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1. H OH In an acid solution H is greater than 1 0 X 10 Mina basic 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 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 QOx10 1010 me En x 10 12 an DH 0 010 LUN MTM OH gt HY LOX 1093 10 io T HH 2 56x10 M i OH 1810 H gt OH 16 4 The pH Scale 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 pH 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 H5O than pure water so2. H ag HSO3 ag Ka 17 X 107 HSO 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 Because of electrostatic attractions we would expect a positively charged proton to be lost more readily from the neutral H SO molecule than from the negatively charged HSO ion This observation is general 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 Kay K 2 K 43 Ascorbic HoC4HgOy 80105 16 x 10 2 Carbonic HCO 43 x 1077 56 x 1074 Citric H3C amp H50 74 x 107 1 x 10 gt 40 x 1077 VI H C 0 59x10 64x 10 Phosphoric H3PO 7 5 x 102 62 x 1075 42 x 107 Sulfurous H 5O3 17 x 107 64 X 10 8 Sulfuric H5S0 Large 12 x 10 Tartaric H C4H40 1 0 x 10 4 6 x 10 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 H55O ag H aq HSO ag complete ionization HSO on the other hand is a weak acid for which K 1 2 x 10 Because K is so much larger than subsequent dissoci
3. we obtain Because this value is greater than 5 of 0 10 M we should work the problem without the approximation using an equation solving calculator or the quadratic formula Rearranging our equation and writing it in standard quadratic form we have This equation can be solved using the standard quadratic formula Substituting the appropriate numbers gives Of the two solutions only the one that gives a positive value for x is chemically reasonable Thus From our result we can calculate the percent of molecules ionized b Proceeding similarly for the 0 010 M solution we have Solving the resultant quadratic expression we obtain The percentage of molecules ionized is HP aq H aq F ag Initial 0 10 M Change x M Equilibrium 0 10 x M x EEI 00 HF 0 10 x x 0 10 x 6 8 x 107 6 8 X 10 6 8 x 10 5x x 6 8 x 10 5x 68 x 10 0 6 5 Xx 10 13 x 8 2 X 10 M _ b Vb 4ac ja 2a Lo 68 x 104 V 6 8 x 105 4 6 8 x 10 2 _ 68 X 10 1 6 x 10 2 rs Ht I 1 783 HM vM concentration ionized Percent ionization of HF x 100 original concentration 79 x 10 M x 100 7 9 010 M 100 7 2 x 268x104 0 010 x RUE H IFT 433510 M 0 0023 M X 100 23 0 010 M s Polyprotic Acids Polyprotic acids have more than one ionizable H atom H SO3 aq
4. H4O HF HC H40 NH Which base is the strongest i Ka NO tr id C2H307 HP eB a for 0 o HCO HC5H30 gt 1 8 1 i 3 3 107 NH 56 10 3 e CO y EP z NH OH i Negligible acidity If the pH 2 for an HNO solution what is the concentration of HNO 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 50x10 e 50x10 e 1 0 x 1072 Copyright 2006 Pearson Prentice Hall In 2 0 X 1 0 a digital pH meter A blue color will result when bromthymol blue is added to an aqueous solution of pH range for color change 2 6 8 10 12 14 NH CI 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 e lt 1 Hypochlorous Chlorous Chloric Perchloric e 1 n T 2 2 H Q H Q Q HOG HOO Q e K 23 0 x10 K 2 11x10 Strong acid Strong acid e gt Increasing acid strength Copyright 2009 Pearson Prentice Hall Inc What is the pH of a 0 010 M HF solution epar RB ccm reall AD aoa Copyright 2006 Pearson Prentice Hall I
5. Ol HCIO O assume that Equation 16 37 is the only source la HO ACIO ag H ag produced by the autoionization of H5O 32 x 104 M a 132 x 1043M 432 X 104 M We now assume a value of x for the initial Equilibrium x 32 x 10 M EN 32 X 10 M 32 X 10 M concentration of ClO and solve the equilibrium problem in the usual way HCIOIITOH 22 x 103 i Kt Olle B2 WY as x 107 CIO x 3 2 x 1074 We now use the expression for the base a dissociation constant to solve for x g 3 2 x19 3 2 x 105 0 31 M lo Pa X x Thus We say that the solution is 0 31 M in NaClO even though some of the ClO ions have reacted with water Because the solution is 0 31 M in NaClO and the total volume of solution 1s 2 00 L 0 62 mol of NaClO is the amount of the salt that was added to the water 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 NHs3 aq H aq NFi3 aq H O NH ag OH aq H5O H aq OH aq _ NH3 H INH NH4 OH NH Kp NH3 H NH4 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 t
6. The pH of the solution is above 7 because we are dealing with a solution of a base 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 N C HO a CH aq OH ag 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 Na and ClO ions The Nat ion is always a spectator ion in acid base reactions The ClO ion is the conjugate base of a weak acid hypochlorous acid Consequently the ClO ion acts as a weak base in water ClO ag H O J 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 sodium 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 We can calculate OH by using the right pOH 14 00 pH 14 00 10 50 3 50 Equation we will use the latter method here OH 103599 32 x 107 M This concentration 1s high enough that we can a 2
7. 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 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 is 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 1onized in a solution that is a 0 010 M b 1 0 X 10 M Answers a 3 9 b 12 a Calculate the pH of a 0 020 M solution of oxalic acid H5C50 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 n 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 quinolin
8. at equilibrium we can construct the following table HCOOH aq H aq HCOO aq P H HCOO HCOOH log H 2 38 H 10 79 42 TU M HCOOH aq H aq HCOO ag Notice that we have neglected the very small 0 10 M Eeexge eg concentration of H aq that is due to the 0 E i 1073 10 3 pasar autoionization of H O Notice also that the amount of 4 2 x 10M 4 2 x 10M 4 2 X10 M 42x 107 HCOOH that ionizes is very small compared with the Equilibrium 0 10 42 x 10 M initial concentration of the acid To the number of significant figures we are using the subtraction yields 0 10 M We can now insert the equilibrium centrations into the expression for K 42 x 10 M 2x 10 M 0 10 4 2 x 107 M 0 10 M _ 4 2 x 10 4 2 x 1077 1 8 x 107 0 10 K 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 oy 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 equilibrium x 100 HA initial Percent ionization For example a 0 035 M solution of HNO contai
9. its pH is gt 7 Solution Type H M OH M pH Value Acidic gt 1 0 x 10 1 0 x 1077 7 00 Neutral 1 0 x 1077 1 0 x 107 7 00 Basic 10x 10 gt 1 0 x 10 7 00 What happens to the pH of a solution as we make the solution acidic An acidic solution is one in which H 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 76 103 8 1 7 x 10 M Gastric juice Wine Tomatoes Banand c c cz2 Black coffee Human blood tears Egg white seawater Baking soda Dope e Milk of magnesia m Lime water E z Household ammonia gt Household bleach NaOUH UTIM H M Lemon juice 1x10 Cola vinegar Em 1x10 1x107 1x107 1x10 77 1x107 io 1x107 1x10 1x10 4 1xJ9 1x10 4 pH 4 0 5 0 6 0 7 0 8 0 9 0 10 0 11 0 12 0 13 0 14 0 pOH 10 0 9 0 8 0 7 0 6 0 5 0 4 0 3 0 2 0 1 0 0 0 OH M Lp rete A A TLE 1510 H 1x107 1x10 1x107 1x10 7 1x10 1x10 1x10 1x10 1x10 1x10 1 1x10 These are the pH values for several common substances at 25
10. represent the same thing namely the hydrated proton that 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 i i CI H UH C ie Mins 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 NH3 aq H20 lI NH4 ag OH aq 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 aci
11. that when dissolved in water increases the concentration of hydroxide ions OH NaOH Na aq OH aq 16 2 Bronsted Lowry Acids and Bases 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 g 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 O H H O H H Chemists use H ag and H4O aq interchangeably to
12. the EN reaction HCN aq Bay CN aq oom o 0 Equilibrium 020 x M Next we tabulate the concentration of the species involved in the equilibrium reaction letting x H at equilibrium Substituting the equilibrium concentrations from the table into the equilibrium constant expression yields We next make the simplifying approximation that X the amount of acid that dissociates is small compared with the initial concentration of acid that 1s Thus Solving for x we have A concentration of 9 9 X 10 Mis much smaller than 5 of 0 20 the initial HCN concentration Our simplifying approximation is therefore appropriate We now calculate the pH of the solution x x 10 Ka 0 20 x i 0 20 x 0 20 3 10 49 x 10 0 20 gt M x 0 20 4 9 x 10719 0 98 x 107 x V0 98 x 10 9 9 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 aq such as electrical conductivity and rate of reaction with an active metal are much less evident for a solution of weak acid than for a solution of a strong acid of the same concentration 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 cont
13. C The pH of a solution can be estimated using the benchmark concentrations of H and OH corresponding to whole number pH values In biological systems many reactions involve proton transfers and have rates that depend on H Because the Speeds of these reactions are crucial the pH of biological fluids must be maintained within narrow limits For example human blood has a normal pH range of 7 35 to 7 45 Illness and even death can result if the pH varies much from this narrow range 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 10717 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 1x105 56x105 15x107 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 oolution From Equation we have pH log H 75 A log H 376 To find H we need to determine the antilog of 3 76 Scientific calculators have an antilog fun
14. Le Chemistry The Central Science 11th edition 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 Le Building 05 Office AA53 Tel 014674198 Fax 014675992 Web site http fac ksu edu sa aifseisi E mail anmad3qel yahoo com aifseisi ksu edu sa _ 2 Al asl cn an ee isla ni amp inll 3lg nll Blaretexvi pos 8 23 Advanced Materials King Saud University 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 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
15. ains 1 M CHCOOH the one on the right contains 1 M HCI with the same amount of magnesium metal b when the Mg is dropped into the acid H gas is formed The rate of H formation is higher for HCl on the right Eventually the e 3 same amount of H forms in both cases d wr As the concentration of a weak acid increases the equilibrium concentration of H aq increases as expected However the percent ionization decreases as the concentration increases Thus the concentration of H aq is not directly proportional to the concentration of the weak acid For example doubling the concentration of a weak acid does not double the concentration of H aq 6 0 5 0 T L IN ce 40 The effect of concentration on is ionization of a weak acid The 3 percent ionization of a weak acid z decreases with increasing 59 concentration The data shown are for acetic acid 1 0 0 0 05 0 10 0 15 Acid concentration M ee Sample Exercise 16 13 Using 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 a The equilibrium reaction and concentrations are as follows equilibrium The equilibrium constant expression 1s When we try solving this equation using the approximation 0 10 x 0 10 that is by neglecting the concentration of acid that 1onizes in comparison with the initial concentration
16. are HSO acid and SO conjugate base and H O base and H4O conjugate acid b HSO aq H9O l H 503 aq OH aq The conjugate pairs in this equation are H O acid and OH conjugate base and HSO base and H SO conjugate acid 100 H O ionized in Acid strength increases L je D p yY Z ACID HCI HS0 HNO H30 aq HSO HPO HF HC5H40 H PO NH HCO HPO H O OH H CH BASE CI HSO NO H O SO H PO CoH30 HCO HS HPO NH3 CO PO OH O H CH Negligible Weak Hn al A Base strength increases 00 gt protonated in H 0 Acid and Base Strength Strong acids are completely dissociated in water Their conjugate bases are quite weak Weak acids only dissociate partially in water Their conjugate bases are weak bases Substances with negligible acidity do not dissociate in water Their conjugate bases are exceedingly strong Example CH contains hydrogen but does not demonstrate any acidic behavior in water Its conjugate base CH54 is a strong base some acids are better proton donors than others likewise some bases are better proton acceptors than others The more easily a substance gives up a proton the less easily its conjugate base accepts a proton Similarly the more easily a base accepts a proton the less easily its conjugate acid gives u
17. are on nitrogen atoms The other bases listed are anions derived from weak acids Sample Exercise 16 15 Using to Calculate OH Calculate the concentration of OH in a 0 15 M solution of NH Solution We first write the ionization reaction and the corresponding equilibrium constant K expression We then tabulate the equilibrium concentrations involved in the equilibrium We ignore the concentration of H O because it is not involved in the equilibrium constant expression Inserting these quantities into the equilibrium constant expression gives the following Because K is small we can neglect the small amount of NH that reacts with water as compared to the total NH concentration that is we can neglect x relative to 0 15 M Then we have H3 aq H9O I NH aq OH aq NH4 IOH NH3 NH3 aq K 18 10 gt H O NH a OH aq miam os sm x _ NH OH wa T OS CS d b XC J i NH d ea AA 2 18 x10 i15 718 x10 0 15 1 8 X 10 2 7 x 10 v NH 0H V27 X 10 L6 x 10 M Comment You may be asked to find the pH of a solution of a weak base Once you have found OH you can proceed as in Sample Exercise 16 9 where we calculated the pH of a strong base In the present sample exercise we have seen that the 0 15 M solution of NH contains OH 1 6 x 10 M Thus pOH log 1 6 x 10 2 80 and pH 14 00 2 80 11 20
18. asic vae rA 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 demonstrates the influence of several salts on pH Salt Indicator Estimated pH NaN O5 Bromthymol blue 7 0 Ca NO3 Zn NO3 Al NO3 3 Bromthymol Methyl Methyl blue red orange 6 9 We can summarize the chapter as follows For strong acid and bases they will be completely ionize to 100 For weak acids and bases we can use the dissociation constants A and to 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 If 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 re
19. ate K for the conjugate acid NH K 10 14 IC LE TET d RN 18x10 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 when 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 aq 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 a 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 CH3 COO aq H9O I CH3COOH ag OH ag 3 A cation that is the conjugate acid of a weak base will decrease the pH 4 Cations of the strong bases will not affect the pH A H O AOH H NH ag H2O l NH3 aq H3O aq 5 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 b
20. ation 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 HO CO aq H20 1 H yCO3 aq H COs aq H aq HCO ag Solution Initial 00037 M o o Proceeding as in Sample Exercises 16 12 and 16 13 we can write the equilibrium reaction and equilibrium concentrations as follows What is the pH of a 0 0037 M solution of H5CO Equilibrium 0 0037 x M v M t M The equilibrium constant expression is as follows K H HCOX x x 43 x 107 LZ 1m gt o Solving this equation using an equation solving calculator we get H3CO4 0 0037 x x 40 x 10 M Alternatively because K is small we can make the simplifying i approximation that x is small so that Thus 0 0037 x 0 0037 Solving for x we have x x gt 43x 107 The small value of x indicates that our simplifyin
21. cid Is a a proton donor D a proton acceptor c an electron pair donor d an electron pair acceptor A Lewis acid is 2 Oo O 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 HPO 2 Q0 O0 9 What is the conjugate acid of SO a H SO 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 0 c 2 30 d 12 30 What is the pH of a 0 0400 M aqueous solution of KOH a 12 60 b 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 M 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 Which 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 HIO HBrO HCIO Which base is the weakest F NH OH o0 e C2HS07 Which acid is the strongest H O
22. ction sometimes Fe A HP sA labeled INV log or 105 that allows us H m antiogt 5 76 10 L7 X 1g AM 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 because the number of decimal places in the pH is two pOH and Other p Scales bl m 2 e he p in pH tells us to take the negative base 10 logarithm of the quantity in this case hydronium tons 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 z 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 The device is a millivoltmeter and the electrodes immersed in the 7 solution being tested produce a dL voltage that depends on the pH of the solution 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 ee 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 val
23. d 2 pOH log 0 028 1 555 pH 1400 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 is 2 X 0 0011 M 0 0022 M Method 1 Den H o A55 x10 M pH log 55 X 10 11 34 Method 2 pOH log 0 0022 2 66 pH 14 00 pOH 11 34 Although all the hydroxides of the alkali 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 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 generali
24. d in one reaction and as a base in another For example H5O is a Brensted 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 ee Conjugate Acids and Bases HX aq H2O I X aq H3O ag In the forward reaction HX donates a proton to H5O Therefore HX is the Brensted Lowry acid and HO is the Brensted Lowry base In the reverse reaction the H4O ion donates a proton to the X ion so H4O 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 Similarly every base has associated with it a conjugate acid formed by adding a proton to the base Thus H4O is the conjugate acid of HO and HX is the conjugate acid of X e The term conjugate comes from the Latin word conjugare meaning to join
25. dic behavior of these compounds is due to the hydrogen atoms attached to oxygen atoms Calculating A 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 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 The first step in solving any equilibrium problem is to write the equation for the equilibrium reaction The ionization of formic acid can be written as follows The equilibrium constant expression is From the measured pH we can calculate H We can do a little accounting to determine the concentrations of the species involved in the equilibrium We imagine that the solution is initially 0 10 M in HCOOH molecules We then consider the ionization of the acid into H and HCOO For each HCOOH molecule that 10nizes one H ion and one ion HCOO are produced in solution Because the pH measurement indicates that H 4 2 x 10 M
26. e has the following structure Its conjugate acid is 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 following 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 NH4NO 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 in water see Table 16 3 for data Answer acidic
27. ely as ions For example 0 20 M solution of HNO3 aq HNOx ag H O I 9 H3O aq NOs4 ag complete ionization HNO3 aq gt H ag NO3 ag H NO 7 0 20 M 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 Check Because H lies between 1 X 107 and 1 X 107 the pH will be between 2 0 and 1 0 Our calculated pH falls within the estimated range Furthermore because the concentration has two significant figures the pH has two decimal places 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 10 44 H EET 357 x10 M pH log 3 57 x 1079 12 45 Metho
28. equilibrium will lie to the right favoring products that is K gt 1 H5O aq CO3 aq SO4 aq HCO aq K 1 Acid Base Conjugate base Conjugate acid Comment Of the two acids in the equation HSO and HCO the stronger one gives up a proton more readily while the weaker one tends to retain its proton Thus the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base 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 m s 6 H7 H5O H20 H30 aq OFT 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 H O molecules and is an extremely poor conductor of electricity Nevertheless the autoionization of water is very important The lon Product Constant of Water H5O 4 H O H30 aq OH aq H5O H ag OH aq The equilibrium expressi
29. g assumption was justified The pH is therefore 0 0037 4 3 x 1077 1 6 x 107 j Comment If we were asked to solve for CO4 we would need to use x H HCO4 V16 x 10 40 x 10 M Kp Let s illustrate that calculation Using the values of HCO and x H calculated above and setting CO y we have the following pH log H log 4 0 X 10 7 o 4 40 initial and equilibrium concentration values HCO aq H aq CO3 ag Assuming that y is small compared to 4 0 X 10 we have The value calculated for y is indeed very small compared to 4 0 X 10 showing that our assumption was justified It also shows that the ionization of HCO gt is negligible compared to that of H5CO as far as production of H is concerned However it is the only source of CO4 which has a very low EE Es concentration in the solution Our calculations thus tell us that in a solution of H COS 4 0 x 10 Jy carbon dioxide in water most of the CO is in the form of CO or H5CO a a2 7 HCO 7 ES 40 x 1075 small fraction ionizes to form H and HCO and an even smaller fraction EN ionizes to give CO Notice also that CO is numerically equal to K 56x10 5610 M COS lt 16 7 Weak Bases Many substances behave as weak bases in water Weak bases react with water abstracting protons from HO thereby forming the conjugate acid of the base and OH ions Bases react
30. he acid dissociation constant for an acid and the base dissociation constant for its conjugate base equals the ion product constant for water Ka X Kp Ky As the strength of an acid increases larger A the strength of its conjugate base must decrease smaller A so that the product 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 Ka Base Strong acid NO 6 8 x 104 T 1 8 x 10 CyH 0 gt 43 x 107 HCO 56 x 10 10 NH 5 6 x 10 CO Negligible acidity 0 Ky Negligible basicity 1 5 x 10 5 6 x 1019 2 3 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 107 We can use Equation 16 40 to calculate K for the conjugate base F K 14 L0 x 107 1 5 x 10 1 K 68 xig 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 calcul
31. ich 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 strong 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 H PO We make use of the relationship shown in Equation 16 40 Ka X Ky Ky We want to know K for the base HPO knowing the value of K for the conjugate acid HPO 7 Ky HPO47 x RPO Ky 10 x 1074 Because K for H PO is 6 2 X 10 8 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 a
32. nc the strongest IS Which aci H S HF HCI HBr HI eese esee ee afe fn esee aleejert vsenpejesfea pen eeoeelealespeett spalel Copyright 2006 Pearson Prentice Hall Inc e ClO BrO BrO Which base is the strongest IO e 10 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 J OH aq OH aq OH is the conjugate acid of the base O OH is also the conjugate base of the acid H5O 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 H5 O l HPO aq OH aq b NH4 ag OH aq NHi ag H3O l Answers a left b right Calculate the concentration of OH ag 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 10 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 10 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
33. ns 3 7 x 10 M H aq Thus the percent ionization is H equilibrium 37 X 10 M Percent ionization X 100 x 100 11 HNO dJinitial 0 035 M Sample Exercise 16 11 Calculating Percent lonization A 0 10 M solution of formic acid HCOOH contains 4 2 X 10 MH aq Calculate the percentage of the acid that 1s 10nized Solution i IET kaunoa 42 X 107 M Percent ionization HCOOH x 100 o10M x 100 4 2 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 _s a ccm le AU CHCOOH CH3COO ag CH COOH aq H aq 0 30 M oe txM Equilibrium 0 30 x M x M IH NCH COO GX9 ie xo CH3COOH 0 30 x 0 30 x 0 30 x K 21 8x10 0 30 x 0 30 1 8 x 10 5 4 x 10 x V5 4 x 10 23 x 10 H x 223 X 10 M pH log 2 3 x 107 2 64 Percent ionization of CH43COOH 0 0023 M 0 30 M n X 10094 0 77 a 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 ag CN aq Writing both the chemical equation for the H ICN ionization reaction that forms H aq and the K 49 x 10 1 equilibrium constant K expression for
34. on for this process Is Ke H O OH This special equilibrium constant is referred to as the ion product constant for water Km 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 10 4 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 Thus this Equation is taken to be valid for any dilute aqueous solution and it can be used to calculate either H if OH is known or OH if H is Known A solution in which H 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 107 1 10 0 x 1 0 x 107 M
35. p a proton In other words 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 H5O I gt H30 aq CI ag HO 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 H50 H30 ag CH3COO aq Acetate is a stronger base than H O so the equilibrium favors the left side K lt 1 From these examples we conclude that in every acid base reaction the position of the equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base As a result the equilibrium mixture contains more of the weaker acid and weaker base and less of the stronger acid and stronger base 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 lies predominantly to the left that is K lt 1 or to the right K gt 1 HSO ag CO aq SO 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 7 will remain mostly unprotonated The resulting
36. sult 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 neutral a Ba CH COO b NH Cl c CH NH Br 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 CH4COOH 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 HCI 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 wh
37. together Reactions between acids and bases always yield their conjugate bases and acids remove H HNO aq T H5O I NO aq T HO aq Acid Base Conjugate Conjugate base aci add H add H NH3 aq H5O I E NH aq 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 CIO The other conjugate bases are HS PH and CO b CN plus one proton H is HCN The other conjugate acids are HSO H O and H5CO 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 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 HSO ag H9O I 9 SOs aq H3O aq The conjugate pairs in this equation
38. ues pat Litmus paper Red paper turns blue above pH 8 e Blue paper turns red below pH 5 ra E Jes gee a Universal indicator papers Papers impregnated with several indicators evs Ge 24 7 ea TEST PAPERS Universal d pH 1 pH 14 Indicator Paper o9H1 pH14 10 Books each 20 strips Total 200 strips Methyl violet Thymol blue Methyl orange Methyl red Bromthymol blue Phenolphthalein Alizarin yellow R DH range for color change b 8 10 12 Yellow a Violet Red Yellow Yellow Blue Red Yellow Red BE Yellow Yellow aME Blue Colorless m Pink Yellow MM nea 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 These 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 entir
39. with water to produce hydroxide ion B ag H OW HB aq OH aq The equilibrium constant expression for this reaction can be written as _ BH OH7 7 B The constant K is called the base dissociation constant The 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 NH aq OH ag Base Acid Cony ge Conjugate aci base remove H The equilibrium constant expression for this reaction is NH 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 H Notice that in the neutral molecules in the Table the lone pairs
40. zed acid dissociation HA aq H9O I HO aq A aq or HA ag H aq A aq Either of the following ways depending on whether the hydrated proton is represented as H4O aq or H aq _ H30 A HA IH JE amp 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 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 HE p HF aq H O I H30 ag F ag 68 x 104 Nitrous HNO H O N O NO HNO ag H9O I H30 ag NO aq 45 x 107 j Benzoic CgHsCOOH H O C CsHs COO C HsCOOH aq H O 3x 10 H30 aq CgHsCOO aq io Acetic CH3COOH H O C C H CH4COO CH COOH aq H5O I Is H3O aq CH3COO aq 1 A peque CIO HClO HO HO ag Cl a 30x 10 Hydrocyanic HCN H C N CN HCN aq H O I H30 aq CN aj 49 x 10 Phenol HOC Hs O CH50 HC H O aq H O gt 13x10 9 H30 aq CgHsO 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 aci
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