HCO H CO K 4.7 3(aq) (aq) 3(aq) HCO H HCO starting 0.00 equilibrium Ka1 = = = 4.4 = 1 5 ph= 4.68 第二階段可以忽略 K a 較小 影響小 Weak bases and their equilibrium constants 弱鹼分兩類 1. molecules; e: NH 3. Anion; 為弱酸之陰離子 : 3(aq) () l 4(aq) (aq) NH HO NH OH (aq) () l (aq) (aq) F HO HF OH E 13-: 水解方程式 a) NO b) Na CO 3 c) KHCO 3 (aq) (aq) (aq) a)no HO HNO OH 3(aq) 3(aq) (aq) b)co HO HCO OH 3(aq) 3(aq) (aq) c)hco HO HCO OH - 7 -

8 B (aq) HO HB(aq) OH(aq) K pk b [HB][OH ] = [B ] =logk b b 3 b 5 b 3 b 5 b NH K = 1.8 pk = 4.74 CHO K = 5.6 pk = 9.5 Calculation of [OH ] in water solution of a weak base E 13-11: ph of 0.M NaF K b for F = F HF OH satrting 0. equilibrium Kb = = = [OH ] = = 1. M poh = 5.9 ph= 14 poh = Relation between K a and K b (aq) (aq) (aq) a HB H B K (aq) (aq) (aq) b )B HO HB OH K (aq) (aq) = a b = w HO H OH K K K K HNO NO 4 11 K K = = o C K =

9 BrФnsted-Lowry acid: a) strong acid: HClO 4, H SO 4 are stronger proton donors than the H 3 O ion b) weak acid: HF, HI are weaker proton donors than the H 3 O ion c) species, such as C H 5 OH; which are weak proton donor than H O molecule. BrФnsted-Lowry base: a) strong base (H, ), are stronger than proton acceptor than OH ion. b) weak base (F, ),are weaker than proton acceptor than OH ion. c) anions of stronger acid (ClO 4, NO 3, ) weaker proton acceptor than H O. strong acid: stronger base: 4(aq) () l 3 (aq) 4(aq) HClO HO H O ClO (aq) () l 3 (aq) (aq) HCl HO H O F 3(aq) () l 3 (aq) 3(aq) HNO HO H O NO (aq) () l (aq) (aq) H HO OH H (aq) () l (aq) (aq) O HO OH OH 5 (aq) () l (aq) 5 (aq) CHO HO OH CHO H, O, C H 5 O 在水溶液中不存在 13-6 Acid-base properties of salt solutions Table 13-5 Acid-base properties of ions in water solution. Anion Cation spectator Cl, Br, I ClO 4, NO 3 Basic C H 3 O, CO 3 F, PO 4 3 many others Acidic Li, Na, K NH 4, Al 3, Mg Ca, Sr, Ba transition metal ions - 9 -

10 E 13-1: Consider water solution NH 4 I, Zn(NO 3 ), KClO 4, Na 3 PO 4 為酸鹼或中性 Salt Cation Anion Solution of salt NH 4 I NH 4 (acidic) I (spectator) acidic Zn(NO 3 ) Zn (acidic) NO 3 (spectator) acidic KClO 4 K (spectator) ClO 4 (spectator) neutral Na 3 PO 4 Na (spectator) PO 3 4 (basic) basic 通則 : K a > K b 酸性 NH 4 F, K a < K b 鹼性 NH 4 ClO K = 5.6 a,nh b,f 4 K = 1.4 K = 5.6 a,nh b,clo K =