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1. 0 10 0 20 0 010 e 0 020 0 0010 If the pH 10 for a Ca OH solution what is the concentration of Ca OH e 1 0 x 107 e 50x10 e 50x 10 e 1 0x 10 2 Copyright 2006 Pearson Prentice Hall In 2 0 X 0 a digital pH meter A blue color will result when bromthymo blue is added to an aqueous solution of pH range for color change 2 6 8 10 12 14 e NH Cl 4 Methyl violet Yellow Wj Violet e K H S O Thymol blue Red Yellow Yellow 3 Blue Methyl orange Red M Yellow AICI 3 Methyl red Red IP Yetiow e KH PO 2 4 Bromthymol blue Yellow Gg Blue o NI a IH PO Phenolphthalein Colorless WH pink Alizarin yellow R Yellow Wi Red Copynght 2009 Pearson Prentice Hall inc What is the pH of a 0 010 M HCIO solution gt _ lt 1 Hypochlorous Chlorous Chloric Perchloric 1 n 0 2 H HO O H O CQ O H G a CX e NI K 3 0x10 K 11x10 Strong acid Strong acid Increasing acid strength Copyright 2009 Pearson Prentice Hall Inc What is the pH of a 0 010 M HF solution eppure ese pera e Mmmm cafes afd morona epesetiesent Copyright 2006 Pearson Prentice Hall Inc the strongest IS Which aci H S HF HCI HBr HI D eront Emm I m ebrei a cfeealalae spaas 6A 16 Copyright 2006 Pearson Prentice Hall Inc e ClO BrO BrO Which base is the strongest lO e 102. When lithium oxide L1 O is dissolved in water the solution turns basic from the reaction of the oxide ion O5 with water Write the reaction that occurs and identify the conjugate acid base pairs Answer O aq H O OH aq OH aq OH is the conjugate acid of the base O OH is also the conjugate base of the acid H O For each of the following reactions use Figure 16 4 to predict whether the equilibrium lies predominantly to the left or to the right a HPO aq H5O0 l H3PO aq OH aq b NH4 aq OH aq NH3 aq H5O Answers a left b right Calculate the concentration of OH aq in a solution in which a H 2 X 10 M b H OH c IH 100 X OH Answers a 5 X 10 M b 1 0 X 107 M c 1 0 X 105 M a In a sample of lemon juice H 1s 3 8 x 10 M What is the pH b A commonly available window cleaning solution has OH 1 9 X10 M What Is the pH Answers a 3 42 b H 5 3 X 10 M so pH 8 28 A solution formed by dissolving an antacid tablet has a pH of 9 18 Calculate H Answer H 6 6 X 10 19 M An aqueous solution of HNO has a pH of 2 34 What is the concentration of the acid Answer 0 0046 M What is the concentration of a solution of a KOH for which the pH 1s 11 89 b Ca OH for which the pH is 11 68 Answers a 7 8 X 102 M b 2 4 X 102 M
3. Chemistry The Central Science 11th edition ix Theodore L Brown H Eugene LeMay Jr Bruce E Bursten Catherine J Murphy Chapter 16 Acid Base Equilibria Ahmad Aqel Ifseisi Assistant Professor of Analytical Chemistry College of Science Department of Chemistry King Saud University P O Box 2455 Riyadh 11451 Saudi Arabia tt Building 05 Office AA53 a PE Ar fie ol 7 Tel 0146741 98 Fax 014675992 mE Og snisinll 1lg nll 940233 Alo Web site http fac ksu edu sa aifseisi Advanced Materials King Saud University E mail anmad3qel yahoo com aifseisi ksu edu sa Acids and bases are important in numerous chemical processes that occur around us from industrial processes to biological ones from reactions in the laboratory to those in our environment The time required for a metal object immersed in water to corrode the ability of an aquatic environment to support fish and plant life the fate of pollutants washed out of the air by rain and even the rates of reactions that maintain our lives all critically depend upon the acidity or basicity of solutions Indeed an enormous amount of chemistry can be understood in terms of acid base reactions 16 1 Acids and Bases A Brief Review a Arrhenius Definition An acid is a substance that when dissolved in water increases the concentration of hydrogen ions H HCI 7 H ao T Cl A base is a substance that when dissolved in water increas
4. H53O Il HB aq OH aq The equilibrium constant expression for this reaction can be written as BH OH B The constant K is called the base dissociation constant Ihe constant K always refers to the equilibrium in which a base reacts with H O to form the corresponding conjugate acid and OH add H NH3 aq H5O I gt NH aq OH aq Dase Acid d Conjugate aci base remove H The equilibrium constant expression for this reaction is INH T OH Ko INH Lists the names formulas Lewis structures equilibrium reactions and values of K for several weak bases in water Base Ammonia NH Pyridine CsHsN Hydroxylamine H NOH Methylamine NH CH Hydrosulfide ion HS Carbonate ion CO Hypochlorite ion CIO Lewis Structure wm H Q Conjugate Acid NH CsHsNH gt 5H si H3NOH NH4CH HS HCO HCIO Equilibrium Reaction NH H O NH OH CsHsN H O CsSHsNH OH H NOH H O H4NOH OH NH5CH H O NH CH OH HS H30 HS OH CO4 H O HCO OH CIO H O HCIO OH Ky 1 8 x 10 17 x 10 L1 x 107 44 x 10 18 x 1077 18 x 107 33 x 107 These bases contain one or more lone pairs of electrons because a lone pair is necessary to form the bond with Ht Notice that in the neutral molecules in the Table the lone pairs are on nitrogen atoms The other bases
5. Sample Exercise 16 8 Calculating the pH of a Strong Acid What is the pH of a 0 040 M solution of HCIO Solution The pH of the solution 1s given by pH log 0 040 1 40 Strong bases are the soluble hydroxides which are the alkali metal group 1A hydroxides Na and K and heavier alkaline earth metal group 2A hydroxides Ca Sr and Ba Again these substances dissociate completely in aqueous solution otrong bases are strong electrolytes existing in aqueous solution entirely as lons For example 0 30 M NaOH consists of 0 30 M Na ag and 0 30 M OH aq Sample Exercise 16 9 Calculating the pH of a Strong Base What is the pH of a a 0 028 Msolution of NaOH b a 0 0011 Msolution of Ca OH Solution a NaOH dissociates in water to give one OH ion per formula unit Therefore the OH concentration for the solution in a equals the stated concentration of NaOH namely 0 028 M Method 1 1 0 x 107 H T 357 x10 7M pH log 3 57 x 107 12 45 Method 2 pOH log 0 028 1 55 pH 14 00 pOH 12 45 b Ca OH is a strong base that dissociates in water to give two OH ions per formula unit Thus the concentration of OH aq for the solution in part b 1s 2 X 0 0011 M 0 0022 M Method 1 0 x 14 II 455 x 101 M pH log 4 55 X is So ae Method 2 pOH log 0 0022 2 66 pH 14 00 pOH 11 34 Although all the hydroxides of the alkali
6. 0 20 4 9 x 10719 0 98 x 10 x V0 98 x 10 7 2 99 x 10 M H pH log H log 9 9 x 10 5 5 00 The properties of the acid solution that relate directly to the concentration of H ag The Figures compare the behavior of 1 M CHCOOH and 1 M HCl The 1 M CHCOOH contains only 0 004 M H aq whereas the 1 M HCI solution contains 1 M H aq As a result the rate of reaction is much faster for the solution of HCI a the flask on the left contains 1 M CHCOOH the one on the right contains 1 M HCI with the same amount of magnesium metal into the acid H gas is formed The rate of H it formation is higher for HCI Aum EE WC E on the right Eventually the m SEE same amount of H forms in both cases a b b when the Mg is dropped Sample Exercise 16 13 Using A to Calculate Percent lonization Calculate the percentage of HF molecules ionized in a a 0 10 M HF solution b a 0 010 M HF solution Solution Initial Change a HF aq H aq F aq 0 10 M Equilibrium 0 10 x M HJF xx a 5 y43 iii m AUTEM Gee x 8 2 X 10 M x 0 10 x 6 8 x 1075 6 8 X 10 6 8 x 10 5x x 6 8 x 10 5x 68 x 10 0 _ b Vb 4ac i 2a Q2 768 X 10 V 6 8 x 10 4 6 8 x 107 b 2 _ 6 8 X 10 1 6 x 10 2 rs H EF xX T0 M Sebbene concentration ionized Percent ionization of H
7. O HOt aq OH aq H5O l H aq OH aq The equilibrium expression for this process Is Ke H30 OH This special equilibrium constant is referred to as the ion product constant for water Kp The term H5O is excluded from the equilibrium constant expression because we exclude the concentration of pure solids and liquids K H O OH H OH 1 0 x 10 4 at 25 C ee K H O OH H OH 1 0 x 104 at 25 C What makes this Equation particularly useful is that it is applicable to pure water and to any aqueous solution Although the equilibrium between H aq and OH aq as well as other ionic equilibria are affected somewhat by the presence of additional ions in solution A solution in which H 2 OH is said to be neutral In most solutions H and OH concentrations are not equal As the concentration of one of these ions increases the concentration of the other must decrease so that the product of their concentrations equals 1 0 x 10 In acidic solutions H exceeds OHT In basic solutions OH exceeds H Sample Exercise 16 4 Calculating H for Pure Water Calculate the values of H and OH in a neutral solution at 25 C Solution We will represent the concentration of H and OH in neutral solution with x This gives HHOH x x 1 0 x 10 H x01 x 1 0 x 107 M H OH In an acid solution H is greater than 1 0 X 10 M in a bas
8. to realize that proton transfer reactions are generally very rapid As a result the measured or calculated pH for a weak acid always represents an equilibrium condition sample Exercise 16 10 Calculating A from Measured pH A student prepared a 0 10 M solution of formic acid HCOOH and measured its pH The pH at 25 C was found to be 2 38 Calculate A for formic acid at this temperature Solution HCOOH aq H aq HCOO ag H HCOO HCOOH pH log H 2 38 log H 2 38 I 104 42 X 10 M HCOOH aq c H ag HCOO ag 42 x 10 M 42 x 10 M 4 2 X 102 M Equilibrium 0 10 42 x 10 M 42 X 10 M 42 x 10 M 0 10 4 2 x 10 M 0 10 M 42 x 105 42 x 107 0 10 K cER 0 ee Percent lonization The magnitude of K indicates the strength of a weak acid Another measure of acid strength is percent ionization which is defined as EM concentration ionized Percent ionization X 100 original concentration The stronger the acid the greater is the percent ionization For any acid the concentration of acid that undergoes ionization equals the concentration of H aq that forms assuming that the autoionization of water is negligible H lequilibrium x 100 HA Jinitial Percent ionization For example a 0 035 M solution of HNO contains 3 7 x 10 M H aq Thus the percent ionization is H equilibriu 37 X 10
9. x 107 7 00 Neutral z10 X 107 z10 x 107 7 00 Basic lt 10 107 gt L0OX 10 gt 7 00 What happens to the pH of a solution as we make the solution acidic An acidic solution is one in which H gt 1 0 x 10 M because of the negative sign in Equation pH log H the pH decreases as H increases Examples H 1 0 x 10 M calculate pH OH log 1 0 x 10 3 00 3 A sample of freshly pressed apple juice has a pH of 3 76 calculate H pH log H 3 76 Log H 3 76 H antilog 3 76 10 8 1 7 x 10 M H M pH pOH OH M Gastric juice m gt d m l A i d a Tart TARDY one A E 7 A TY N 22221 gt S 24 Iz p E 1 M 3 PIA d J 1 T H rf n P i ES a 4 4 i Lu 1 4 d i Lemon juice 4 Cola vinegar EE These are the pH values for several Wine s l Tomatoes 1x107 4 0 100 1x10 common substances at Banana A Black coffee 1x105 50 90 1x10 25 C cc NN 1x10 60 80 1x10 liil sea Milk The pH of a solution Human blood tears 1x10 7 0 7 0 1x10 Egg whi ater can be estimated using UU Tn PH 1x108 80 60 1x10 Baking soda the benchmark orgy enti 1x10 9 0 50 1x10 concentrations of H Milk of magnesia 4 439 10 499 40 1x10 and OH corresponding S Lime water z i to wh
10. F x 10095 original concentration 79x10 M X 9 0 10 M x _ 68x10 0 010 x SACS IN IE FT 23x10 M 0 0025 M X 23 0 010 M i a Polyprotic Acids Polyprotic acids have more than one ionizable H atom H SO3 aq H ag HSO3 ag Ka 1 7 X 10 HSO3 ag H ag SO3 ag Kn 64 x 10 The acid dissociation constants for these equilibria are labeled Ka and K2 In the preceding example A is much smaller than K It is always easier to remove the first proton from a polyprotic acid than to remove the second oimilarly for an acid with three ionizable protons it is easier to remove the second proton than the third Thus the A values become successively smaller as successive protons are removed TABLE 16 3 Acid Dissociation Constants of Some Common Polyprotic Acids Name Formula Kat Ka K Ascorbic H5C amp 4H amp 4Og 8 0 x 107 16 x 1072 Carbonic H CO4 4 3 x 1077 5 6 x 1071 Citric H3CCH507 74 x 107 1 7 x 107 40 x 1077 Oxalic H5C504 5 9 x 10 64 x 10 gt Phosphoric H3PO 75 x 10 62 x 1073 42 x 10732 Sulfurous H5504 17 x 10 6 4 X 107 Sulfuric H550 Large 1 2 x 107 Tartaric H5C4H40g 1 0 x 10 4 6 X 10 gt oulfuric acid is strong acid with respect to the removal of the first proton Thus the reaction for the first ionization step lies completely to the right H5SO ag H aq HSO4 ag complete ionization HSO on
11. M Percent ionization x 100 x 100 11 HNOp initial 0 035 M Sample Exercise 16 11 Calculating Percent lonization A 0 10 M solution of formic acid HCOOH contains 4 2 X 10 M H aq Calculate the percentage of the acid that 1s 10nized Solution El Lossitiboums 42 x 10 M Percentionization x 10095 IO 4296 ercent 1iOn1Zatiorn HCOOH ne 010M Calculating pH from K Calculate the pH of a 0 30 M solution of acetic acid HC5H4O at 25 C H ag CH3COO aq CHCOOH ag K for acetic acid at 25 C is 1 8 x 10 H CH3COO E a OT 1 XK OID CHCOOH CH3COO ag CH COOH aq H aq 0 30 M oe txM Equilibrium 0 30 x M x M IH 7FCHSCOO 96 yg ag CH3COOH 0 30 x ee 0 30 x 0 30 x 5 Ky 1 8 X 107 7 0 30 x 0 30 1 8 x 10 5 4 x 10 x V5 4 x 10 23 x 10 H x 2 3 X 10 M pH log 2 3 x 107 2 64 Percent ionization of CH43COOH 0 0023 M X 100 0 77 0 30 M Sample Exercise 16 12 Using to Calculate pH Calculate the pH of a 0 20 M solution of HCN K 9 9 x 10 Solution HCN aq H aq CN aq H CN ke HCN aq c H ag t CN aq Mu 0548 xli CN x x 10 Ko 449 X 10 020 x 020 x 020 2 X 49 x 10 19 0 20
12. Niacin one of the B vitamins has the following molecular structure O oy 0 H N A 0 020 M solution of niacin has a pH of 3 26 What 1s the acid dissociation constant K for niacin Answers 1 5 X 10 A 0 020 M solution of niacin has a pH of 3 26 Calculate the percent ionization of the niacin Answer 2 7 The Ka for niacin Practice Exercise 16 10 1s 1 5 X 10 What is the pH of a 0 010 M solution of niacin Answer 3 41 The percent ionization of niacin K 1 5 X 10 in a 0 020 M solution is 2 71 Calculate the percentage of niacin molecules ionized 1n a solution that is a 0 010 M b 1 0 x 103 M Answers a 3 9 b 12 a Calculate the pH of a 0 020 M solution of oxalic acid H C O See Table 16 3 for K and K b Calculate the concentration of oxalate ion C O 7 in this solution Answers a pH 1 80 b C 0 7 6 4 X 10 M A solution of NH in water has a pH of 11 17 What is the molarity of the solution Answer 0 12 M a Which of the following anions has the largest base dissociation constant NO PO or Ny b The base quinoline has the following structure Its conjugate acid 1s listed in handbooks as having a pK of 4 90 What is the base dissociation constant for quinoline Answers a PO K 2 4 X 10 b 7 9 x 10 In each of the following indicate which salt in each of the followin
13. PO 2 Q0 O 9 What is the conjugate acid of SO a HSO b HSO c SO d H4SO The stronger the acid the X Its conjugate base Acids and bases generally react to form their Y conjugates a X stronger Y stronger D X stronger Y weaker c X weaker Y stronger d X weaker Y weaker What is the pH of a 0 0200 M aqueous solution of HBr a 1 00 b 1 70 C 2 30 d 12 30 What is the pH of a 0 0400 M aqueous solution of KOH a 12 60 D 10 30 c 4 00 d 1 40 The A of HF is 6 8 x 10 What is the pH of a 0 0200 V aqueous solution of HF a 1 0 b 2 43 c 3 1 d 12 30 The A of HF is 6 8 x 10 What is the pH of a 0 0400 M aqueous solution of KF a 2 28 b 2 43 c 6 12 d 7 88 Whicn choice correctly lists the acids in decreasing order of acid strength a HCIO gt HCIO gt HBrO gt HIO b HCIO gt HBrO gt HIO gt HCIO c HIO gt HBrO gt HCIO gt HCIO d HCIO gt HIO gt HBrO gt HCIO Which base is the weakest F NH OH o0 e C2HS07 Which acid is the strongest H O H4O HF HCHO NH Which base is the strongest i Ka NO tr id C2H307 7 a i 07 o HCO HC5H30 gt 1 8 1 i 3 3 1077 2 e EP uc 56 10 1 NH 2 OH Negligible acidity If the pH 2 for an HNO solution what is the concentration of HNO
14. T OH aq Base Acid Conjugate Conjugate aci base remove H Sample Exercise 16 1 Identifying Conjugate Acids and Bases a What is the conjugate base of each of the following acids HCIO H S PH HCO b What is the conjugate acid of each of the following bases CN SO H O HCO Solution a HCIO less one proton H is ClO The other conjugate bases are HS PH and CO b CN plus one proton H is HCN The other conjugate acids are HSO H O0 and H CO3 Notice that the hydrogen carbonate 10n HCO is amphiprotic It can act as either an acid or a base Practice Exercise Write the formula for the conjugate acid of each of the following HSO F PO CO Answers H SO HF HPO 7 HCO Sample Exercise 16 2 Writing Equations for Proton Transfer Reactions The hydrogen sulfite ion HSO 1s amphiprotic a Write an equation for the reaction of HSO with water in which the ion acts as an acid b Write an equation for the reaction of HSO with water 1n which the ion acts as a base In both cases identify the conjugate acid base pairs solution a HSO3 ag H2O l S03 ag H30 aq The conjugate pairs in this equation are HSO acid and SO conjugate base and H O base and H O conjugate acid b HSO aq H90 l H5503 ag OH aq The conjugate pairs in this equation are H O acid and OH conjugate base and HSO base and H5SO conjugate
15. acid ee Acid and Base Strength ACID BASE o HCI CI D Strong acids are completely E HSO Hso amp dissociated in water HO mno No 2 Their conjugate bases are quite weak H4O ag H O HSO So s Weak acids only dissociate partially HPO HPO in water HF F z Their conjugate bases are weak bases HCHO CHO E u Eco HCO x Substances with negligible acidity 34 us tX E do not dissociate in water P HO HPOJ Their conjugate bases are exceedingly z NH NH3 strong HC CO z HPO PO Example 1 73x EE CH contains hydrogen but does not Ts Hie gt loe demonstrate any acidic behavior in n a H protonated Water Its conjugate base CH is a Y cu cu HOo Strong base E l A The stronger an acid the weaker is its conjugate base the stronger a base the weaker Is its conjugate acid In any acid base reaction the equilibrium will favor the reaction that moves the proton to the stronger base HCl g H O gt H30 aq CI ag H O is a much stronger base than Cl so the equilibrium lies so far to the right that K Is not measured K gt gt 1 CH3COOH aq H2O H30 aq CH3COO aq Acetate is a stronger base than H O so the equilibrium favors the left side K lt 1 Sample Exercise 16 3 Predicting the Position of a Proton Transfer Equilibrium For the following proton transfer reaction use Figure 16 4 to predict whether the equilibrium
16. at is responsible for the characteristic properties of aqueous solutions of acids Because the emphasis in the Br nsted Lowry concept is on proton transfer the concept also applies to reactions that do not occur in aqueous solution In the reaction between HCI and NH for example a proton is transferred from the acid HCI to the base NH3 1 CI H 7 Cl m as H H Lets consider another example that compares the relationship between the Arrhenius definition and the Brensted Lowry definitions of acids and bases an aqueous solution of ammonia in which the following equilibrium occurs Ammonia is an Arrhenius base because adding it to water leads to an increase in the concentration of OH aq It is a Brensted Lowry base because it accepts a proton from H O The H O molecule in the equation acts as a Brensted Lowry acid because it donates a proton to the NH molecule NHs ag H5O I NH4 ag OH ag An acid and a base always work together to transfer a proton In other words a substance can function as an acid only if another substance simultaneously behaves as a base o be a Brensted Lowry acid a molecule or ion must have a hydrogen atom that It can lose as an H Ion To be a Brensted Lowry base a molecule or ion must have a nonbonding pair of electrons that it can use to bind the Ht ion Some substances can act as an acid in one reaction and as a base in another For example H5O is a Brenst
17. dium hypochlorite NaClO to enough water to make 2 00 L of solution has a pH of 10 50 Calculate the number of moles of NaClO that were added to the water K 3 3 X 10 solution pOH 14 00 pH 14 00 10 50 3 50 DH 210 232 x10 M CIO ag H0 HcClO OH ag mtia m 0 9 anra E EA Equilibrium x 32 x 10 M ET 32X10 M 3 2 x 10 M HCIOI OH 1074 TIL LJ NC L Ben WT Y 33 x 107 CIO 7 x 32 x 10 4 oo Ig G2 x 1075 T 3 2 x 107 0 31 M Du X a 16 8 Relationship Between A and A To see if we can find a corresponding quantitative relationship lets consider the NH and NH conjugate acid base pair Each of these species reacts with water NH ag NH3 aq H aq NH3 aq H9O l 9 NH aq OH aq H5O H aq OH aq _ NH3 H INH NH4 OH NH Ky NH3 H NH 4 OH NH NH3 H OH Ky K x ky K and K are related in this way Therefore if you know one of them you can calculate the other This relationship is so important that it should receive special attention The product of the acid dissociation constant for an acid and the base dissociation constant for its conjugate base equals the ion product constant for water K X Kp Kw As the strength of an acid increases larger Kj the strength of its conjugate base must decrease smaller A so that the produc
18. ecause the number of decimal places in the pH 1s two a pOH and Other p Scales Cl m I e he p in pH tells us to take the negative base 10 logarithm of the quantity in this case hydronium ions e Some similar examples are pOH log OH px logx pK z log K Larger the x smaller the px Because H 0 OH K 1 0 x 10 14 we know that log H O log OH log K 14 00 or in other words pH pOH pK 14 00 Measuring pH For accurate measurements one uses a pH meter which measures the voltage in the solution n The device is a millivoltmeter and the electrodes immersed in the 7 solution being tested produce a ui voltage that depends on the pH of om the solution ARA A voltage in millivolts which varies with the pH is generated when the electrodes are placed in a solution This voltage is read by the meter which is calibrated to give pH For less accurate measurements one can use Indicators special chemicals that change color if there is a change in the pH caused by adding an acid or alkali Indicators change color at different pH values Litmus paper e Red paper turns blue above pH 8 Blue paper turns red below pH 5 Universal indicator papers RUE Epp es Papers impregnated with several indicators cee S L TEST PAPERS Universal d pH 1 pH 14 Indicator Pap
19. ed Lowry base in its reaction with HCl and a Brensted Lowry acid in its reaction with NH a substance that is capable of acting as either an acid or a base is called amphiprotic An amphiprotic substance acts as a base when combined with something more strongly acidic than itself and an acid when combined with something more strongly basic than Itself e g HCO HSO H O EN Conjugate Acids and Bases HX aq H5O I lt X aq H3O ag In the forward reaction HX donates a proton to H5O Therefore HX is the Brensted Lowry acid and H O is the Brensted Lowry base In the reverse reaction the H4O ion donates a proton to the X ion so H4 O is the acid and X is the base When the acid HX donates a proton it leaves X which can act as a base Likewise when H O acts as a base it generates H4O which can act as an acid An acid and a base such as HX and X that differ only in the presence or absence of a proton are called a conjugate acid base pair Every acid has a conjugate base formed by removing a proton from the acid for example OH is the conjugate base of HO and X is the conjugate base of HX Every base has associated with it a conjugate acid formed by adding a proton to the base Thus H O is the conjugate acid of HO and HX is the conjugate acid of x remove H HNO aq H5O I NO aq H30 aq Acid Base Conjugate Conjugate base acid add H add H NH3 aq H5O I NH aq
20. er o9H1 pH14 10 Books each 20 strips Total 200 strips y manutschore variety of on indicetor papes coena ee west postie nore d apicmmon in educicem ere medcree ang indost Methyl violet Thymol blue Methyl orange Methyl red Bromthymol blue Phenolphthalein Alizarin yellow R pH range for color change 4 b 8 10 12 Yellow Violet Red Yellow Yellow E Blue Red Yellow Red BE Yellow Yellow aM Blue Colorless mg Pink Yellow E Rea The pH ranges for the color changes of some common acid base indicators Most indicators have a useful range of about 2 pH units Methyl orange changes color over the pH interval from 3 1 to 4 4 Below pH 3 1 It is in the acid form red In the interval between 3 1 and 4 4 it is gradually converted to its basic form which has a yellow color By pH 4 4 the conversion is complete and the solution is yellow 16 5 Strong Acids and Bases e he seven most common strong acids are HCl HBr Hl HNO HCIO and HCIO monoprotic and H5SO diprotic hese are by definition strong electrolytes and exist totally as ions in aqueous solution e For the monoprotic strong acids H O or H acid otrong acids are strong electrolytes existing in aqueous solution entirely as ions For example 0 20 M solution of HNO3 aq HNOx ag H5O I H3O ag NOs4 ag complete ionization HNO3 aq 9 H ag NOs aq H NO4 0 20 M
21. es the concentration of hydroxide ions OH NaOH Na aq OH aq 16 2 Bronsted Lowry Acids and Bases ee Bronsted Lowry Definition The Arrhenius concept of acids and bases while useful has limitations For one thing it is restricted to aqueous solutions Breonsted Lowry concept is based on the fact that acid base reactions involve the transfer of H ions from one substance to another Brensted Lowry An acid is a proton donor A base is a proton acceptor HCl 2 H5O I H30O ag CI ag When a proton is transferred from HCI to H O HCI acts as the Brensted Lowry acid and H O acts as the Brensted Lowry base What happens when an acid dissolves in water Water acts as a Br nsted Lowry base and abstracts a proton H from the acid As a result the conjugate base of the acid and a hydronium ion are formed An Ht ion is simply a proton with no surrounding valence electron This small positively charged particle interacts strongly with the nonbonding electron pairs A Brensted Lowry acid must have a removable acidic proton A Brensted Lowry base must have a pair of nonbonding electrons Water molecules to form hydrated hydrogen ions For example the interaction of a proton with one water molecule forms the hydronium ion H4O aq H 0 H H O H Chemists use H ag and H4O aq interchangeably to represent the same thing namely the hydrated proton th
22. g pairs will form the more acidic or less basic 0 010 M solution a NaNO or Fe NO b KBr or KBrO c CH NH Cl or BaCl d NH NO or NH NO3 Answers a Fe NO b KBr c CH NH4CI d NH NO Predict whether the dipotassium salt of citric acid K HC H O will form an acidic or basic solution n water see Table 16 3 for data Answer acidic
23. ic solution H is less than 1 0 X 1077 M Practice Exercise Indicate whether solutions with each of the following ion concentrations are neutral acidic or basic a H 2 4 X 10 M b OH 1 X 107 M c OH 7 X 10 M Answers a basic b neutral c acidic Sample Exercise 16 5 Calculating H from OH Calculate the concentration of H aq in a a solution in which OH is 0 010 M b a solution in which OH is 1 8 X 10 M Note In this problem and all that follow we assume unless stated otherwise that the temperature is 25 C Solution a Using Equation we have This solution is basic because b In this instance This solution is acidic because H OH 1 0 x 10 4 01079 L05 c10 E x 10 12 d OH 0 010 d E Mo OH gt H LO X 107 10x to T HY _ gt 5 6 X 10 M ee OH 1 82 107 H gt OH 16 4 The pH Scale ee pH pH is defined as the negative logarithm in base 10 of the concentration of hydronium Ion pH log H 0 or pH log H In pure water K H O OH 1 0 x 10 14 Since in pure water H 0 OH H O0 N 1 0 x 10 14 1 0 x 10 e Therefore in pure water OH log 1 0 x 10 7 00 e An acid has a higher H O than pure water so its pH is 7 e Abase has a lower H O than pure water so its pH is gt 7 Solution Type H M OH M pH Value Acidic 1 0 x 107 1 0
24. l a Ba CH COO b NH Cl c CH4NHBr d KNO e AI CIO Solution a This solution contains barium ions and acetate ions The cation Ba is an ion of one of the heavy alkaline earth metals and will therefore not affect the pH The anion CH4COO is the conjugate base of the weak acid CH COOH and will hydrolyze to produce OH ions thereby making the solution basic b This solution contains NH and CF ions NH is the conjugate acid of a weak base NH and is therefore acidic Cl is the conjugate base of a strong acid HCl and therefore has no influence on the pH of the solution Because the solution contains an ion that 1s acidic NH and one that has no influence on pH CI the solution of NH4CI will be acidic c This solution contains CH4NH and Br ions CH4NH is the conjugate acid of a weak base CH4NEL an amine and is therefore acidic is the conjugate base of a strong acid HBr and is therefore pH neutral Because the solution contains one ion that is acidic and one that is neutral the solution of CH NH Br will be acidic d This solution contains the K ion which is a cation of group 1A and the ion NO which is the conjugate base of the strong acid HNO Neither of the ions will react with water to any appreciable extent making the solution neutral e This solution contains AP and ClO ions Cations such as AP that are not in groups 1A or 2A are acidic The ClO ion is the conjugate base of a stro
25. lies predominantly to the left that is K lt 1 or to the right K gt 1 HSO aq CO aq SO4 aq HCO aq Solution CO appears lower in the right hand column in Figure 16 4 and is therefore a stronger base than SO CO therefore will get the proton preferentially to become HCO while SO will remain mostly unprotonated The resulting equilibrium will lie to the right favoring products that is K gt 1 H5O aq CO3 aq SO4 aq HCO aq Kc Acid Base Conjugate base Conjugate acid 16 3 The Autoionization of Water a Autoionization of Water Depending on the circumstances water can act as either a Bronsted acid or a Bronsted base water is amphoteric In the presence of an acid water acts as a proton acceptor in the presence of a base water acts as a proton donor In fact one water molecule can donate a proton to another water molecule H O H H9 t 4 0 H7 H H H5O H O H O aq OH aq This is referred to as autoionization of water In pure water a few molecules act as bases and a few act as acids At room temperature only about two out of every 10 molecules are ionized at any given instant Thus pure water consists almost entirely of HO molecules and is an extremely poor conductor of electricity Nevertheless the autoionization of water is very important ee The lon Product Constant of Water H O 4 H
26. listed are anions derived from weak acids eee Sample Exercise 16 15 Using K to Calculate OH Calculate the concentration of OH in a 0 15 M solution of NH Solution NH3 aq H5O Il NH4 aq OH aq NH4 OH NH3 NH3 q H00 NHy q OH aq Ke 18 10 gt Equilibrium 0 15 2 M Ll m ee NH OH 3x 1 x 105 Ky iNIH 015 1 8 x 10 2 5 ELE x 015 1 8 x 10 0 15 1 8 X 10 9 0 70 x 1079 x NH 0H V27 x 10 16 x 10 M Types of Weak Bases Weak bases fall into two categories The first category contains neutral substances that have an atom with a nonbonding pair of electrons that can serve as a proton acceptor Most of these bases contain a nitrogen atom These substances include ammonia and a related class of compounds called amines H H N CHa ag t H O ai aq OH aq H H The chemical formula for the conjugate acid of methylamine is usually written CH4NH The second general category of weak bases consists of the anions of weak acids For example NaClO dissociates to give Nat and ClO ions Na ion is a spectator ion CIO ion is the conjugate base of a weak acid hypochlorous acid Consequently the CIO ion acts as a weak base in water CIO ag H5O HClO ag OH ag Ky 3 3 x 107 Sample Exercise 16 16 Using pH to Determine the Concentration of a Salt A solution made by adding solid so
27. metals group 1A are strong electrolytes LiOH RbOH and CsOH are not commonly encountered in the laboratory The hydroxides of the heavier alkaline earth metals Ca OH Sr OH and Ba OH are also strong electrolytes They have limited solubilities however so they are used only when high solubility is not critical Another strong bases include the oxide ion lonic metal oxides especially Na O and CaO are often used in industry when a strong base is needed The O reacts with water to form OH leaving virtually no O remaining in the solution O ag H5O l 2 OH aq Thus a solution formed by dissolving 0 010 mol of Na O s in enough water to form 1 0 L of solution will have OH 0 020 M and a pH of 12 30 16 6 Weak Acids ee Dissociation Constants Most acidic substances are weak acids and are therefore only partially ionized in aqueous solution We can use the equilibrium constant for the ionization reaction to express the extent to which a weak acid ionizes e For a generalized acid dissociation HA aq H5O I H30O aq A aq or HA ag H aq A aq Either of the following ways depending on whether the hydrated proton is represented as H O aq or H aq _ HOA HA IAAT Ka r Ka HA This equilibrium constant K is called the acid dissociation constant Because H O is the solvent it is omitted from the equilibrium constant expression The magnitude
28. n salts dissolve in water they are completely dissociated Consequently the acid base properties of salt solutions are due to the behavior of their constituent cations and anions Many ions are able to react with water to generate H ag or OH aq This type of reaction is often called hydrolysis The pH of an aqueous salt solution can be predicted qualitatively by considering the ions of which the salt is composed Effect of Cation and Anion in Solution An anion that is the conjugate base of a strong acid will not affect the pH 2 An anion that is the conjugate base of a weak acid will increase the pH B HO BH OH CH4COO a9 H5O I CHCOOH ag OH ag A cation that is the conjugate acid of a weak base will decrease the pH Cations of the strong bases will not affect the pH A HLO AOH H NH ag H2O I NH3 aq H3O aq When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base the affect on pH depends on the Kand K values If Ka gt Ky the ion will cause the solution to be acidic If K gt Ka the solution will be basic f dom b e rd Salt solutions can be neutral acidic or basic These three solutions contain the acid base indicator bromthymol blue a NaCl solution is neutral pH 7 0 b NH CI solution is acidic pH z 3 5 c NaCIO solution is basic pH 9 5 This Figure demon
29. ng acid HCIO and therefore does not affect pH Thus the solution of AI CIO will be acidic Sample Exercise 16 19 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na HPO will form an acidic solution or a basic solution on dissolving in water Solution HPO 4 aq H aq PO ag 16 45 HPO aq H O HPO aq OH aq 16 46 The reaction with the larger equilibrium constant will determine whether the solution 1s acidic or basic The value of K for Equation 16 45 is 4 2 X 10 We must calculate the value of K for Equation 16 46 from the value of K for its conjugate acid H5PO We make use of the relationship shown in Equation 16 40 Eo E GE We want to know K for the base HPO knowing the value of K for the conjugate acid HPO Ky HPO4 A K H5PO X 10 x 10 14 Because K for H PO is 6 2 X 105 we calculate K for H PO to be 1 6 X 107 This is more than 10 times larger than K for H5 PO thus the reaction shown in Equation 16 46 predominates over that in Equation 16 45 and the solution will be basic A Br nsted Lowry acid Is a a proton donor D a proton acceptor c an electron pair donor d an electron pair acceptor A Lewis acid Iis oO 0 9 a proton donor a proton acceptor an electron pair donor an electron pair acceptor What is the conjugate base of HPO HPO HPO PO H
30. of K indicates the tendency of the acid to ionize in water the larger the value of K the stronger the acid Hydrofluoric acid HF for example is the strongest acid listed in Table and phenol HOC H is the weakest Structural Conjugate Equilibrium Acid Formula Base Reaction K Hydrofluoric HF A E HF ag H O l H30 aq F aq 68 x 107 Nitrous HNO H 0 N 0 NO HNO ag HO I H30 ag NO aq 45 x 107 Benzoic CgHsCOOH H 0 C CsHsCOO C H5COOH ag H O 3 x 10 H30 ag CgHsCOO aq B Acetic CH3COOH H O C C H CH4COO CH3COOH ag H5O I EP AUE H3O ag CH3COO aq Hypochlorous HCIO H 0 CI ClO HClO ag H O H3O ag ClO ag 3 0 x 105 Hvdrocyanic HCN H C N CN HCN ag H5O0 l H30 ag CN ag 49X 10 10 Phenol HOCH i o C H O7 HC H O aq H O 13 x 10 H3O ag CeH3O aq The proton that ionizes is shown in blue In almost all cases the hydrogen atoms bonded to carbon do not ionize in water instead the acidic behavior of these compounds is due to the hydrogen atoms attached to oxygen atoms eee Calculating K from the pH In order to calculate either the A value for a weak acid or the pH of its solutions we will use many of the skills for solving equilibrium problems In many cases the small magnitude of A allows us to use approximations to simplify the problem In doing these calculations it is important
31. ole number pH E 1x10 11 0 3 0 JxI10 values e iw z Household ammonia 1x107 2 12 0 2 0 1x107 2 Ss Household bleach NaOH 0 1 M 1x10 130 10 1x10 1x10 140 00 1 1x1079 Sample Exercise 16 6 Calculating pH from H Calculate the pH values for the two solutions described 1n Sample Exercise 16 5 Solution a In the first instance we found H to be 1 0 X 107 M pH log 1 0 x 1071 12 00 12 00 Because 1 0 x 107 has two significant figures the pH has two decimal places 12 00 b For the second solution H 5 6 x 10 M Before performing the calculation it is helpful to estimate the pH To do so we note that H lies between 1 X 10 and 1 X 10 gt 1x 10 56 x 10 lt 1 10 gt Thus we expect the pH to lie between 6 0 and 5 0 pH log 5 6 x 1079 5 25 Sample Exercise 16 7 Calculating H from pH A sample of freshly pressed apple juice has a pH of 3 76 Calculate H Solution From Equation we have pH log H 275 inns log H 376 To find H we need to determine the antilog of 3 76 Scientific calculators have an antilog function sometimes amp yeu P en labeled INV log or 10 that allows us H entilege 5 76 10 LZ ea to perform the calculation Comment Consult the user s manual for your calculator to find out how to perform the antilog operation The number of significant figures in H is two b
32. strates the influence of several salts on pH Salt NaN O5 Ca NO3 gt Zn NO4 Al NO4 4 Indicator Bromthymol Bromthymol Methyl Methyl blue blue red orange Estimated pH 7 0 6 9 5 5 We can summarize the chapter as follows gt For strong acid and bases they will be completely ionize to 100 gt For weak acids and bases we can use the dissociation constants A and Ato find the amount that has been dissociated gt For salts when they dissolve in water H OH they can produce acidic or basic solutions based on the type the reaction of the anion and the cations of the salt with H or OH of the water If the anion in the salt is a conjugate base of strong acid such as HCl the acid will not form in this direction and consequently the H will not form and no change in pH will result f the anion salt is a conjugate base of weak acids such as acetic acid the acid will form and the hydroxide ions will form as well OH giving basic solution For the cation in the salt if it is a cation of the 15t or 2 9 A groups they will not affect the pH but if they are transition metals they will abstract the OH ions from water and result in formation of H ions leading to acidic solution Such effect will depend on the dissociation constants Sample Exercise 16 18 Determining Whether Salt Solutions Are Acidic Basic or Neutral Determine whether aqueous solutions of each of the following salts will be acidic basic or neutra
33. t K x K equals 1 0 x 10 14 at 25 C Remember this important relationship applies only to conjugate acid base pairs Last Equation can be written in terms of pA and pA by taking the negative log of both sides pK pK pK 14 00 at25 C Some Conjugate Acid Base Pairs Acid HNO HF HC5H340 HCO NH HCO OH K Base Strong acid NO 6 8 x 10 F 1 8 X 10 CoH30 43 x 1077 HCO 5 6 x 10 NH 5 6 x 10 CO Negligible acidity O Ky Negligible basicity 15 x 10 5 6 x 107 23 x 10 1 8 x 10 1 8 x 1074 Strong base Sample Exercise 16 17 Calculating K or K for a Conjugate Acid Base Pair Calculate a the base dissociation constant K for the fluoride ion F b the aciddissociation constant K for the ammonium ion NH Solution a K for the weak acid HF is given in Table 16 2 and Appendix D as K 6 8 X 104 We can use Equation 16 40 to calculate K for the conjugate base F Ky 1 iia Um si TS Los T gb T K 68x10 b b K for NH is listed in Table 16 4 and in Appendix D as K 1 8 X 10 gt Using Equation 16 40 we can calculate K for the conjugate acid NH Ky 1 0 x 10 s 86 w 10717 Ky 118x105 16 9 Acid Base Properties of Salt Solutions lons can also exhibit acidic or basic properties Salt solutions can be acidic or basic Because nearly all salts are strong electrolytes we can assume that whe
34. the other hand is a weak acid for which K 1 2 x 10 Because K is so much larger than subsequent dissociation constants for these polyprotic acids most of the H aq in the solution comes from the first ionization reaction As long as successive K values differ by a factor of 10 If the difference between the K for the first dissociation and subsequent K values is 10 or more the pH generally depends on y on the first dissociation Sample Exercise 16 14 Calculating the pH of a Polyprotic Acid Solution The solubility of CO in pure water at 25 C and 0 1 atm pressure is 0 0037 M The common practice is to assume that all of the dissolved CO is in the form of carbonic acid H CO which is produced by reaction between the CO and H O j a Sf ES What is the pH of a 0 0037 M solution of H CO Ros False Solution H2COx a9 e HH aq HCO ag 0 0037 M x AO x 107 M 0 0037 x 0 0037 n Pee x x T cdd eg x 0 0037 4 3 x 1077 1 6 x 107 x H HCO VL6 107 40 10 M pH log H log 4 0 x 10 440 HCO aj H ag O 7 ag 2 5 _ HCO _ 40x 105 7 HCO 40 x 105 y 560010 A COS 16 7 Weak Bases Many substances behave as weak bases in water Weak bases react with water abstracting protons from H O thereby forming the conjugate acid of the base and OH ions Bases react with water to produce hydroxide ion B aq
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