【文章內(nèi)容簡(jiǎn)介】 base equilibrium（ continued） Method 2 Proton reference level –PRL 質(zhì)子參考水準(zhǔn) ?Choise of PRL a．溶液中大量存在的 b．參與質(zhì)子轉(zhuǎn)移反應(yīng) ?Make out PRL 等式左邊 ——得質(zhì)子后產(chǎn)物 等式右邊 ——失質(zhì)子后產(chǎn)物 根據(jù) 質(zhì)子得失相等原則 列出質(zhì)子條件式 Acidbase equilibrium（ continued） 3[ ] [ ] [ ]H O O H B???? － Acidbase equilibrium（ continued） ? For weak acid HB 23H B H O H O O H????2 2 3H O H O H O O H????1. Chose HB and H2O as proton reference level 2. Gaining proton product: H3O+, 3. Losing proton product: B, OH PBE: For NaHCO3 solution, PRL: PBE: For NaNH4HPO4 solution, PRL: PBE: 23 ,H C O H O?22 3 3[ ] [ ] [ ] [ ]H H C O C O O H? ? ?? ? ?+42 24,N H H P O H O?3234 3 4 4[ ] [ ] 2 [ ] [ ] [ ] [ ]H H P O H P O N H O H P O? ? ? ?? ? ? ? ? Acidbase equilibrium（ continued） pH Calculation Strong acids and bases 強(qiáng)酸強(qiáng)堿溶液 P179181 1. Strong acid HX (c mol/L) PBE: [H+] = [OH] + [X] = [OH] + c [OH] =Kw/ [H+] Equation 1 [H+]2 – c [H+] – Kw = 0 accurate equation Equation 2 [H+] = c simplest equation 2. Strong base MOH (c mol/L) Equation 3 [OH]2 – c [OH] – Kw = 0 accurate equation Equation 4 [OH] = c simplest equation pH Calculation 1. When the concentration is ―high‖ (≥106mol/L), the pH has that value we would calculate by just considering the added H+ or OH. That is, the pH of mol/L KOH is － 用最簡(jiǎn)式 2. When the concentration is ―low‖ (≤108mol/L), the pH is . we have not added enough acid or base to significantly affect the pH of the water itself. 3. At intermediate concentrations(~106 to 108mol/L), the effects of water ionization and the added acid or base are parable. Only in this region is it necessary to do a sysmetic equilibrium pH Calculation Monoprotic weak acids and bases 1 Monoprotic weak acids HA (c mol/L) PBE: [H+] = [OH] + [A] [][][ ] [ ][ ] [ ]awawK HA KHHHH K HA K????????[ ] [ ]aH K H A? ?2[ ] ( [ ] )[ ] [ ] 0aaaH K c HH K H K c??????? ? ?aa K [ ] K 1 0 , [ ] [ ]wW h e n H A c K H A c H ?? ? ?且[]c H c???2 4[] 2aaaK K c KH ? ? ? ??[] aH K c? ? ,1 0 / 1 0 0a w aK c K c K??10awK c K?If, 1 0 / 1 0 0a w ac K K c K ?[] awH K c K? ??? For weak bases, using [OH] and Kb substitute [H+] and Ka, respectively. pH Calculation（ continued） Example 2F i n d t h e p H o f 0 . 1 0 m o l / L C H C l C O O H3 2 . 8 6 1 2 . 6 01 . 4 0 1 0 = 1 0 , 1 0 1 0a a wK c K K? ? ?? ? ? ?1 . 8 6 2 . 0 02 . 8 60 . 1 0/ 1 0 1 0 0 ( 1 0 )10acK ???Solution 2 . 8 6 5 . 7 2 3 . 8 621 0 1 0 4 1 0[ ] 1 . 1 1 0 ( / )2H m o l L? ? ???? ? ? ?? ? ?1 . 9 6pH ? Polyprotic acids and bases ? For c mol/L H2A ? PBE: [H+] = [HA] + 2[A2] + [OH] 21221[][][][ ] 2[ ] [ ] [ ]aaawH A K KHH A K KHH H H??? ? ?? ? ? ? pH Calculation（ continued） 2212[ ] [ ] ( 1 )[]aawKH H A K KH??? ? ?222 1 11[ ] 1 0 , 0 . 0 5[] aaa a waKKH A K c K KH cK?? ? ?21[ ] [ ] aH H A K? ? 1[ ] ( [ ] )aH K c H????2112111[ ] [ ] 04[]2aaaaaH K H c KK K c KH???? ? ?? ? ??So for many protic acids and bases, which have big difference in their Ka1,Ka2 value, we can treat them as monoprotic acid. But if their difference is not significant, then we will use other method that we don’t discuss here. 1[] aH c K? ?/ 1 0 0acK21110 , 5/ 10 0aawaaKc K KcKcK?21110 , 5aawaKc K KcK? pH Calculation（ continued） ? If the dissociation tendency for H2A is small, then [H2A ] = c – [H+] ≈ c Mixture of strong acid and weak acid HAc (c mol/L) + HCl (ca mol/L) PBE: [H+] = [OH] +[Ac] + ca Acidic solution, [OH] is neglectable 2[][]( ) ( ) 4( )[]2 20[ A c ] t he n [ ]aaaa a a a a aaaKH c cHKc K c K c c KHi f c H c????????? ? ? ? ??? pH Calculation（ continued） Example F i n d t h e p H o f 0 . 1 0 m o l / L H A c + 0 . 0 1 0 m o l / L H C lSolution 54520 . 1 0 1 . 8 1 0[ A c ] = = = 1 . 8 1 0 [ ] 1 . 8 1 0 1 . 0 1 0aacKKH???? ? ??? ?? ? ? ?2[ ] 2 0 [ A c ] , [ ] 1 . 0 1 0 , 2 . 0 0H H p H? ? ? ?? ? ?3H C l I f c = 1 . 0 1 0 , p H = ???23[ ] 1 . 0 1 0 , [ A c ] = 1 . 7 6 1 0 , [ ] 2 0 [ A c ]HH? ? ? ? ? ?? ? ?23( ) ( ) 4( )[ ] 1. 9 102 a a a a a ac K c K c c KHpH??? ? ? ? ?? ? ?? Mixture of weak acids cHA mol/L HA + cHB mol/L HB PBE: [H+] = [OH] +[A] + [B] Acidic solution, so [OH] is neglectable [ ] [ ][ ] = + [ ] [ ]I f tw o a c id s is w e a k e r , th e n [ ] , [ ][ ] =H A H BH A H BH A H BH A H BH A K H B KHHHH A c H B cH c K c K??????? pH Calculation（ continued） Amphoterism (acid salt, salt of weak acid and weak base, zwitterion) Acid salt NaHA (c mol/L) PBE: [H+] = [OH] +[A] [H2A] 21[ ] [ ] [ ][ ] = + [ ] [ ]awaH A K H A HKHH H K? ? ???? ? pH Calculation（ continued） 121( [ ] )[ ] =[]a a waK H A K KHK H A?????[]H A c? ?121()[ ] = a a waK c K KHKc? ??1211 1 2 1 2 1 0 [ ] =1 1 0 , 1 0 [ ] = p H = ( )2aaawaa w a a a a ac K Kc K K HKcc K K c K H K K p K p K???? pH Calculation（ continued） Example 2 4 F i n d t h e p H o f 1 . 0 1 0 m o l / L N a H P O??3 8 1 31 2 3S o l u t i o n : 7 . 6 1 0 , 6 . 3 1 0 , 4 . 4 1 0a a aK K K? ? ?? ? ? ? ? ?2153 w 281 . 0 1 04 . 4 1 0 1 0 K , / 1 06 . 3 1 0aac K c K????? ? ??8 1 3 2 1 42326 .3 1 0 ( 4 .4 1 0 1 .0 1 0 1 .0 1 0 )()[ ] =1 .0 1 0a a wK c K KHc? ? ? ???? ? ? ? ? ?? ??103 . 0 1 0 / 9 . 5 2m o l L p H? ????For