Example 2 Proton-accepted zero proton-lost products standards products Na2C03/H20 +H+ -H* H;O+ H20 OH system +H+ HCO3 C032- +2H+ H2C03 C032 H01+HC03}+2H2C03=[OH
zero standards Na2CO3/H2O system H3O+ H2O OH- -H+ +H+ CO32- CO32- HCO3- +H+ +2H+ H2CO3 [H3O+] + [HCO3-] + 2[H2CO3] = [OH- 质子条件式 ] Example 2 Proton-accepted products proton-lost products
Exercise 1.Calculation of pH of strong acid (base)solution E.g.:A solution of a mol.dm3 HCI/H2O Proton-accepted zero proton-lost products standards products -H+ HCIH,O HCI Cr system +H+ -H* H0* -H,O OH Proton conservation equation:[H3O+]=[CI-]+[OH-]=a+[OH-] concentration at equilibria then: [HO]=a+K0/H,O*] [H,O]2-a[H,O]-K=0 [HO*]= a+va+4K0 …accurate expression 2 [HO*]=a.....approximate expression(condition:a>105)
[H O ] [H O ] 0 θ 3 w 2 3 − − = + + a K 2 θ w 3 4 [H O ] accurate expression 2 + aa K + + = ⋅⋅⋅⋅⋅⋅⋅⋅⋅ 5 3 [H O ] approximate expression(condition 10 ) a a + − = ⋅⋅⋅⋅⋅ > ： [H O ] /[H O ] 3 θ 3 w + + = a + K HCl/H2O system H3O+ HCl ClH2O OH- -H+ -H+ +H+ Proton conservation equation：[H3O+] = [Cl -] + [OH-] = a +[OH-] E.g.: A solution of a mol.dm-3 HCl/H2O then： Exercise 1. Calculation of pH of strong acid (base) solution concentration at equilibria zero standards Proton-accepted products proton-lost products
For a strong base,e.g.NaOH OH ] b+1b2+4K0 ........accurate expression 2 OH ]=b.approximate expression(condition:a>10 E.g.:pH of a solution of 1.0X 10-8mol.dm-3 HCI/H,O Anwser:a=1.0X10-8<10-5 pH=g.0x10-8=8 H01=a+G+4K_10x10+00x10y+4x10 2 2 H30*]=1.05×10-7mol.dm3 pH=6.98
E.g.: pH of a solution of 1.0 ×10-8mol.dm-3 HCl/H 2 O Anwser: a = 1.0 ×10-8 < 10-5 2 1.0 10 ( 1.0 10 ) 4 10 2 4 [ H O ] θ 8 8 2 14 w 2 3 − − − + × + × + × = + + = a a K [H 3 O +]=1.05 ×10-7mol.dm-3 pH=6.98 pH=-lg1.0x10-8 = 8 2 θ w 4 [OH ] accurate expression 2 − bb K + + = ⋅⋅⋅⋅⋅⋅⋅⋅ 5 [OH ] approximate expression(condition 10 ) b a − − = ⋅⋅ > ： For a strong base, e.g. NaOH
5.3 Equilibrium of ionization for weak acids and bases Ionization of water is ALWAYS involved in acid-base equilibria） 5.3.1 Equilibrium of ionization for monoprotic acids and bases 5.3.2 Equilibrium of ionization for polyprotic acids -5.3.3 Acid-base equilibrium in salt(!?) solution
§ 5.3 Equilibrium of ionization for weak acids and bases 5.3.3 Acid-base equilibrium in salt(!?) solution 5.3.2 Equilibrium of ionization for polyprotic acids 5.3.1 Equilibrium of ionization for monoprotic acids and bases (Note! Ionization of water is ALWAYS involved in acid-base equilibria)
5.3.1 Equilibrium of ionization for monoprotic acids and bases 1.Equilibrium of ionization for monoprotic acids Exercise:Calculate the pH of a solution of HAc with C0.IM HAc(aq)+H2O(I)-H;O*(aq)+Ac-(aq) [H3O+]=[OH-]+[Ac]! Ac=?1 K (HA)=[H'IIA-] [HA] [H,0]= Ko K[HA] [HO][H,O] [HO*]=K+K[HA].....accurate expression
1.Equilibrium of ionization for monoprotic acids 5.3.1 Equilibrium of ionization for monoprotic acids and bases [HA] [H ][A ] (HA) θa + − K = [H O ] [HA] [H O ] [H O ] 3 θ a 3 θ w 3 + + + = + K K HAc(aq)+H2O(l) H3O+(aq)+Ac－(aq) [H3O+] = [OH-] + [Ac-] ! θ θ 3 wa [H O ] [HA] accurate expression K K + = + ⋅⋅⋅⋅⋅⋅⋅⋅⋅ Exercise: Calculate the pH of a solution of HAc with Cinitial = 0.1M Ac- =?