mofddoun.org www.nfm8mwohriwvrifnww.shboovndlb 1. A.NÑW MY 1.1 Cao -11-120 > Ca④H) , ca" -102- + Ht -1-01-1 - • f f ' OH - Lola 2882 Ca→CaH+2i GO 26 Otze- → 02- 1. 2 502+1-120 ' Hzso} sulfurous acid 2502+02 > 250 } SO , + tho s 1-1,504 sulfuric acid 1 .} 1-12104-12NaOH ' Naik -121-120 Hzso¢+e - → HSOI Hsocite- → so; - 21-1+-1 so + 2Nd-1+201-1- l ' 1-1,0-1102 1.4 Narco, +LCH, COOH > ICH, COO Nat two, acetic in Wat-1105-+24-1,100- -1-4--1 - - j : - " +. iii.iii. . / a- E- i :&:| t, " . 11 f) Fc 1,6-1 1,1-1--10=1 C- il n - O H → Htte2.1 pH= - log 11-1,0+1 pH : 3.5 ↳ itirslunjntnsiovoinfns - log[1-1,0-1]=3-10.5+1-7 : = 4- 0.5 i-4loglo-logt.li = - logio-4-logs.tl/ogIHz0+I---Yogl3.1xlo--l 11-1,0+1=3.1×10-4 moi/Lim rivals [1-1,0-1]=2×10-5 M pH= ? pH= - logo-1,04 = - log(2×10-5) = - log 2-logic-5 = - loge -15104yd = -0.3011-5--4.699
11-1,0+1=-1×10-9M www.ntgm8nnwohriwwrifnww.9/1bNVnd1bpH--- logltlgot] = - log (1×10-9) 0 = - Kogi - logo 's = gloglo =g 2.2 ✓ non ins -711ns Not a → us 2.3 NaCl -11-120 > NaOH -11-14 Nat -14-+1-1+-1 OHstrong strong base acid * iinniioriibrionnhnnonbbritivntbon' Jvwid1WibÑUnN1s Tahrir hydrolysis 2.4 Mgso¢ → My" -150¢ " Ca" -110,1- → Caco, is ) IÑninhvÑn Cally "" > [a " -1 Lei Cao → Cu" -10 " Copper CI) oxide ( v20 → Cut -102- Copper III oxide 2.5 HCl 7. o no/ /L Johnno 150 mL USUI moi HCI son : not ,%. - 1%5,0, . -0.15m" son: mole 1m01 ✗ 150 mL ✗ It 100am, =0-15 ma / 7L
Unit 10 non - BUÑ ACID-BASE vii.Invasionbunt 0186%88 nonacid ;a^ ) knowinwww.nifbonnoioluvtlt hydrogen iont 1. minimal Binary acid ) : " Wasim;D, " (g) hydrogenfluoride hydrogensulfide ( agg ) HE - - hydrofluoric acid aqueous HCl : hydrochloric ñ aquarii Hzs . - hydrosulfuric ñ HAHA) > Htcaq] -1AÑbbMnÑWÑH+bÑuo 1mW → monoprotieñ HF Httf - hadrofluioricñ fluoride ion www.ni-ve-lpoi-e-pair = 7-8-0=-1 ~ H - ÷: . ' 1-1++1 :# :] - Ha > 1-1++05 *msn.ms * hydrochloric ñ chloride ion HBR ' H'- + Br - HIN - ' H-ttcvhydrocyanic.ci cyanide ion hydrommonniooi bromindion HI > 1-1++5 5- i - E hydroiodioñ Iodide ion H - CEN : 1-1++1:P _=Ñ :] - b 4- L - § 11 2. nTdddnÑ Oxyacid ioxo acid 㱺 0,1-1,X " roadway - Écarbonyl - OH hydroxyl Formal charge HNO, Nitric oi Fc , Ni 5-0-4=+1 " 1+5+18=24 N ( N - a) 0--6-6-7 = -1 H - É " " §: (No -1-10=6-4-2--0 HNOZ Nitrous of ln= 0--6-4-2=0 - , 1+5+12--18 H , Nig. AXLE V-shape - - Nishant, 8mmol riwwrifnww.sn bound 16
- - g ' . 11 1-11104 perchlorica H - E -CI - ci : " " y " " 7+7-124--32 - :p. : - - j . HU0¢→H++ao¢- f.ci?ii-?ci. ↳ perchlorate ?/lnydrodgenion ' ' :&. :-O , 1-140 , chloric of proton ion 1-1402 Chlorouscan 1-140 hypochlorousoi CH}C0OH acetic a^ - pm > 4¥ . "iÉ:&? / + µ+ It n-o-itite-H-E-iiii.ci , e-ip.io ?-H:p fi H n:O e=O (1-1,1001-1 > ( Hzcooiaq, -1 Htcaqi Hzloz - ' 1-1+-11-105 ii. . - j . - a- -É -i - H ( H - &. -É - i \ | . - :] Hcoj i ' Httco} - :O: 2- I :O:-c" .ci :| , - 1-131004 1-131004 - ' 1-1+-1 Hzpqi :O: " Hzpoci . ' H" -114M¢ " H - Q - Y - fi. -H Hpoci - i ' 1-1+-1 POI - :O : it www.nfm8mwohriwvrifnww.shboovndlb
BASE hydroxide ion NaOH → Nat -101-1 - I:&. - HI ' KOH → Kt -101-1 - Ca( OHI, > call -1201-1 - m base NH , -11-1,0 - ' NHI -101-1 HE ' ti e ' 1-1++1:&- H ] - tho ← 1-1++01-1 - Nazca} ' 2Na++COg 21-1++105 Hug weakñ Nat -101-1 - > NaOH hydrolysis / strong base hydrolysis ✗ co, " -11-120 . ' Hcoj -101-1 ' base, acid, acid, base, MoTdÑoUNHNbÑÑD Pro SRIOH)z ssr" -1201-1- 1-1404 ' Ht -140¢ ' strontium base perchlorica hydrogen ion perchlorate ion hydroxide strong acid acid lncrnbbn' ) HNO , -11-120 > 1-1,0+-1 Noj Nitric acid Nitrate ion LIOH > Lit -101-1 - acid lithium hydroxide hydroxide ion H++N0j strong base countout , base qH+ Hzsoqttho > HSOI -11-1,0+-10 Sulfuric of hydronium ion C strong acid) hydroxonivm ion acid HSOI -11-1,0 , ' 1-1,0+-1504" ② weak acid acid sulfate ion (nineteen) sulphate ion Nishant, 8mmol riwwrifnww.sn bound 16
10.1.2 mw§nohbud bUÑsÑbMn-ñÑ Bronstect - Lowry Theory non Irishwoman " Proton donor" !H+ Not ÉUTWrnoW " Proton acceptor " pit n=o E- 0 Ht Haig, + ÑH > (g) > NHI -14 In - - (+1%+0*1) H - Ii : > H-1-i.ci : H - N - H -11-1-1 , I hydrogen chloride gas " t, ' H y l-Klcgt-NHzc.gs > Ntklks ) Ammonium acid acid µ base Ammonium chloride I Hllcg) -11420cL, > Hzotcaqltcliaq, acid base acid base I Ntlzigitthocl] , ' NHjcaqitot-icaqlbasei.it#,.acid acid base Hit > ntirbboantwinoon (Amphoteric substance) www.owidwroinCAmphiprotic Substance) now Hllcaqg)tNH}caq1 ' NH¢4C9q) ☒ Ht Hclcapqitte, tcaqptctcapq) acid base NI-hlaqpttlaqy.it/NcicaqItHTeiH4caq1tNHzlaq)7NH4Cl Caq) 1-14+1-520 > 1-120++0 NHI -11-120 i ' NHI' -101-1' HCHNH}-121-120 > NH¢U -121-120 Mom .ÑdUNHWÑHd 10.7 CH}C00HNa "' ↳ CH}(00 - + Nat sodium acetate / µ .¥iiÉ:&:-P (1-13100 Na-11-120 ' CHINH 1- NaOH (Hzcoo - + Nat-114++01-1' acetic ai sodium hydroxide tweak acid) (strong base) Nishant, 8mmol riwwrifnww.sn bound 16
CH}co0Hiii.ñnN§nrvñ8In8iÑd (hydrolysis) CH}GO0 - +1-120 , ' (Hzcoolt -101-1' Ht -101-5 base ✗ H++i > Ha NH¢Cl "" > NH¢+tci strongan ji acid NHI -11-120 , ' HÑtNH} CH}CO0NH¢ > CH, COO - tNHq+ CH}C00--11-120 , ' CH}( OOH -101-1 - base Kb -5.5×10-10 NI-14+-11420 \ ' H}O++NH } acid Kai 5.56×10-10 ka=k6 ' MNM Nall > Nattct Nat -101-5 > NaOH strong base 1-1++4' > HCl strong acid Nation: Ci Istoria Hydrolysis reaction www.wievirtwnow 10.1.3 MVÑnTAbUdÑoÑd Lewis Theory nrntun.je-wntiainjetklion-N.FI > (g) > Nikita- * + 4- µ . ,¥y, ( H - ) acid ' H revivebase F HE NH, -1131=3 > NHyBF, HN - B - F H' ¥, 't, Blip 14 fbase acid MTWÑdUnnNÑVO 10.8 it H 1. if. . Al t p a- t.lyI , 't, > 9 - Al - p ' - H il 'm o o 2. [ H - I:] - to"=c= :O. > [ H-c.i.0-do.ci. :] ' acid " base OH - CO2 base acid Nishant, 8mmol riwwrifnww.sn bound 16
Nishant, 8mmol riwwrifnww.sn bound 16
10.2 Conjugate acid - base pairs njnon -but HF -1-1-120 , ' 1-130+-1 Facid base acid base n!nraÑiimw Htannniinhiuann's CH}C00 - + tho , ' CHZCOOH + OH - base, acid, acid , base , HI NH, + H2O - ' NHI -101-1 ' base , acid, acid, base , MMdÑoUnnNlV%d 1- - HNO} -11-12/0 , ' 1-1,0-1 + No, - N'Nmialdv HNO, > 1-1+-1 No - I 1-120+14-1 > thot } acid, base, acidz base , 2. NH} - ' H+ -1MHz - weak ñ strong base NHI -11-1,0 > NH, -101-1 - base, acid acid, base , 1. H}P04 -11-120 - ' 1-1,0+-11-121004' acid , base, acid, base , HZPOI -11-120 - ' 1-1,0++1-11004" acid , base , acid, base, HPOIÉ -11-120 , ' Hyo-1 -1100¢ } - acid, base , acid, base , 4. S " -11-120 , ' 1-15-101-1 - base acid acid base Hs + 1-120 , ' His + OH - base acid acid base Con Acid Con Base - 14,100¢ HZPOI 1431004 142100¢ H POI - - Has 1-15 His HSS2- Nishant, 8mmol riwwrifnww.sn bound 16
SCI-PKC-CH30224-004-02 : page 1 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. Lesson 10 : Acids and Bases Unit 2 : Conjugate Acid-Base Pairs Part 4 : Conjugate Acid-Base Pairs Conjugate Acid and Base Definition Conjugate acids and bases are Bronsted-Lowry acid and base pairs, determined by which species gains or loses a proton. When a base dissolves in water, the species that gains a hydrogen (proton) is the base's conjugate acid. Acid + Base ⇌ Conjugate Base + Conjugate Acid (1) In other words, a conjugate acid is the acid member, HX, of a pair of compounds that differ from each other by gain or loss of a proton. A conjugate acid can release or donate a proton. A conjugate base is the name given to the species that remains after the acid has donated its proton. The conjugate base can accept a proton. (Helmenstine, 2019) The hypothetical steps are useful because they make it easy to see what species is left after an acid donated a proton and what species is formed when a base accepted a proton. We shall use hypothetical steps or half-equations in this section, but you should bear in mind that free protons never actually exist in aqueous solution. Suppose we first consider a weak acid, the ammonium ion. When it donates a proton to any other species, we can write the half-equation: NH4 ⇌ H + NH3 (2) But NH is one of the compounds we know as a weak base. 3 In other words, when it donates a proton, the weak acid NH4 is transformed into a weak base NH . 3 Whenever an acid donates a proton, the acid changes into a base, and whenever a base accepts a proton, an acid is formed. An acid and a base which differ only by the presence or absence of a proton are called a conjugate acid-base pair. Thus NH3 is called the conjugate base of NH4 , and NH4 is the conjugate acid of NH3 . A strong acid like HCl donates its proton so readily that there is essentially no tendency for the conjugate base Cl– to reaccept a proton. Consequently, Cl is a very weak base. A strong base like the H ion accepts a proton and holds it so firmly that there is no tendency for the conjugate acid H to donate a proton. 2 Hence, H is a very weak acid.(Anonymous, 2019) 2 suttikan Kimprakhoh 27212 5/1 16 q UP hill www.oihin downhill N%Viñ
SCI-PKC-CH30224-004-02 : page 2 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. Lesson Discussion and Summary Topic Main idea Interesting vocabulary Vocabulary Translations of vocabulary Vocabulary Translations of vocabulary EXAMPLE : Conjugate Acid-Base Pairs 1. What is the conjugate acid or the conjugate base of a. HCl is a …………………….. When it ……………. a proton to any other species,we can write the half-equation : …………………………………………… Thus ……. is called the conjugate base of …….., and …… is the conjugate acid of……. b. CH NH 3 2 is a …………………….. When it ……………. a proton to any other species,we can write the half-equation : …………………………………………… Thus ……. is called the conjugate base of …….., and …… is the conjugate acid of……. c. OH is a …………………….. When it ……………. a proton to any other species,we can write the half-equation : …………………………………………… Thus ……. is called the conjugate base of …….., and …… is the conjugate acid of……. d. HCO3 is amphoteric. Its conjugate acid is ………… and Its conjugate base is ………… write the half-equation when it accepts a proton: …………………………………………… write the half-equation when it donates a proton: …………………………………………… 2. Write a balanced equation to describe the reaction which occurs when a solution of potassium hydrogen sulfate, KHSO4, is mixed with a solution of sodium bicarbonate, NaHCO3. 3. What reactions will occur when an excess of acetic acid is added to a solution of potassium phosphate, K3PO4? Svttikan Kimprakhoh 27272 5/1 16 Conjugate Acid - Base Pairs determined ÑluwÑNU The hypothetical steps ÑwmowwNÑ gain %Ñ&U suppose tNNÑ ' compounds ntvtbinou differ linn'nHÑW release Vodou absence via remain eisai pair nj tendency www.WN essentially Ñuañn strong acid donate HCl > Httci c, - acid acidHub"" Ha [1- weak base accepted at,%ih+ii¥ :(Hiiiii; CH}NH}+ CHbNH2 ( Hzntlz EH}NH}+ weak base accepted base acid acid 01-1-+14-1 > tho OH1--120 tho OH - bÑwiñn%ninwibV Hz[o } CO} ' HCO} - + H+ ' Hfo} 1-1105 > 1-1++105 salt KHs04 ' K'- + Hsoi 1-1504--11%05, - Halo,-15042 ' 2 NAHH } > Nat -11-1105 HSOÉ -11-1105 , ' Hzsoqtcoz" salt K,PO¢ > 3kt + pq¢ } - i.win CH} COOH -1100¢ " . ' CH }COo-1-1-11004 " i. MIN CH } COOH -11-110042- , - ctzcoo-1-1-121005 CH} [OOH 1- Hzpogi , - [ Hzcoo-1-1-131004
SCI-PKC-CH30224-004-02 : page 3 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. What is the conjugate acid or the conjugate base of (a) HCl; (b) CH3NH2; (c) OH–; (d) HCO3–. Solution: a. HCl is a strong acid. When it donates a proton, a Cl– ion is produced, and so Cl– is the conjugate base. b. CH3NH2 is an amine and therefore a weak base. Adding a proton gives CH3NH3+, its conjugate acid. c. Adding a proton to the strong base OH– gives H2O its conjugate acid. d. Hydrogen carbonate ion, HCO3–, is derived from a diprotic acid and is amphiprotic. Its conjugate acid is H2CO3, and its conjugate base is CO32–. The use of conjugate acid-base pairs allows us to make a very simple statement about relative strengths of acids and bases. The stronger an acid, the weaker its conjugate base, and, conversely, the stronger a base, the weaker its conjugate acid. EXAMPLE 11.12.211.12.2 : BALANCED EQUATION Write a balanced equation to describe the reaction which occurs when a solution of potassium hydrogen sulfate, KHSO4, is mixed with a solution of sodium bicarbonate, NaHCO3. Solution The Na+ ions and K+ ions have no acid-base properties and function purely as spectator ions. Therefore any reaction which occurs must be between the hydrogen sulfate ion, HSO4– and the hydrogen carbonate ion, HCO3–. Both HSO4– and HCO3– are amphiprotic, and either could act as an acid or as a base. The reaction between them is thus either HCO−3+HSO−4→CO2−3+H2SO4 HCO3−+HSO4−→CO32−+H2SO4 or HSO−4+HCO−3→SO2−4+H2CO3 HSO4−+HCO3−→SO42−+H2CO3 Table 11.12.111.12.1 tells us immediately that the second reaction is the correct one. A line drawn from HSO4– as an acid to HCO3– as a base is downhill. The first reaction cannot possibly occur to any extent since HCO3– is a very weak acid and HSO4– is a base whose strength is negligible EXAMPLE 11.12.311.12.3 : REACTION PREDICTION
SCI-PKC-CH30224-004-02 : page 4 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. What reactions will occur when an excess of acetic acid is added to a solution of potassium phosphate, K3PO4? Solution The line joining CH3COOH to PO43– in Table 11.12.111.12.1 is downhill, and so the reaction CH3COOH+PO3−4→CH3COO−HPO2−4 CH3COOH+PO43−→CH3COO−HPO42− should occur. There is a further possibility because HPO42– is itself a base and might accept a second proton. The line from CH3COOH to HPO42– is also downhill, but just barely, and so the reaction CH3COOH+HPO2−4→CH3COO−+H2PO−4 CH3COOH+HPO42−→CH3COO−+H2PO4− can occur, but it does not go to completion. Hence double arrows are used. Although H2PO4– is a base and might be protonated to yield phosphoric acid, H3PO4, a line drawn from CH3COOH to H2PO4– is uphill, and so this does not happen. Anonymous, (2019,June 6) Re: Conjugate Acid-Base Pairs [Online]. Retrieved from https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11Re actions_in_Aqueous_Solutions/11.12%3A_Conjugate_Acid-Base_Pairs [4 November 2019] Helmenstine, A.M. (2019,March 6) Re: Conjugate Acid-Base Pairs [Online]. Retrieved from https://www.thoughtco.com/definition-of-conjugate-acid-605846 [4 November 2019]
10.3 msn.nnmddvotnsa-l.vn 10.3.1 mofbnnoifvdsnrnbbri-buntl.br' 1-14+1-1,0 > 1-1,0++4- 5 1.OM - O O Initaili -1 - t + INJUN 1.0M - 1.0M 1. UM Éw§n o - 1.0M 1.0M Final NaOH > Nat -101-1 - base I 1. OM O O IN 1.0M - 1.0Mt 1.0Mt F o 1.0M 1.0M (alot), ' ca" -1201-1 - I 1.0M . 0 0 + JW 1.0M 1.0Mt 2.OM F O 1.0M 2.0M Ex.lt/Br(g) "" > Htcaq)tBriaq, iiyoyTwoHBr = Imo/ HBR ✗ 0.br#+x2X=l.zmolHBir1moXH+ it Ex.} sr(01-11 , > Sr" -1201-1 - V.4L I.0m01 /L d 2L 0.2 iiwowfwnsrH= imolsr" ✗motor, - ✗ 107¥14- ✗ 0.AM/--A1moIsr212L.:molo1-i.-0.2mo10HMJIdntoUnnNl "vYId 1. HI -11-120 > 1-1,0+-11 - 500mL 0.2 Mol /L moltlyotilmoltlyot ✗ 0.2m¥ ✗ ask = 0.1 moi thot 1m¥- It 2. KOH > Kt -101-1' 250mL 0.5mW /L 1.WWTNbdf.nl KOH- -39+161-1--56 glmol 2. Win KOH = 5681×01-1 ✗ IMOTKOH ✗ 0.5m¥ ' ✗ V. 25K =7gK0H 9mkoH 1m¥- 1¢ www.nfmifmwohriwhkifnww.shboovndlb
10.3.2 MibbMnÑoVdgnrnidy HF +1-120 thot + FIuitail 0.5 M - o o change o.AM - 0.02+0.02-1 Equilibrium 0.48M - 0.02 0.01 % msn.mnMda HF -- 0.02×100 =y?g= 4% 0.5 Kid-0212.12×10-42=4×10-4 5×10-1 g× , , - , = 0.8×10 ' } 0.5 14250¢ -11-120 > 1-1,0++1-150,5 IM 1M IM 1-1504-+1-120 1-1,0-1+5041 - I 1M - Ot Ot C ✗ - - X X F 1-✗ - ✗ ✗ Ka- - l ✗ lot Ka- - [ tho-4150¢'T ¥81800M 11450¢'T X2 e- 1.8×10-5 1×10-1=1/2 0.50 - ✗ 1-✗ ✗2=10.50 - ✗ 711.8×10-51 ×? 9×10%4.8×10" / ✗ Ftl.8HO-51×-19×10-9=0 ✗ = - (1.8×10-5)=1/11.8×10-55-41111-(9×104) Lllloinniousrn = -1.8×10-5-1 ✓ 3.24*-10-136×10" 2 = -1.8×10-5 I ✓ 3.6×10-6 2 = -1.8×10-5 -1-6×10-3 2 = -0.9×10-5 -1-3×10-1 - 0.0000091--3×10-7 = -0.009×10-1 -1-7×10-3 = 3×10-7 g3×10-1 Nishant, 8mmol riwvrifnww.sn bound 16
piraino NH}FHz0 = NH¢+-101-1 - I 0.2M - - - C - y + ✗ + ✗ f- 0.2 - X ✗ X Kb :[Ntki' ][ 01-1-1 = ✗ 1 = 1.8×10-5 LNH,] 0.1¥ ¥71000 ✗ 2 = 1.8×10-5 0.2 ✗ 2--0.36×10-5 ✗2--3.6×10-6 ✗ = 1.9×10-3 10.3.3oiovninwbbnnoiimohdntowno.nwl.info HA -11-120=1-40-1 -1A - Yo , [Ht] ✗ too 0.1M ? C 10% 11-17--10×0.2=0.02 M ¥ HB -11-120=1-1,0++13- 0.02M ? 11-1+1=20×0.02--0.004 M 20% 1¥ HA bÑwnoannmh HB bwoiobioinoioannnoi 10.3.4 nñbbMnÑIVd9ÑY The self ionization of water Hike, -11--120 (e) 7- Hzoiaq, -101-1 - lag) P. 27 Kw - - [1-1,0+5101-1]=1×10-14 [1-40-1]=[01-5]--1.0×10-7M MY H2O Ñ d--1.0 glml (tho ]i= ? [1-120]--1 Mol ✗ 1,8m , ✗ 1000mL = 95.56 M moth 188 1L Mw : 18 glmol Nishant, 8mmol riwvrifnww.sn bound 16
10.3.5 nhaofwntboo.in'M ka Kb bid# Kw ① It>COOH -11-120 (Hyatt -11-1,0-1 Ka=[ (1-1,100-111-1,0) [4-4001-1] ② OH}C00 - +1-120=-04,0001-1+01-1. Kb :[CH}CO0H][OH] ICH,COOT ① + ② 21-120=-14,0++01-1- kaKb - -111-1,100111-1,0+3 ✗ [01-1,001-11101-1] [11-11004] [(Hscoo - ] Kwi [1-130+1101-1] Kw - KaKb bbUUÑnÑN 10.3 t.BA/0Hlzs-Ba'-t2OH100Mt 1.0m01 /L MWBACOHIL -737-121161-12111=177 glmol i. Woo BANH)z = 171g Balotllz ✗ lmolBalotlhxlmolOH-rllggm.pl/oomL-- 8.55g (8) 1m01 Balon)z ZMOIOH' 7L Natural ~ [1-1,0+111-1505] 2. 1-12503+1-120=1-130++1-1505 Kai [ Huo, ] 1-1505-+1-120 = 1-40+-1505 Kaz _- [thot ][Soi] [Hsoj] 11-1,0+1711-15051 > [Soi] Nitrous 3. HNOZ -11-120=1-40+-1 Noi ka. -5.62*10-4 HF + 1-1,0 = 1-1,0+-1 F- Kai 6.31×10-4 [ Hit] Mn HF > HNOL 4. [ OH - 1- ✓ Kbc =/ 1.0×10-4×2 sixto-2--1.4×10-2 5. ✗2=1.1×10 , 90=11-1+1×100 0.5 c ✗ = ✓ 1.1×10380.5 = 1.5×10-2×102 = 1.5×10" 0.5 =3 Yo 6. %,F%fG ÷:E, i ÷g : 1.264 Nishant, 8mmol riwhkifnww.sn bound 16
7. % :) ✗ 100 42×10"=[¥g ✗ 102 142×10-4?([¥g) " K -14.2*185×0.8 = 14.1×10-6 = 1.4×10--5 8. Hunt 4- Rbi Kw Ka = 10-14 2×10-6 = 5×10-9 9. Kwikoikb Kaikw Kb = yo-14 5×10-10 = 10×10-15 5×10-10 = 2×10-5 www.nfiafmwohriwwonifnww.sn bound 16
SCI-PKC-CH30224-005-02 : page 1 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. Lesson 10 : Acids and Bases Unit 3 : Acids and bases ionization Part 5 : Acids and bases ionization Acids and bases ionization constant The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The acid-ionization constant The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka . For the reaction of an acid HA: HA(aq)+H2O(l) ⇌ H3O+(aq)+A−(aq), (1) the acid ionization constant is written Ka = 3 [H O ][A ] [HA] (2) where the concentrations are those at equilibrium. To illustrate this idea, three acid ionization equations and Ka values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order CH3COOH < HNO2 < HSO4 −: CH3COOH(aq)+H2O(l) ⇌ H3O+(aq)+CH3COO−(aq) Ka = 1.8×10−5 (3) HNO2(aq)+H2O(l) ⇌ H3O+ (aq)+NO2 −(aq) Ka =4.6×10−4 (4) HSO4 −(aq)+H2O(aq) ⇌ H3O+ (aq)+SO4 2−(aq) Ka =1.2×10−2 (5) The percent ionization of a weak acid Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture: % ionization= 3 eq 0 [H O ] [HA] ×100 (6) where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, [A−] = [H3O+]). Unlike the Ka value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. The base-ionization constant Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant, Kb in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B: B(aq)+H2O(l) ⇌ HB+(aq)+OH−(aq), (7) the ionization constant is written as Kb = [HB ][OH ] [B] (8) where the concentrations are those at equilibrium. suttikan kimprakhoh 27212 5/1 16
SCI-PKC-CH30224-005-02 : page 2 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. Inspection of the data for three weak bases presented below shows the base strength increases in the order NO2 − < CH3COO− < NH3. NO2 − (aq)+H2O(l) ⇌ HNO2(aq)+OH−(aq) Kb = 2.17×10−11 (9) CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq) Kb = 5.6×10−10 (10) NH3(aq)+H2O(l)⇌NH4 +(aq)+OH−(aq) Kb = 1. 8×10−5 (11) As for acids, the relative strength of a base is also reflected in its percent ionization, computed as % ionization= eq 0 [OH ] [B] ×100 (12) but will vary depending on the base ionization constant and the initial concentration of the solution. https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases [17 November 2019] https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases [8December2022] Example: Calculation of Percent Ionization 1. Calculate the percent ionization of a 0.10 M solution of acetic acid (a weak acid) 2. Calculate the percent ionization of a 0.125 M solution of nitrous acid with a pH of 2.0. Step 1: Write the balanced acid dissociation reaction. Step 2: Write the expression for Ka Step 3: Use an ICE table to determine algebraic expressions for the equilibrium concentrations in our Ka expression: Equation Initial Change Equilibrium Step 4: Calculate percent dissociation Step 1: Converting the provided pH to hydronium ion molarity yields. Step 2: Calculate percent dissociation. suttikan kimprakhoh 27212 5/7 16 CH}C00H -11-110=0-1,100--11-1}o+ pH=-log[ Hzot] ptt-2.o-oizloglo-logiiloglol-loglka.CC/-hC00-JlH,o+jlCHzC00HJpH= - logs - logio-Z-loglth.lt ]= - logllxlo" ) ( 1-1310014+1-1,0 (14>COO - -11-130-1 11-130+1=1×10-2M 0.1 - - 0 0 yo= 1×1074×10040=840 ✗ + ✗ + 0.115M ✗ - or - ☒ - X X P. , ^ Kait ' ✗ ii.34×10-3M 0.1 - ✗ 1.8×10-5 _-☐¥* ii.8×10-6 g. = 1.34×10/-3×1100=1.3470 1/0-1
SCI-PKC-CH30224-005-02 : page 3 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. 3. Calculate the percent ionization of a 0.50 M solution of ammonia. 4. Calculate the percent ionization of a 0.25 M solution of ammonia with a pOH of 5.0. Step 1: Write the balanced acid dissociation reaction Step 2: Write the expression for Kb Step 3: Use an ICE table to determine algebraic expressions for the equilibrium concentrations in our Ka expression: Equation Initial Change Equilibrium Step 4: Calculate percent dissociation Step 1: Converting the provided pOH to hydroxide ion molarity yields. Step 2: Calculate percent dissociation suttikan Kimprakhoh 27212 5/7 16 NH , -11-120 NHI' -101-1 - 1001-1--5 / ogio - logs - boycott]= - log , - logo-5 ky=CNHÑI0H' ] - logloti]= - log (1×10-5) ( NH} ] [01-1-5=10-5 Ntt} -11-120 . ' NHÉTOH ' 0.5 - O O - X - TX TX yo . - [-01-1]×100 c 0.5- x x x = 10-5×102 Kiki g- ✗ --1.8×10-5 0.25--10-5×102 25×10-2 ✗ = ✓ 0.5×1.8×10-5 = 3×10" = 4×10-3 Yo to : 7×10 ' } 0.5 ✗ 100--0.690
SCI-PKC-CH30224-005-02 : page 4 Chemistry for World-Class Standard School: Created by ; Mr.Suthat Chanprakhon ; since 2022 Name……………………………………………..…Id………………..Class………..Ordinal……….. 5. 3.00 moles of N2 gas and 1.00 mole of H2 gas are combined in a 1 L reaction vessel. At equilibrium 0.663 moles of H2 remain. What are the resulting concentrations? 6. Phosphorus pentachloride decomposes into Phosphorous trichloride and Chlorine gas. 0.500 moles of pure Phosphorus pentachloride is placed in a 2.00 L bottle. What are the resulting concentrations? Step 1: Write the balanced acid dissociation reaction Step 2: Write the expression for Kc Step 3: Use an ICE table to determine algebraic expressions for the equilibrium concentrations in our Ka expression: Equation Initial Change Equilibrium Step 1: Write the balanced acid dissociation reaction Step 2: Write the expression for Kc Step 3: Use an ICE table to determine algebraic expressions for the equilibrium concentrations in our Ka expression: Equation Initial Change Equilibrium Suttikan Kimprakhon 27212 9/1 16
10.4 dwÑÑnrnbudvogl.info Nall (5) > Natcapqltlicagg) strong Nat -101-1- > NaOH base Idw Hydrolysis grid]Ñ Hi , ' 1-1++01-1- H+-1cL - > 1-14 strong acid * neutral inn) Ntkclcs) > Ntkicapyltcicapg) NHI' -101-1 NH¢0H weak base Hydrolysis reaction NHÉ, -11-120=1-1,0+-1 NH } acid base acid base Ka UH NH¢+= 5.56×10-10 CHOONa > (Hzcoo- + Nat CH,COOH weak acid r NaOH strong base ✗ Hydrolysis (Hzcoo- + H2O CH> 1001-1-1 OH - base acid acid base Kb i 5.56×10-10 mrnionntou Universal indicator fun pH 1-14 NH¢NO}cs ) ' Ntkicaqltivojcaq) NH¢+tOHNH¢OH weak base Ht + NO HNO, strong acid Hydrolysis NHI -11-120 NH, -11-1,0-1 Ka - 5.56×10-10 acid Natllo, > Nat -11-1005 Nat -101-1- > NaOH str base 1-1+-1 Hcoj . ' Hzco, weak acid ~ 1. 1-1105-11-120 , ' 1-1,0++6,2- acid Ka - -468×10-11 M 2. Hcoj -11-120 , ' Hzco} -101-5 base Kb= 2.24×10-8 Amphoteric Kb > Ka=> Base CH}C00/NH¢ > Cthgcoo- + NH¢+ CH}C0OH weak acid NH¢OH weak base 4-1,00-+1-120 - ' ( Hzcootl + OHbase Kb= 5.56×10-10 www.nfm8nnwohriwwonifnww.sn bound 16
NHI -11-120 1-1,0+-1 NH} acid Ka= 5.56×10-10 Ka - - Kb 㱺 neutral crow) 1. Ka > Ky 㱺 acid 2. Kb > Ra 㱺base 3. Ka - Kb 㱺 neutral ↳ no@ blritluifbbn' ↳ nsaeiow-woteiowka-kbl.lv VÑN 10.4 1.1 KCN → K'- +CÑ 2.1 HCOOKC base) K+-101-1 - → KOH strongbasex HCOO-ttkot-HCOOH-OHH-1-CN-E.HN weak acid - base acid acid base CN-1-1-120 HINT OH' base 2.2 NH¢NOz lalidt Ka>Kb+ Ntki -1420¥ NH,-11-1,0 Ka 1.2 (Hylooli ' CH, COO- +Lit Noi -11-120 7- HNO, -101-1 - Kb lit-104 ' ' LIUH strong base ✗ CH , lot -11-1+-711-41004 weak acid / ethno- +H2o CHINH-101-1 - base 2.3 NH¢HWg cacid) ka >Kb NHI, -1420¥ NH>1- thot ii. NaCl0¢ ' Nat -1110¢ ' Hcoj -11-120=405-11-1,0-1 Ka Nat -101-1 - → NaOH strong base ✗ 1-1105-11-120=1-12103-101--1 - Kb AOF-11-1+-11-11104 strong acid -1 1.¢ (NH¢)zPO¢→ 3 NHI -1 POI - NH¢+ -101-1- → NH¢oH breach baser Pap- -131-1-1 → H]PV¢ weak acid ✓ Ntkittho NH} -101-1' HzP0¢ -1420¥ Hzpoci -11-1,0-1 1.5 Naito} → Hat + coj - Nat -101-1- → NaOH strong base ✗ 105-121-1-1 → Hzco, weak acid ✓ co} -1-1-120 Hcoj -1101-1 - Hcoj -11-120=1-1200, -10h - 1.6 KNO, → Kt +No} - ¥+0k- → KOH strange base ✗ Noj -11-1-1 → HNO} strange acid ✗ www.nfiafmwohriwakifnww.sn bound 16
10.5 pH VdÑWNJdWÑNbbnibU " Power of Hydrogen" pH - - - 1og[Hit] H2O Hyo-1+014- Kw - - [thot][01-5]=7.0×10-14 [ Hyo-11=[01-5]--1.0×10-7M ptt-togc-th.ci] pot-1=-108101-1 ' ] pH=-log[thot] = - toga. 0×10-7 - - logi.o-loglo-7ilogxh.hr/ogX--7log1o-1og1iloglo-- I = 7 - o logi -0 = 7 POH - - 7 pH -1pct-1--14 Ex. 7 CH > COOH 0.5M I pH= ? CHZCOOH -11-120 , ' Ctlzcoo-1-1-1,0-1 [1-1,0-1]=1- Ka - C i ¥21000 = ✓ 1.8×10-5×0.50 = 3×10-3M pH - - - logos ✗ 10-7 =3 - log} =3 - 0.477 = 2.923 # pH - -2.523 i:B [1-1,0-1]=2. M pH = 2-10.523+1-1 = 3-0.477 =3 loglolog} = - loglo-3-log3-loglth.lt ]= - log (3×10-3) 11-130+7=3×10-3 M Ex-8 GWANG pH 10 Ñ [01-1]--2. M POH -1111-1=14 pot-1--14-10--4 pot-1--410810 - log' = - logio-4-logl-logCOHT-togcixlo.tt ) [01-5]=1.0×10-4 M. Eng A :[1-1,0+1=2.0×10-2 M 㱺 pH B :[01-1-1=2.0×10-2 M 㱺 pot } Ur pH C : [1-1,0-1]=5.0×10-9 M 㱺 pH www.nfmifmwohriwvonifnww.sn bound 16
① pH V09 A -- together] - - - logczxlo" ) pH A) C) B = zlogio - loop = 2- 0.301=1.699 ② POH Vos 13--1.699 pH = 14-1.699--12.7301 ① ptlvosc = 9-logs = g- 0.699 i 8.301 Ex. pH - -8.301 0 :Ñ [01-5]--2. M Pott 14-8.301 = 5.699 = 6-0.301 = 610810 - login - logfoti] = - log (2×10-6) CAHI = 2×10-6 M Ex-10 1) pH 㱺 [1-130-1] 㱺 insane thot , Voronov hythz 4th pH⇐[Hzo" ]⇐ HCl pH -2 Vi 100mL CÑOV;) ① Ui [1-130-1] in pH - -2 pH= 2-0 㱺loglo - logl =- loglo-1-logi-logcth.FI - logano-3 [1-1,0+1=1×10-2 Molk ④mastoid thot = 1×10-2 Mol 1010-00 my ✗ tooth = 10-3 mol 2) bÑdbÑW HCl pH -1 Withhold 100 mL ?⃝ all [1-40-1]=1×10-1 M ② Jiwani and Hyo! 1×10 " Mol ✗ 100mL 7000mL = 1×10-2 ma/ ÑIWswTWdoIN= 1810-3+7×10-2 = 0.1×10-2-1 1×10-2 = 10 " (1+0.1) = 1.1×10-2 Mol V9WINTON 100-1100=200 ML [1-1,0-1] treat = 0.0mmol * 1000mL 200mL 1L = 0.055 moth www.nfa8mwohriwwonifnww.sn bound 16
our pH law'= - log [1-130-1] = - log (5.5×10-2) = 210810-1085.5 = 2- 0.44 = 126 3) boisrwiYWIWIMJloomlnsfbntio.no#oo1VHC1IwTo I i. [1-1,0-1]=1×10' mol ✗ 1000Mt 100-1100mL il base = 5×107 M Ex-12 6m8cm NaOH pH-1001N NaOH pH - - - logos ✗ 10-3) phiz 10mL bÑwÑTbn'll@ = 310GW -1085 =3 - 0.7 I. NaOH →Nat -101--1' ' moi - -1×10-4 .no/ ✗ Aml =L.} philo 1000mL 11-409=101%10-1081 EX.tl 1. bacterial HCl pH } Nimmi 1 nini = 1×10-10 M ?⃝ [1-1,0-1]=1×10-3 M [01-1]=10-4 M ② iminium Hyo-1=1×10" no / ✗ IL IL pl-1--12 = 1×10-3 mol 11-40%-1210910-1091 = 1×10-1'M 2. HCl pH 5 Urwin 400mL lÑoÑ, ) [01-5]--10-2M ?⃝ [1-1,0-1]=1×10-5 M Mo / [1-40-1]=1×10- 2m01 ✗ 10mL ① jmwTWn Hyo-1=1×10-5 Md ✗ 400 mL 1000mL 1000mL = 1×10-4 no/ = 4×10- b Mot Ul A 3. ÑlwoÑWdÑ£Ñd9bÑNbÑN 1×10-4 Mol ✗ A ml = 1×10-4 mol (1*10-3) - (4×10-6) not 1000mL 0.001-0.00*004 mcl IOTA -- 10-4 Ñnhw%qNn Ai 1000mL i. = 0.001 bÑWw%w'd 1000-10--990 ml 4. Ulmanis HCI 0.10M ÑñogbÑwiÑw = 1000mL ✗ 0.001 Mol 0.10mW IN,nÑ HCl V0.10 = 10mL Md1dÑdUnnWbVT1D 1. Brod"lÑU pH v09 HNOZ HNO, 4-1,1001-1 0.1M Kai? ① [thot] V09 HNOZ - ✓ 0.1M ✗ 5.62×10-4 = 9.20×10-7 M pH - - - log( 7.20×10-31=310910-10872 =3 -0.86--2.14 • • • pH Cthcootl > HNOZ>HNO} Ka HNG > HNOZ> CHYCOOH ② [tho" ] vosHNGlbbnnoii.NO/o1--o.lMpH---logax1o-'1--llog1o-1og1-- I KAXpµL ?⃝ [1-130-1] Vos CHHOOH-ilo.IM ✗1.80×10-5=1.34×10'M pH = -10811.34×10-31=310%10 - logi .}4--3-0.13=2.87 motion;a8mwÑ riwvrifnww.sn wound 16
2. pH V09 CH}NHz 1.0M bÑW HCl pH 3 100mL Nsfw CH}NHz-11-120 ( H,NH, -1+0%1" " ↳ Ky [OH] - ✓ 1.0M ✗ 4.57×10-4 = 2.14×10'M pot-1=-1 gC2-14×10-4=2- log2.14 = 1.67 pH -14-1.67=12.33 pH nos NH, 1.0M Nth -11-120 # NHI -101-1' Kb [OH] = f- 1.0×7.8×10-5 = 4.24×10 -3M pot = - log (4.14×10-9) Mol .tw - -1×10-41 -11×107=11×104--7.1×1 o→md = 310g 10 - log 4.24 40×10-4 =) - 0.63 = 2.37 Visa -200 mL pH : 14-2.37=11.63 11-1,0+1 Iasi = 1.1×10-3 not ✗ 100mL . -5.5×10"M 200mL 1L 10.5.2 Moriah pH IWÑvdid1VniNbunt pH . - - log 15.5*10" ) 1. pH ÑBMDÉ =3 logic - log5.5 2. Swain lad} aka 㱺 Kzn pH :3 - 0.74=2.26 ~ Hinttho In ' -11-1,0-1 acid form base acid [HIN] > to 1ñÑvosgVnra [In ' ] [In'T > to IÑÑvosWw CHIN] Kai [Hit][In] [ HIN] [1-1,0-1] - - Kzn ✗ [ HINT [ In- ] - loglthott-logken-logIH.tn] [ In] pH=pK±n - log [ H1n1 [In] pH - - pk-tn-lagl.INI [ HINT mouths pH vosmiwofiqwnfdoyofonl.no} pH range _-pKIn -1-1 bits pK±n - 5 pHrange -- 5-1-1 = 4-6 I long - falafels ) i. pH 24 Minions pH 76 shirttails pH - -4-6 Keshia www.nfm8mwohriwvonifnww.sn bound 16
Exits ÑwÑbAbMoÉ ntwoiñvA B town @ ddbiwÑ 74.4 74.4 TuÑN1nNoDUnf 16.0 77.6 Ñwo¥ÑNIÑW 28.3 18. } pH range - -4.4-6.0 4.4-7.6 Acid Acid neutral Base bbUWÑnÑN 10.5 11 [Hzot] i 3×10-4 pH = 4- log} = 4-0.477 = 3.523 㱺 acid pot-1--14 - pH i 14-3.523 = 10.477 2) Calmly ' Ca" -1201-5 0.2g 200mL [01-5]--2maloti ✗ rimoltattlz ✗ 0.29mHz ✗ 1000Mt = 0.027M 1 Catty 7490-112 200mi/h K NsÑWbNRd 40-121167+2111 -74g/Mol POH = - log (2.7×10-2)=2-1082.7--1.57 pH = 14-1.57--12.43 3) HCl -11-120 > 14,0+-14- 12M 10mL 㱺 500mL ?⃝ FINNISH 1-111=12 Malta ✗ 10mL 1000mL = 1.2×10-1=0.12 me / ?⃝ bÑWW%UÑVd9ñVÑW9MMo 500mL [1-14]--0.12 not ✗ 1000mL -0.24M . 500mL 1L ① +② [1-14]=12 molt-14 ✗ 10mL ✗ 1000mL 1000mL 500mL 1L = 0.24M pH : -10g( 0.241 = 0.62 4) NaOH > Nat -101-1' ?⃝ pH -11 JENNI 500 mL ptltpol-1=14 pot-1--14-11--3 ur [OH - Join pot =3 [01-1]=1×10- P -01-1=1*-3 µ TINNIN OOH - =/ ✗ lotmol ✗ 500mL = 5×10-4mW 1000mi www.nfmifmwohriwwonifnww.sn bound 16
① pH-40 V8 Nino 100 mL potti 14-10--4 EOH' ]: Hyo-4 tinsmith 01-5=1×10-4mW 1000mL ☒ 100Mt = 7.0×10-5 ma) ① w9ÑloNiñVn%ÑdswdwÑy Ñ7YsrÑNN = 5×10-4 -17×10-5 ma) VENIN 500+700 ML [01-5] noir = 10-415+0.1 ) moi ✗ 1000mL e- 8.5×10-4 M 6×102 ML 7L [OH] - -8.5×10-4M POH = - log 18.5 ✗ 10-41 = 410g 10 - log8.5 - - ¢ - 0.97=3.07 pH : 14-3.07 - 10.93 www.nfa8mwohriwvonifnww.sn bound 16
pH - - plea + log [In] CHIN] 2.) = - log 15×10-81 + log [In-1 0.699 CHIN] 2.3--8 - togs + Iog[In] CHIN] log[In-1=-5 CHIN] tgct-n.li#og1o-5) [ HINT [In - I = %bUÑ%* CHINA ¥5 =/toooo - - In - /HIn= 1:10 pH- - pkatlog /In - ] CHIN] 7-- plea -1108,1-0 7-- plea -7 pkai 8-0 Kai to-8 Kyi 10-6 TO :| pH=pkb -110g [In - ] [ HIN] 7-- pkbtlogloipkb -17 Pkb:b Kyi 10-6 Kaito-8 www.nfm8mwohriwwonifnww.sn bound 16
HIN Nano, > Nat -1N 0-21In-1 ptlipkatlagcivoit CHNO, ] = - log 15.6×10-4) 1-logo. 2,25 -1410910-1095.611-1080.225 -14-0.751+1-0.651 = 2.60 pH= 21-0.61-1-1 i 3-0.4 = 310810 - log ) - log [1-40-1]=-10813×10-1 ) [1-1,0-1]=3×10-3M www.nfiafmwohriwakifnww.sn bound 16
10.6 Vytifeihbnsfoituiisnrarifi.vn 1-14-1 NaOH 9 Nall -11-120 strñ str base salt Neutralization reaction " Ugiriorvlntibñro" Nat .ci isiwin Hydrolysis 㱺now pH7 [1-1,0-1]=[01-1]--1×10> M CH]C00H + NaOH > CH]C0ONq -11-120 CH,C00Na > CH , COO - 1- Nat m CH}CO0- +1-120 Cthcootl -101-5 base onto NH¢N0, > NH¢t+Noj Hydrolysis:NI-14++1-120 7- NH ,-11-130-1 pHl7 Ex. NaOH 0.1M 10 mL > 500mL NaOH > Nat -101-1- i. Jimmerson OHi 1m01 OH- ✗ 0.1mW NaOH ✗ 10 mL Imo/ NaOH 1000mL = 1×10-3 Mol 2. Soil :[01-1-1=1×10-1 Mol 900mL ☒ 1000mL 1L = 2×10- ' M so/ 2 : [01-1]--1 maloti ✗ A.1mW NaOH × 10mL 1000mL = 2×10 " Imo / NaOH 1000mL 500mL 1L 3. POH = - log (2×10-3) = 3 - log 2 =3 - 0.301--2.699 pH : 14-2.699 = 11.301 www.nfm8mwohriwvrifnww.sn bound 16
10.7 NÑMtnrMnrN - bunt misinformation D. 56 1. tint Natori → NAA -11-120 104.22 glmol Mi ? 0.204 GHA 8 mL 12mL nil ① tinsmith HA - - l MOIHA ✗ 0.204g = 1×10-3 mol 204.22g NwTNbNqd HA ?⃝ [ NaOH] = Imo/NaOH ✗ 1×10 " MOIHA IZMLHA ✗ 1000mL 1m01 HA 12mLHA 8 MLNAOH the - - 0.129 M , Cmol /L) ?⃝ 1-② N2ÑNbNRd [NaOH] - -1mWNaOH ✗ Imo/ HA 204.zzg ✗ 0.20¢ JHA ✗ 12mL HA ILMLHA gmLNagµ ✗ 1000mL Imo/ HA 1L Winslow = 0.125 Molk - ÑWinVn%ÑW M= ? - nhodiñVW7M9I7W - nnIdUnnNÑNÑWM → Tiwari wanna 㱺 titrate igwoi.mn,o&wÉvw%, - qnYñ end point goin - i ,nNNi Equivalent point Yanina_=wI㱺 moln-cid.no/base 2.tiii+NaoÑ > Nacl -11-120 Mi ? 0.2M 10mL tritrate 㱺 10mL Mo /Acidimol Base ?⃝ i1WiyTwN NaOH = 0.2mW ✗ 10mL = 2×10-3 Mol 1000mL ② [1-14]=2×10' } not ✗ 1000mL 10mL 7L = 0.2M ① +② [1-14]: lmoltlcl ✗ 0.2mWNaOH ✗ 10mLNaOH ✗ 1000mL Imo/ NaOH 1000mL NaOH 10mL 1-14 7L www.nfm8mwohriwwonifnww.sn bound 16
2. ci + Ba+18H52 > BaCk + 2H28 n = ? 0.2(md)(2) 10 mL tritrate => 1mL ① viwwe BaCOH)2 = 0.2 mal x10 mL = 2x10 " molB = morA boat casing 1800mL ②CHC1J = 2mo/AX(x10" molB=A x1000mL I MOB IOMLA IL = 0.4M v 5M5 =cavaqe (A= 2x0.2x18 3 In", dawh 10 = 0.4M EmoIT?3:mostensmermprionat = 0.4 m///L, M Conversion factor 008sWnr 10.7 -.- 1. HNOs + KOH < kNOz + H2O In CVA = CVB M= 2 0.1M (A = 0.1528 25 mL tritrate = 20 mL 25 mor Acid = mo. Base = 0.08M ①Viwswed KOH = 0.1md 120 mL = 2010 - mor 1000mL ②CHNOS] = 2x18" ma x1000mL = 0.08 M 25mL 12 * CHN0z] = molA x 0.1 molB X20 mLB x 1000mL = 0.08M 1md)B 1000MLB 25MLA 12 2. NaOr+riCT > NaCl+428 1.0M 0.002M 100mL 1429 4u8d ↓ uvd 000] [Hy0t] 1118d) ①Vies68N0 H2 = 0.002 mol +100 mL 1000mL = 2x15-4 mat ②dWiNVvSNaOHIyea l1mLconval = TomoximL +1yrd = 500mete 9vizzwowwes HCI nisew molAcid-mclBase 2x154ma -5x155mol 20154m01- 0.5x154ma = 1.5410mot wishoonwa inworth wishcount16
④ renminbi"NÑwvo Hellmuth =L. 5×10- 4m01 ✗ 1000mL 100mL 1L :[1-1,0+7=1.5×10-3 moi /L pH = - log (1.5×10-3) i 7- logl.5 =3-0.18=2.82 14 A) moi NaOH : 1.0mW ✗ 1mL 1000mL 20µg@ ✗ 449N = 2×10 - 4m01 NaOH 11-1305=1×10- P" 2×10-4 Mol Hll g1×10-7 into imnwinhriwnntasiinhrninnonowjnienwooin.HU"winwwiñoÑbÑwnñs pH -7 pH=7 - o Yog 30+1--71%10- tog - - Fogel ✗ 10-7 [Hz 7=1×10-7M =[ OH] 16 UGA ) = 10m01 1mL 1000mL 2oyqg× GUNN = 3×10-4 MOI NaOH 2×10-4 Mol HU insular NaOH vÉbuÑd -7×10-4 - 2×10-4 = 1×10-4 Mol i. [01-5]=7×10 " Mol ✗ 1000mL = 1×10'M 400mL ' IL [ 1-1,0+3=7 [1-1,0+1101-5]=1×10" M [1-1,0-1]=1×10-14=1×10" M 7×10-1 pH= - logo ✗ 10-111 = 11-1081 = 11 NH } -11-14 > NH¢Cl 25mL 0.90M town 32.40mL NINTH tonga NH} - -14-11137 -77µmol TOWN = 17g NH} ✗ Imo/ NH} ✗ 0.5 no/ Hcl g) 2.40mL1-148100 1 no/ NH} Imo/ HCl 1000mL 25 MLNH, = 7.1 Yo hvv www.nfiafmwohriwvonifnww.sn bound 16
I.Window bring (61-1806--6112) -11181-16116) -476 glmol 2. %W1w= 1768161-1800 ✗ 1m01 ( 61-1806×0.02 MOINAOH ✗ 15.2mL NaOH ✗ 100 lmolcotlgob 2m01 NaOH 1000mL NaOH 0.1 gvo = 27% w/w 10.8 Ññd¥dWÑWoNo§ MIMI 10.5 ?⃝ ( 1-13001-1 ÑU CH}COONa CH}cooNa→CH}Coo- + Nat CHzlOOHICHXOO-boiwthlctlzcoo-ttlzote.CH >COOH -11-120 - both OH' CH, COOH -1-01-5 , ' ( Hgcoo-1-1-120 ② H]P0¢ÑUNaHzP0¢ HzP04/HzP0F - bÑN Hzot Hzpoci -11-1,0-1=1-1,1004-11-120 - bÑN OH' HgP04-101-1- I HzP0¢_ +1-1%0 ⑥ Ntkclnv NH} NHI' / NHS - bÑN thot Nth -11-110+5 Ntkittho - void OHNI-14++01-1 - 㱺 NH} -11-120 MJWdoUnnNbV%o unit 68 1. CH} COOH / CHzC00Na - both Hyo-1 C.Hgloo- -11-130 CH} COOH -11-120 -briar OHCHycooH-IOH-T-CHfoo.tl-120 2.11-14-1 NaOH 㱺 IN'bÑwVWoWo! • strong' 2.2 Natlcoj bin: Naito} www.nfmifmwohriwakifnww.sn bound 16
Natlco} ' Nat-11-1105 Nazca}→2Na+-1cg" ' tkoi1co} U.IM 0.1M 2.3 CH>COONA ÑUCHCOOH Cthlootlrctlzcoo2.4 H,PO¢n% Northport NaHzPO4→Na+tHzPO¢ ' thpoclthpoi 0.5M 0.5M 3. Ntkcl /NH) NHI -11-40=1-40+-1Nth pH. - pka - log [NHI] [Nth] ① INNIT :[Nth] pH -- pka - logs pttpka = -10g 15.56×10-11 = 1010g - log5.56--10-0.745--9.255 ② [ NHI] > [NH} ] 2M IM plt.pk/a-log2,--pka -0.301 pH< pka pl-1--9.2505-0.301--8.95<47 pka pH ③ [NHI' ]C[ NH}] IM 2M pH - - pKa - logtz = g. 255+0.301 pH - - 9.556 pH > pKa I. 4-131001-1 + NaOH ' CHzC0ONa -11-120 0.7M 0.1M }1ÑÑwVWbÑdÉ 10mL 10mL JudiNIU 㱺 now CHyC00Na → ( Hgcoo- + Nat 11-13100-+1-120=4-41001-1-1I base 2. Ñwm%wd CHYOONaniiriniiw-lmolcthlOONaxo.IM/CHzC00HxlomL lmolctblootl 1000mL - -1×10-7 Mol www.nfm8mwohriwvonifnww.sn bound 16
2. [ Cthcoo' ] = 1×10-1 not ✗ 1000mL 20mL 1L = 5×10-1M CHHOO.tt/zO--CHzCoOHtOHsxio-2COH-J-- ✓ 5×10-1×5.56×10-10=5.3×10-6 M 3. POH = - log 15.3×10-4 = blog to - log 5.3 - 6- 0.724=5. 276 pH ⇐ 14-5.276 = 8.724 Base MoidddUNnNbÑW Unit 69 1. thpoc 1- NaOH > Na>Poa 0.1M 0.1 M= ? 10mL 5mL Mento uwn㱺Ivn4uwnW9was M? i. • www.awt/,po4n-HiiwmririnU.griovi--1MoIHyP04 ✗ 0-1 MOI NaOH ✗ 5mL = 5 ✗ to -4mW 1m01 NaOH 1000mL 2. imdb.NO v09 It]P0¢ÑÑoÑ, 10.1M 10mL) = 0.1 Mol 1000mL ✗ 1° Ml = 1×10-3 my i. Jinshan typo, viii.unto 1×10-3 - S ✗ lo - ¢ mel 10×10-4 - 5×10-4=5×10-4 mol 4. (1-1,1004)=5×10-4 Mol ✗ 1000mL 15 mL 1L i 0.033 M s.ilwowTNNWINatkpoyndwint.is = 1m01 Natlzpoq ✗ 0.1 moi NaOH 5mL -- 5×10-4 Mol 1m01 NaOH 1000mL 6. CHZPOLI ] -5×10-4 Mol ✗ 1000 mL = 0.033 M 15mL 7L a. ptl-pka-logcth.PE [Hzpoci] pH- - pkatogl = - log(6.92×18)--3/ oglo - log 6.92=3-0.84--2.16 Nishant,M§mwÑ riwakifnww.sn bound 16
*ÑvdiNiÑWbWo&n%wmo Henderson - Hassel batch 4-1,1001-1 /CH ,COOCH,cooH+Hzo=+ѧcoo' ka -11-1,0+110-1,100 ] [ CH} COOH] [Hit] - ka ☒ [ CH, HOH] [ Ctb Coo' ] - log[1-1,0-1]=-1ogkalog [CH}COOH ] [(Hgcoo- ] ptl-pka-logcctlzcooi.tl ] [ CH, Coo ] nnwÑwWwÉ pH Kupka ① [EH}COOH ] - - ((Hzcoo] pH =p Ka - logs pH - - pKa ② [4-1,1001-1] > [( Hzcoo- ] 2M 1M pH - - pKalog} pH : pKa-0.301 pH ( pKa ⑦ ICH} COOH ] ( [Chito] 1M 2M pH - - pKa -10Gt =pKa - Hagi -10gal ptt-pka-o.vn pH > pKa ② 4-1300-+1-1,0=4-1,1001-1-1 Kyi (4-131001-17101-5) [ CH} COO] [ OH]=Kb[Ctlzcoo' ] [CHZCOOH- ] - logcoti] = - logkb - 1og[ G-blot] [ (tycoon] pbH=pKb - loglctbcooI [CHZCOOH ] pH + pot-1--14 pH - -14 - POH pH = 14 - pkb-logcc.tl, 00-3 Nishant, 8mmol riwvrifnww.sn bound 16 ( CH} COOH ]
HCOOH -11-120=1-130++1-1100 - 3.0×10-3M 1.WoÑWbnqn 1-11004=1123-112+16121 _- 46g/Md 2.nnWbÑWÑWVd9Ñ1Fn¥d1V HCOOH , [1-11001-1] moth = IMOIHCOOH ✗ 23g ✗ ↳= 0.05M 46g 3. Ñoevni WIND bbolnoif = 100% 0.05MHz, -1 ✗ ''0×10 ' > MHzO+= 0.3 = 6% 0.05 4. ntingnimobiornois. Kai [1-1,0+111-1100] = 17×10-32 0.05- g. go} = 9*106=1.91×10-4 [1-11001-1] 0.047 NH} -1420¥ NHI +01-5 4.3g 4.2×1074 4.2×10-3M 250mL 1.NsÑNbNJi NH, -14+7341=17 g /MOI 2.( NH, ] = 1m01 ✗ 4.3g ✗ 1000mL = 1.01M 17g 250mL 7L 3. %msbbMnÑo= look 1.01M OH - ✗ 42×10%01-5=0.4290 ¢. Ky=[NH¢+t[OH ' ] =/4.2×10'T 1.01 -0.0042--17-64×106=1.80×10-5 ( NH}] CbH}l00H -11-120=1-130++161-15100 0.20M Ka -- 5.75×10-5 Kai [1-1,0*1141-15100] = ✗ 2 0.20 - ✗ ' ¥-71000 f- GOHSCOOH ] ✗ = [1-1,0+1 = 0.20×5.75*10-5 = ✓7.15×10-5 = ✓ 17.5×10-6 =3. 39×10-3 M pH = - log[1-1,0-1]=-1084.39×10-3 =)/ oglo -1083.39--3-0.53=2.47 pot-1--14-2.47--11.53# www.nfm8mwohriwakifnww.sn bound 16
CoHsNHz+HzO CcHsNHz+ OH - 0.20M Ky = 7,47x78-10 -x + X + X X =SOHT = 1 0.28 x 9.47+18-1 pH+pOH = 14 = 1.22x40- pOH=-log[0H-] 5N pH =74-4 = - log(1.22x(55) = 5/og 10 - 1097.22 = 5 - 0.886 = 4.914 pH = 2.45 CHOOH + H20 = H30 + + CH3C08- 4.W/V = 2+0.45+ 1 - 1 |a= 1.5x155 3.5x18 - 3M = 60g x 0.68 mot x100 =3 - 8.55 Ka=CHy8+][CHy(00] 1m0) 1000ML =310910- 1083.5 CCHyCOOH] = 4.08% W/ Nog(H30+] =- Ng (3.5 x 10") [Hy8+] =3.3x183N 1.8x155 = (3.5x183,2 (X3.5x183) 1- 3.5x13: (3.5x153)2 1.8x10- 5 X = 0.68 + 0.8835 CCHCOOH] = 0.68 mol/L wonterion City280H=((72) + 4(1) + 2(76)=60g/mol 3.1 Cacl -> ca" ++225&wird HydrolysisI CaCt +20H-> CaCOH)orford H++2i -> HCI Modoir 5156:0121881519 5.2k2203 - 2k+ + 2032- k+ + 01 - -> 40H 18s(r(03+ H20*H(85 + 0H - H + + cO2-->H905Asdore 5.3 LiN -> (i + + sN- (i + + 0H - ->Li0H(soon H + + CNH(N CNTH20 = HCN +OH10d wishoonwa inworth wishcount16
5.4NH¢NO> → NHj-NOIH-l-l.NO, - → HNO, nsnbbn ' NHj-OH-T-NH4OHbvntdoWNI-kt-HLOT-Hzot-NHr.MN 5.5 lNH¢)zP0¢→3NH¢4Po¢* NHI' -11-120=1-1,0+-1 NH, Ka _- 5.56×10-10 Pop-1-1-120=-1-11004--101-1 - Ry - -2.09×10-2 Ky >Ka Base NaOH -11-14 → NaCl -11-120 4. [ NaOH ] ndiuwdo 0.10M 0.10M = 5×10- 4m01 ✗ 1000mL 25mL 20mL ] 25-120 Mol 7L bunko UWNYV = 0.01 Mol /L ,M ntirrinuwnnbknns 1.Ñ1WJÑNn 1-14--0.10molxzoml 5. NaOH → Nat -104' 1000mL [01-1]=0.01 M = 2×10-3 not pOH= - log (1×10-2) = 210810-1081--2 2.ÑlwzwÑWD NaOH - -0.10mW ✗25mL pH= 14-2--12 1000mL = 2.5×10-3 Mol 3. ÑmwTWdVd9 NaOH ÑiuN% = 2.5×10-3 -2×10-1=5×10-4 mol pH 25.4 pH - - - logano5) = b- logic -10Gt = bVis pH vo9mrbVoÉvwÑ 3.8-5.4 MN%⑤s%iW NaOH → Nat -101-1- is 1.0×10 ' }M potti - logano-3 =3 / oglu -10Gt =3 pH -44-3=17 www.nfa8nnwohriwwonifnww.sn bound 16
1. Calo, -11-111 → lack -1 Hzcoz wsootiwunnan b 1. Calo} - -401-12-13176) (02-11-120 too glmol 2. Mgco, -121-14 → Mcgclz -1102+1-120 2. Mgcoz -24-1-12+31167 3. Mgo-11-14 → Mgck -101-1,1%-1-1 = 84 grmol ?⃝MgO - -24-116--40 4. Mq(OHI, -121-14 → Mgclz -121-120 4. Mgcol-172=24+21167-12111 - -58 glmol NaOH -114,50¢ → Naz.SO¢ -1214,0 0.10M 0.04M 50mL V9W1MiVd9 NaOH -1000 MLB ✗ Lma/13×0.04 mola ✗ 50mL 0.10 mall} Imola 1000mL i40 MLNAOH G.Hgcootlt NaOH → GH5C0ONa -11-120 CbHqC00Na→ Cfthgcoo- + Nat Hydrolysis 161-1510071-120=-41-156001-1 Base www.sina.nl (61451001-1--71121-1611) -12116) - - 122g/Mol mdAimoIB 1.ÑmwtwNVdI (61-151001-1 _- 1m01 ✗1.248=0.01 mol 122g 2.VINI Mintoninner NaOH _- 1000mL ✗ 0.01 Mol = 55.56mL 0.18mW VINI,MrvoJntwimv NaOH 0.78 Molk = 1000 MLB ✗ 1mn01 B ✗ G.18mn13 I MOIA ftp.qg?-xl.24gA - -56.47mL NaOH Wishon;M§mwÑ ÑwU8ÑnW v.911 bound 16
21-14-1 Mglotll, > Mgclz -121-120 0.10M eudonia 10mL 0.10g wdntwbdnanmglotllz.LI -111161-12111=58 glmol Yowlw - -58g Mglotllz imomgcoi.pt 1M¥"" ✗ "10m¥ ✗ 10 ,!Y%gg, ✗ look 2¥44 1000m¢ = 29 % WIN www.nfiafmwohriwakifnww.sn bound 16
UNIT II ↓HEN4 ↳Versewi 1. Na H2 N2 = 0 2. H20 HNO3 25588:8 3.N=oSUewISW-Verrin 4. NazO Nals - Nat,gi+e 5.faceisendeavorwage =rewinders IA e,IA, IHA Nat Mg2+13t 6. NaO --I peroxide +1 - 1 >NaO2 O: I superoxide +2x - 1) 8. OF2 8 = + 2 9.H12trd= + 1 Natt Tourna H= - 1 oN NHNOs apttoIn EstaIn + 2P 4- · 41st + [Fe)(N)s] F(NT · 3k+ (Fe(CN)o]? - mometer 11.I IUCSs +(usOpCag) a InSOpsapl+ (uC3) 2- cuso4Cs) -> suttap + SO4 [IuCH2O)4] 1. Inis - Insage- =Oxidation over er Zuc+(ntags-Zuagst(us) 2. Cuta + re -> [ucs) =Reduction re- =Redox In ser enter, moronine, Oxnose cutr e- none, assi consisterviewed 2.5 down16
MrW 11.2 Tog: Mg, In, (u Turner des MySO4 -Mg + SOG InsO4-In2++ - 9042- CuSOp->(n2++sOp- It 1. f Misters Mg + Mg 2.M Mg 2n> zu aver chin: Labrondy (n) 2u2t> Mightwe 2. 2n + M2t X g In+2n In + Cu2+ ->2n++(u 3. Cu+Mg Cu + 2n2+ X (u+(u2 + -> - Mis8882212,5768 P.98 1.(UCs + Ag+(ag) - (n2+ + Ag Oxidation Rx (u(5) -> (ntcap) +Le x1 riding Reduction Rx (AgT(ng) +--2Ag(s) *2 Redox CUCS) + 2Agt(98) < Cutag)+2AgIS) 8 AP + 6H+ + 2A! + 3H2 Redox Oxidation Rx Redox A1 - Al+ + - X2 2A1 + 6H+- > 2Al + + 3th Reduction Rx 2H* + 25- -H x3 consisterviewed 2.5 down16
midneuraunderwintere rijis 3 && 2 18 2A9+32n+ . +2A13+32n we 3 ↳ensWigse Ox 11 - Al+- x 2 Red zne + 2n x 3 R2dOX 2Al+ 32n->2Alt + 3zu dee's 4 Wa t +5 +1 + 3 - 4 + 4 -4 3 1A+3HN, +4HCl -> HAUCI+NOs +TH20 ser 3 ↑ * IN0 => IYW H20 Oxidation Aut4HC) -> HAuCl4 +- o Reduction HNO,e ->NOL +H2O x3 Redox Au+3HNO, +4HCl -> NO2+H2O +HAuCl4 Mog's s &16 3d 3 I &135628 + 7 - 8 OOH8H3Z+Mn84--32n +2M + H20 + 00H - ·O se 2 4H20 Meerenszigrie Oxidation 32n - xIn'+ x 3 Reduction 8 + x2 4t4N *28 MnO ->2MnOc +wissemer inwere wise res 4H20
MEG80W&27212418 p.99 rele=6 2. IC, ↑ TESstate Nd5x2 = 6 ↳ minced wine durchs Redox (r20+ + 3H2S + 8H+->2(r+ +35 + 4H20 modernsegrise as Oxidation Rx 3HS-IS *e- x3 niers, end Reduction RxEve- +Cre >2(r3+ + TH20 x1 RedoxCrO+2HS +8H - 33 + 2(r3+ + 7Hz8 Misterendor sensi &13:028 Jan 2 I ↓ 911812 = 6e +IC+3-2(r3++ 3S 20+80H- + 8OH- I u ad3 x 2 = 6 *28 IH20 RedoxCr20g+ 3 H2S + H20 -> 2Cr+ +35+8844- michurdering wigrise Oxidation Rx CH2) - 35 +H+fe- x3 - 2 + 14= 12 RedutErg- + 14Ht 22 + 0 + 140 x7 Redox He 3H2) + (r207- 20 H20 - 35+6da consisterviewed 2.5 down16