"user manual"
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1. H2O H4O aq CoH4O 5 aq for acetic acid at 25 C is 1 8 x 10 Calculating pH from K The equilibrium constant expression IS H4O 5 HO gt H30 Calculating pH from We next set up a table M Change At Equilibrium 0 30 x 0 30 We are assuming that x will be very small compared to 0 30 and can therefore be ignored AA Lx Calculating pH from NOW a 1 8 x 105 0 30 1 8 x 107 0 30 5 4 x 1076 23 103 Calculating pH from K pH log H5O 109 2 3 x 10 3 2 64 SAMPLE EXERCISE 16 11 Using K to Calculate pH Calculate the pH of a 0 20 M solution of HCN Refer to Table 16 2 or Appendix D for the value of Solution Analyze We are given the molarity of a weak acid and are asked for the pH From Table 16 2 K for HCN is 4 9 x 10710 Plan We proceed as in the example just worked in the text writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H is our unknown Solve Writing both the chemical equation for the ionization reaction that forms H aq and the equilibrium constant K expression for the reaction HCN aq H aq CN aq _ K 49 10 HCN Next we tabulate the concentration of the species involv2. stren Acid and Base Strength e Substances with negligible acidity do not 5 Negligible NO H H H dissociate In water ro heir conjugate bases are cuo ME exceedingly strong x HCO HCO x E HS 3 HjPO HPO NH NH3 E 3 2 Oo 1 gt protonated 2 5 Acid and Base Strength In any acid base reaction the equilibrium will favor the reaction that moves the proton to the stronger base HCl aq gt H4O aq CI aq is a much stronger base than so the equilibrium lies so far to the right K Is not measured K gt gt 1 Acid and Base Strength C5H4O5 aq 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 lies predominantly to the left that is lt 1 or to the right gt 1 HSO aq COS aq 047 aq aq q Solution Analyze We are asked to predict whether the equilibrium shown lies to the right favoring products or to the left favoring reactants Plan This is a proton transfer reaction and the position of the e
3. lt 0 1 lt 0 1 M Ba C H 0 PRACTICE EXERCISE In each of the following indicate which salt will form the more acidic or less basic 0 010 M solution a NaNO Fe NO4 4 b KBr KBrO c CH NH Cl BaCl d NH NO NH NO Answers a Fe NO4 b KBr CH4NH5CI d NH NO SAMPLE EXERCISE 16 18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na will form an acidic solution or a basic solution on dissolving in water Solution Analyze We are asked to predict whether a solution of Na HPO will be acidic or basic This substance is an ionic compound composed of Nat and HPO ions Plan We need to evaluate each ion predicting whether each is acidic or basic Because Na is the cation of a strong base NaOH we know that has no influence on pH It is merely a spectator ion in acid base chemistry Thus our analysis of whether the solution is acidic or basic must focus on the behavior of 127 ion We need to consider the fact that HPO can act as either an acid or a base HPO aq 5 H ag PO4 gt aq 16 45 HO HPO aq aq 16 46 The reaction with the larger equilibrium constant will determine whether the solution 18 acidic or basic Solve The value of for Equation 16 45 as shown in Table 16 3 is 4 2 10 13 We must calculate the value of K for Equation 16 46
4. HNO aq aq H3O aq Acid Base Conjugate ase aci add A SAMPLE EXERCISE 16 1 Identifying Conjugate Acids and Bases a What is the conjugate base of each of the following acids H S b What is the conjugate acid of each of the following bases 50 27 2 Solution Analyze We are asked to give the conjugate base for each of a series of species and to give the conjugate acid for each of another series of species Plan The conjugate base of a substance is simply the parent substance minus one proton and the conjugate acid of a substance 1s the parent substance plus one proton Solve a less one proton is The other conjugate bases are HS PH and 2 b CN plus one proton is HCN The other conjugate acids are and Notice that the hydrogen carbonate ion 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 5 Answers HSO HF HPO HCO SAMPLE EXERCISE 16 2 Writing Equations for Proton Transfer Reactions The hydrogen sulfite ion HSO is amphoteric a Write an equation for the reaction of with water in which the ion acts an acid b Write an equation for the reaction of HSO with water in which the ion acts as a base In both cases
5. The pH of a 0 10 M solution of formic acid at 25 is 2 38 Calculate for formic acid at this temperature To calculate K we need the equilibrium concentrations of all three things We can find H5O which is the same as HCOO from the pH Calculating from the pH pH log H4O 2 38 log H45O 2 38 log H20 10 238 10109 H30 H O 4 2 103 H30 Calculating K from pH Now we can set up a table HCOOH M HCOO M Change 4 2 x 103 4 2 x 103 44 2 x 1073 0 10 4 2 x 10 3 4 2 x 10 3 4 2 10 Equilibrium 0 0958 0 10 Calculating K from pH 4 2 10 3 4 2 1073 Ka 0 10 1 8 x 104 Calculating Percent lonization H4O Percent lonization _ 1730 leg x 100 HA initia In this example H3O eqg 4 2 x 1078 HCOOH 1 0 M Calculating Percent lonization 42x 107 Percent lonization LETE x 100 4 2 SAMPLE EXERCISE 16 10 Calculating K and Percent lonization from Measured pH A student prepared a 0 10 M solution of formic acid HCHO and measured its pH using a pH meter of the type illustrated in Figure 16 6 The pH at 25 C was found to be 2 38 a Calculate K for formic acid at this temperature b What percentage of the acid is ionized in this 0 10 M solution Solution Analyze We are given the molar concentration of an aqueous solution of weak a
6. 10M 42 x 10 M 3 3 Equilibrium 0 10 42 x 10 M 42 x10 M 42 X 10 M Notice that we have neglected the very small concentration of H aq that is due to the autoionization of H O Notice also that the amount of HCHO that ionizes is very small compared with the initial concentration of the acid To the number of significant figures we are using the subtraction yields 0 10 M 0 10 42 x 10 M 0 10 M We can now insert the equilibrium concentrations into the expression for K 42 x 10 9 4 2 x 10 K f _ 1 8 10 0 10 Check The magnitude of our answer is reasonable because for a weak acid is usually between 10 3 and 10 19 SAMPLE EXERCISE 16 10 continued b The percentage of acid that ionizes is given by the concentration of or CHO at equilibrium divided by the initial acid concentration multiplied by 100 p x 100 4 2 10 x 100 4 29 PLCEIL lt _ HCHO linitial 0 10 PRACTICE EXERCISE Niacin one of the B vitamins has the following molecular structure O 0 020 M solution of niacin has a pH of 3 26 a What percentage of the acid is ionized in this solution b What is the acid dissociation constant K for niacin Answers a 2 796 b 1 5 x 10 Calculating pH from K Calculate the pH of 0 30 solution of acetic acid at 25
7. CIO x 3 16 x 104 Thus 3 16 x 10 4 3 16 x 10 0 30 3 3 X 10 We say that the solution is 0 30 M in NaClO even though some of the CIO ions have reacted with water Because the solution 1s 0 30 M in NaCIO and the total volume of solution is 2 00 L 0 60 mol of NaCIO is the amount of the salt that was added to the water PRACTICE EXERCISE A solution of NH in water has a pH of 11 17 What 1s the molarity of the solution Answer 0 12 M K and K Acid K Base HNO Strong acid NO3 Negligible basicity HF 6 8 x 107 L5x10 L8 x 107 C gt H309 56 10 43 107 23 107 56 1 8 x 107 5 6 x 101 CO 1 8 x 10 Negligible acidity Strong base are related in this way Ky Therefore if you know one of them you can calculate the other SAMPLE EXERCISE 16 16 Calculating K or K for Conjugate Acid Base Pair Calculate a the base dissociation constant for the fluoride ion F b the acid dissociation constant for the ammonium ion Solution Analyze We are asked to determine dissociation constants for the conjugate base of HF and the conjugate acid of Plan Although neither F nor NH appears in the tables we can find the tabulated values for ionization constants for HF and NH and use the relationship between K and K to calculate the
8. identify the conjugate acid base pairs Solution Analyze and Plan We are asked to write two equations representing reactions between and water one in which HSO should donate a proton to water thereby acting as a Br nsted Lowry acid and one in which HSO should accept a proton from water thereby acting as a base We are also asked to identify the conjugate pairs in each equation Solve a HSO3 aq 037 aq aq The conjugate pairs in this equation are acid and SO conjugate base and H5O base and H4O conjugate acid b HSO aq lt H gt 503 aq OH aq The conjugate pairs in this equation are H O acid and conjugate base and HSO base and H SO conjugate acid PRACTICE EXERCISE When lithium oxide Li O is dissolved in water the solution turns basic from the reaction of the oxide ion 0 with water Write the reaction that occurs and identify the conjugate acid base pairs Answer aq H5O OH aq OH aq OH the conjugate acid of the base O also the conjugate base of the acid H O Acid and Base Strength e Strong acids are 100 one completely dissociated in water gt Their conjugate bases quite Weak acids only dissociate partially in water gt heir conjugate bases are 100 weak bases protonated in H5O Acid strength increases
9. that the ionization of is negligible compared to that of HJCO as far as production of Ht is concerned However it is the only source of CO3 which has a very low concentration in the solution Our calculations thus tell us that in a solution of carbon dioxide in water most of the CO is in the form of CO or a small fraction ionizes to form and and an even smaller fraction ionizes to give CO Notice also that 7 is numerically equal to PRACTICE EXERCISE a Calculate the pH of a 0 020 M solution of oxalic acid H C O See Table 16 3 for K and b Calculate the concentration of oxalate ion C O 7 in this solution Answers pH 1 80 b C 0 7 6 4 x 10 M Weak Bases Bases react with water to produce hydroxide Ion add H NH3 aq T H O ag OH ag Base Acid Con upate E aci base remove H Weak Bases The equilibrium constant expression for this reaction IS _ HB OH W B7 where 15 the base aissociation constant SAMPLE EXERCISE 16 14 Using to Calculate Calculate the concentration of in a 0 15 M solution of Solution Analyze We are given the concentration of a weak base and are asked to determine the concentration of Plan We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids that 1s we write the che
10. we can calculate the percent of molecules ionized concentration ionized Percent ionization of x 100 original concentration 79 X 10 M X 100 7 9 0 10 M b Proceeding similarly for the 0 010 M solution we have Y 68x10 0 010 x Solving the resultant quadratic expression we obtain x H 23 x 10 M The percentage of molecules ionized 18 0 0023 X 100 23 0 010 Comment Notice that if we do not use the quadratic formula to solve the problem properly we calculate 8 2 ionization for a and 26 ionization for b Notice also that in diluting the solution by a factor of 10 the percentage of molecules ionized increases by a factor of 3 This result is in accord with what we see in Figure 16 9 It is also what we would expect from Le Ch telier s principle Section 15 6 There are more particles or reaction components on the right side of the equation than on the left Dilution causes the reacti to shift 1n the direction of the larger number of particles because this counters the effect of the decreasing concentration of particles SAMPLE EXERCISE 16 12 continued PRACTICE EXERCISE In Practice Exercise 16 10 we found that the percent ionization of niacin K 1 5 x 102 in a 0 020 M solution is 2 7 Calculate the percentage of niacin molecules ionized in a solution that is a 0 010 M b 1 0 x 103 M Answers a 3 8 b 12 Polyprotic Acids e Ha
11. 0 1077 7 00 Basic lt 1 0 X 107 gt 1 0 107 gt 7 00 These are the pH values for several common substances H M pH M Gastric juice Lemon juice Cola vinegar Wine Tomatoes S Banana ee ee Black coffee Human blood tears 1 1077 7 0 7 0 1 1077 Egg white seawater Baking soda Borax Milk of magnesia 1x1078 8 0 6 0 1 1079 1 1079 9 0 5 0 1 1077 1 0 0 100 40 1 107 Lime water Household ammonia Household bleach NaOH SAMPLE EXERCISE 16 6 Calculating pH from H Calculate the pH values for the two solutions described in Sample Exercise 16 5 Solution Analyze We are asked to determine the pH of aqueous solutions for which we have already calculated H Plan We can use the benchmarks in Figure 16 5 to determine the pH for part a and to estimate pH for part b We can then use Equation 16 17 to calculate pH for part b Solve a In the first instance we found H to be 1 0 x 10 12 M Although we can use Equation 16 17 to determine the pH 1 0 x 10 12 is one of the benchmarks in Figure 16 5 so the pH can be determined without any formal calculation pH log 1 0 x 10712 12 00 12 00 The rule for using significant figures with logs 18 that the number of decimal places in the log equals the number of signi
12. 5O H5O H4O aq OH aq This is referred to as autoionization lon Product Constant e The equilibrium expression for this process IS H30 OH e This special equilibrium constant Is referred to as the Ion product constant for water At 25 K 1 0 x 1071 A SAMPLE EXERCISE 16 4 Calculating H for Pure Water Calculate the values of H and in a neutral solution at 25 Solution Analyze We are asked to determine the concentrations of hydronium and hydroxide ions in neutral solution at 25 Plan We will use Equation 16 16 and the fact that by definition OH in a neutral solution Solve We will represent the concentration of and OH in neutral solution with x This gives H IOH x x 10 x 10 c 10 x 1 0 x 107 M OH X In an acid solution H is greater than 1 0 x 1077 in a basic solution is less than 1 0 x 07 M PRACTICE EXERCISE Indicate whether solutions with each of the following ion concentrations are neutral acidic or basic H 24 x 10 M D OH 1 x 107 M c OH 2 7 x 10 5 M Answers a basic b neutral c acidic SAMPLE EXERCISE 16 5 Calculating H from OH Calculate the concentration of aq in a a solution in which OH is 0 010 M b a solution in which is 1 8 x 1077 M Note In this problem and all that follow we assume unless stated o
13. Chemistry The Central Science 10th edition Theodore L Brown Eugene LeMay and Bruce E Bursten Chapter 16 Acids and Bases John D Bookstaver ot Charles Community College ot Peters MO 2006 Prentice Hall Inc Acids 8 Bases Acids taste sour litmus turns red Bases taste bitter litmus turns blue feel slippery More than OH acidic More than H basic When they react with each other neutralization occurs Some Definitions Arrhenius Acid Substance that when dissolved in water increases the concentration of hydrogen ions Base Substance that when dissolved in water increases the concentration of hydroxide ions A Some Definitions Br nsted Lowry gt Proton donor gt Proton acceptor A Brensted Lowry acid must have a removable acidic proton A Brensted Lowry base must have a pair of nonbonding electrons If it can be either It IS amphoteric What Happens When an Acid Dissolves in Water Water acts as a Brensted Lowry base and abstracts a proton H from the acid Asaresult the conjugate base of the acid and a hydronium ion are formed H Of Conjugate Acids and Bases From the Latin word conjugare meaning to Join together Heactions between acids and bases always yleld their conjugate bases and acids remove
14. acid Increasing acid strength For a series of oxyacids acidity increases with the number of oxygens Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic X H C C O lt H C C O resonance SAMPLE EXERCISE 16 19 Predicting Relative Acidities from Composition and Structure Arrange the compounds in each of the following series in order of increasing acid strength a AsH HI NaH HO b H SeO H SeO H O Solution Analyze We are asked to arrange two sets of compounds in order from weakest acid to strongest acid Plan For the binary acids in part a we will consider the electronegativities of As I Na and respectively For the oxyacids in part b we will consider the number of oxygen atoms bonded to the central atom and the similarities between the Se containing compounds and some more familiar acids Solve a The elements from the left side of the periodic table form the most basic binary hydrogen compounds because the hydrogen in these compounds carries a negative charge Thus NaH should be the most basic compound on the list Because arsenic is less electronegative than oxygen we might expect that AsH would be a weak base toward water That is also what we would predict by an extension of the trends shown in Figure 16 13 Further we expect that the bina
15. asked to arrange a series of salt solutions in order of increasing pH that is from the most acidic to the most basic Plan We can determine whether the pH of a solution is acidic basic or neutral by identifying the ions in solution and by assessing how each 1 will affect the pH Solve Solution i contains barium ions and acetate ions Ba is an ion of one of the heavy alkaline earth metals and will therefore not affect the pH summary point 4 The anion C H30 is the conjugate base of the weak acid HC H 0 and will hydrolyze to produce ions thereby making the solution basic summary point 2 Solutions 11 and ii1 both contain cations that are conjugate acids of weak bases and anions that are conjugate bases of strong acids Both solutions will therefore be acidic Solution 1 contains NH which is the conjugate acid of NH 1 8 x 10 Solution iii contains which is the conjugate acid of 4 4 x 107 Because NH has the smaller K and is the weaker of the two bases will be the stronger of the two conjugate acids Solution 11 will therefore be the more acidic of the two Solution 1v contains the ion which is the cation of the strong base KOH and the NO ion which is the conjugate base of the strong acid HNO Neither of the ions in solution iv will react with water to any appreciable extent making the solution neutral Thus the order of pH is 0 1 lt 0 1 M
16. cid and the pH of the solution at 25 and we are asked to determine the value of K for the acid and the percentage of the acid that 18 ionized Plan Although we are dealing specifically with the ionization of a weak acid this problem is very similar to the equilibrium problems we encountered in Chapter 15 We can solve it using the method first outlined in sample Exercise 15 5 starting with the chemical reaction and a tabulation of initial and equilibrium concentrations Solve a The first step in solving any equilibrium problem is to write the equation for the equilibrium reaction The ionization equilibrium for formic acid can be written as follows HCHO aq CHO aq The equilibrium constant expression 1s _ H ICHO HCHO From the measured pH we can calculate H pH log H 2 38 log H 2 38 I 10 4 9 X 10 SAMPLE EXERCISE 16 10 continued We can do little accounting to determine the concentrations of the species involved in the equilibrium We imagine that the solution 1s initially 0 10 M in HCHO molecules We then consider the ionization of the acid into and CHO For each HCHO molecule that ionizes one ion and one CHO ion are produced in solution Because the pH measurement indicates that H 4 2 x 10 M at equilibrium we can construct the following table HCHO aq aq aq Initial 0 10 M gt Change 42 x 10 M 42 X
17. concentration of the acid Answer 0 0046 M Other p Scales 6 77 e he p in pH tells us to take the negative log of the quantity In this case hydrogen tons Some similar examples log OH pK log Ky Watch This Because H4O OH 1 0 x 1014 we know that log log OH log K 14 00 or In other words pH pK 14 00 E SAMPLE EXERCISE 16 9 Calculating the pH of a Strong Base What is the pH of a a 0 028 M solution of NaOH b a 0 0011 M solution of Ca OH Solution Analyze We re asked to calculate the pH of two solutions given the concentration of strong base for each Plan We can calculate each pH by two equivalent methods First we could use Equation 16 16 to calculate H and then use Equation 16 17 to calculate the pH Alternatively we could use OH to calculate and then use Equation 16 20 to calculate the pH Solve 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 10H H 3 57 x 10 P pH 1og 3 57 x 10 3 12 45 0 028 Method 2 pOH log 0 028 1 55 pH 14 00 pOH 12 45 b is a strong base that dissociates in water to give two ions per formula unit Thus the concentration of OH aq for the solution in par
18. ed in the equilibrium reaction letting x H at equilibrium HCN aq H a 0 20 M o0 Equilibrium 020 x SAMPLE EXERCISE 16 11 continued substituting the equilibrium concentrations from the table into the equilibrium constant expression yields x x 49x 10710 020 x K 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 0 20 x z 0 20 Thus Y 10 49 x 10 0 20 Solving for x we have x 020449 x 10 9 0 98 10 9 x V0 98 x 10 99 x 10 M H A concentration of 9 9 x 1079 M is much smaller than 5 of 0 20 the initial HCN concentration Our simplifying approximation 1s therefore appropriate We now calculate the pH of the solution pH log H log 9 9 X 1079 5 00 PRACTICE EXERCISE The for niacin Practice Exercise 16 10 is 1 5 x 107 What is the pH of a 0 010 M solution of niacin Answer 3 42 SAMPLE EXERCISE 16 12 Using K 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 Analyze We are asked to calculate the percent ionization of two HF solutions of different concentration Plan We approach this problem as we would previous equilibrium problems We begin by writing the chemical equation for the equilibr
19. ficant figures the original number see Appendix A Because 1 0 10 has two significant figures the pH has two decimal places 12 00 b For the second solution Ht 5 6 x 1079 M Before performing the calculation it is helpful to estimate the pH To do so we note that H lies between x 10 and 1 x 10 gt 1 10 lt 5 6x10 lt 1x10 gt Thus we expect the pH to lie between 6 0 and 5 0 We use Equation 16 17 to calculate the pH pH log 5 6 x 10 5 25 SAMPLE EXERCISE 16 6 continued Check After calculating a pH it is useful to compare it to your prior estimate In this case the pH as we predicted falls between 6 and 5 Had the calculated pH and the estimate not agreed we should have reconsidered our calculation or estimate or both Note that although H lies halfway between the two benchmark concentrations the calculated pH does not lie halfway between the two corresponding pH values This is because the pH scale is logarithmic rather than linear PRACTICE EXERCISE a In a sample of lemon juice H is 3 8 x 1077 M What is the pH b A commonly available window cleaning solution has a of 5 3 x 10 What is the pH Answers a 3 42 b 8 28 SAMPLE EXERCISE 16 7 Calculating H from pH A sample of freshly pressed apple juice has a pH of 3 76 Calculate H Solution Analyze We need to calculate H from pH Plan We will use Equation 16 17 pH log H for the calculati
20. from the value of for its conjugate acid H PO We make use of the relationship shown in Equation 16 40 We want to know for the base HPQ knowing the value of K for the conjugate acid K HPO 2 HPO 1 0 x 10 4 Because for is 6 2 x 107 Table 16 3 we calculate for HPO to be 1 6 x 1077 This is more t 10 times larger than for HPO thus the reaction shown in Equation 16 46 predominates over that in Equation 16 45 and the solution will be basic SAMPLE EXERCISE 16 18 continued PRACTICE EXERCISE 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 7 Y TIN E Factors Affecting Acid Strength 5 The more polar the H X bond and or the weaker the H X bond the more acidic the compound Acidity increases from left to right across a row and from top to bottom down a group Factors Affecting Acid Strength Shift of electron density H In oxyacids in which an OH 1 bonded to 7 another atom Y the more lt electronegative Y IS the more acidic the Acid ENofY acid HCIO 30 3 0 x 10 HBrO 2 8 25 x 10 HIO 2 5 23 101 Factors Affecting Acid Strength Hypochlorous Chlorous Chloric Perchloric sO HO n 6 HG 0 0 0 K 230x10 5 11 107 Strong acid Strong
21. ionization constants for each of the conjugates Solve for the weak acid HF is given in Table 16 2 and Appendix D as 6 8 x 107 We can use Equation 16 40 to calculate for the conjugate base A03 Ig Ky 15 10 68 10 b for NH is listed in Table 16 4 and in Appendix D as 1 8 x 10 Using Equation 16 40 we calculate for the conjugate acid NH Ko LOR 10 K 56 x 10710 K 18x10 SAMPLE EXERCISE 16 16 continued PRACTICE EXERCISE a Which of the following anions has the largest base dissociation constant or b The base quinoline has the following structure Q Its conjugate acid is listed in handbooks as having a pK of 4 90 What is the base dissociation constant for quinoline Answers a 2 4 x 102 b 7 9 10 10 Heactions of Anions with Water Anions are bases As such they can react with water in a hydrolysis reaction to form and the conjugate acid X aq HX aq OH aq Reactions of Cations with Water Weak electrostatic interaction IN 9 0 e Cations with acidic protons uU like NH will lower the pH Weak shift of electron density Of a Solution Strong Most metal cations that are ond hydrated in solution also Q0 lower the pH of the solution Strong shift of electron density Reactions of Ca
22. ium and tabulating the known and unknown concentrations of all species We then substitute the equilibrium concentrations into the equilibrium constant expression and solve for the unknown concentration that of H Solve a The equilibrium reaction and equilibrium concentrations are as follows HF aq 5 H aq F aq Initial 0 10 M Change The equilibrium constant expression 1s HIF 6 8 X 10 010 When we try solving this equation using the approximation 0 10 x 0 10 that 1s by neglecting the concentration of acid that ionizes in comparison with the initial concentration we obtain 8 2 X 10 SAMPLE EXERCISE 16 12 continued 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 0 10 x 6 8 x 1079 6 8 x 10 6 8 x 10 5x 68 x 10 x 68 x 10 0 This equation can be solved using the standard quadratic formula 22 b Vb i 2a Substituting the appropriate numbers gives 6 8 x 1075 V 68 107 4 6 8 x 105 2 6 8 x 10 1 6 x 107 2 Of the two solutions only the one that gives a positive value for x 1s chemically reasonable Thus x H 79 x 10 M SAMPLE EXERCISE 16 12 continued From our result
23. mical equation and tabulate initial and equilibrium concentrations Solve We first write the ionization reaction and the corresponding equilibrium constant expression NHz aq 0 OH ag _ NH 4 OH K 1 8 x 107 NH3 We then tabulate the equilibrium concentrations involved in the equilibrium NH3 aq H5O l aq OH 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 _ _ x x 1 8 107 NH 0 15 x SAMPLE EXERCISE 16 14 continued Because K 1s 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 x 0 159 1 8 x 103 0 15 1 8 36 10 7j 2 7 amp 10 x NH OH V2 7 10 1 6 10 M Check The value obtained for x is only about 1 of the NH concentration 0 15 M Therefore neglecting x relative to 0 15 was justified PRACTICE EXERCISE Which of the following compounds should produce the highest pH as a 0 05 M solution pyridine methylamine or nitrous acid Answer methylamine because it has the largest K value Weak Bases K be used to find OH and through it Lewis Conjugate Base Structure Acid Equilib
24. n these substances dissociate completely in aqueous solution Dissociation Constants e For a generalized acid dissociation HA aq H O A aq H3O aq the equilibrium expression would be HO A7 K RA his equilibrium constant is called the acid dissociation constant Dissociation Constants The greater the value of K the stronger the acid Structural Conjugate Acid Formula Base Equilibrium Reaction Hydrofluoric HF H F HF aq F aq 6 8 1075 Nitrous H U N 0 HNO aq 7 H3O T NO aq 4 5 X 10 4 HNO O Benzoic 0 41 HC HsOx aq H2O 1 H30 aq aq 6 3 10 HC7Hs0 b C H Acetic 0 C a H HC H302 aq H5O l EE H3O C9H3O 1 8 X ig HC5H30 1 H Hypochlorous HCIO 1547 CIO HClO a4 Hx30 aq CIO aq 3 0 x 103 Hydrocyanic H C N HCN aq H gt Q 1 CN aq 49 107 S Phenol H oO m 0 T aq aq 13 X 9 The proton that ionizes is shown in blue Calculating K from the pH The pH of a 0 10 M solution of formic acid at 25 is 2 38 Calculate for formic acid at this temperature We know that _ HOT a HCOOH Calculating K from the pH
25. nalyze We are given the pH of a 2 00 L solution of NaCIO and must calculate the number of moles of NaClO needed to raise the pH to 10 50 NaClO is an ionic compound consisting of and ClO ions As such it is a strong electrolyte that completely dissociates in solution into which is a spectator ion and CIO ion which is a weak base with K 3 33 x 1077 Equation 16 37 Plan From the pH we can determine the equilibrium concentration of OH We can then construct a table of initial and equilibrium concentrations in which the initial concentration of CIO is our unknown We can calculate C107 using the equilibrium constant expression Solve We can calculate OH by using either Equation 16 16 or Equation 16 19 we will use the latter method here pOH 14 00 pH 14 00 10 50 3 50 10 3 16 x 107M This concentration is high enough that we can assume that Equation 16 37 is the only source of OH that 1s we can neglect any produced by the autoionization of H O We now assume a value of x for the initial concentration of ClO and solve the equilibrium problem in the usual way CIO aq HClO aq aq 3 16 x 104M 316 104 3146 x 10 M 3 16 x 107M 3 16 x 107M x 3 16 x 104 M SAMPLE EXERCISE 16 15 continued We now use the expression for the base dissociation constant to solve for x HCIOIOH _ 316 104 5 3 3 X 1017
26. on Solve From Equation 16 17 we have pH log H 3 76 Thus log H 3 76 To find H we need to determine the antilog of 3 76 Scientific calculators have an antilog function sometimes labeled INV log or 10 that allows us to perform the calculation H antilog 3 76 10 3 17 x 10 M g 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 18 two Check Because the pH is between 3 0 and 4 0 we know that H will be between 1 1077 and I x 1077 M Our calculated H falls within this estimated range PRACTICE EXERCISE A solution formed by dissolving an antacid tablet has a pH of 9 18 Calculate H Answer 6 6 1019 M SAMPLE EXERCISE 16 8 Calculating the pH of a Strong Acid What is the pH of a 0 040 M solution of Solution Analyze and Plan We are asked to calculate the pH of a 0 040 M solution of HCIO Because HCIO is a strong acid it is completely ionized giving H CIO 0 040 M Because H lies between benchmarks I x 10 and 1 10 in Figure 16 5 we estimate that the pH will be between 2 0 and 1 0 Solve The pH of the solution is given by pH log 0 040 1 40 Check Our calculated pH falls within the estimated range PRACTICE EXERCISE An aqueous solution of HNO has a pH of 2 34 What is the
27. quilibrium will favor the proton going to the stronger of two bases The two bases in the equation are CO the base in the forward reaction as written and 50127 the conjugate base of HSQ We can find the relative positions of these two bases in Figure 16 4 to determine which 1s the stronger base Solve CO appears lower in the right hand column in Figure 16 4 and is therefore a stronger base than 50 2 CO therefore will get the proton preferentially to become while 50 2 will remain mostly unprotonated The resulting equilibrium will lie to the right favoring products that 15 gt 1 HSO ag CO3 aq 0 47 aq aq K gt 1 Acid Dase Conjugate Conjugate base acid Comment Of the two acids in the equation HSO and HCO the stronger one gives up a proton while the weaker one retains its proton Thus the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base SAMPLE EXERCISE 16 3 continued PRACTICE EXERCISE For each of the following reactions use Figure 16 4 to predict whether the equilibrium lies predominantly to the left or to the right ag ag OH aq b OH aq H2O Answers a left b right Autoionization of Water As we have seen water Is amphoteric n pure water a few molecules act as bases and a few act as acids H
28. rium Reaction Ammonia NH3 1 0 NH OH 1 8 x 107 E 5 Y 9 Pyridine C5H5N a Fl CsHsNH C5H5N H20 C5H5NH OH 1 7 10 ed Hydroxylamine H KN 0OH H4NOH H NOH H O H4NOH 11 107 Methylamine H NH3CH HO NHs3CH3 OH 4 4 107 NH 2CH3 Hydrosulfide ion H St H S HS HS 1 8 107 HS ir Carbonate ion 18 107 ar Hypochlorite ion Lt am HClO CIO HCIO OH 3 3 107 CIO of Basic Solutions What is the pH of a 0 15 M solution of NH OH aq OH INHy 1 8 10 gt gt oH of Basic Solutions Tabulate the data At Equilibrium 0 15 0 18 oH of Basic Solutions x 0 15 1 8 x 107 0 15 x 2 7 x 1076 x 1 6 x 1079 x 1 8 x 107 oH of Basic Solutions Therefore 1 6 x 103 M log 1 6 103 2 80 14 00 2 80 11 20 SAMPLE EXERCISE 16 15 Using pH to Determine the Concentration of a Salt solution made by adding solid sodium hypochlorite NaCIO to enough water to make 2 00 L of solution has of 10 50 Using the information in Equation 16 37 calculate the number of moles of NaCIO that were added to the water Solution A
29. ry hydrogen compounds of the halogens as the most electronegative element in each period will be acidic relative to water In fact HI 1s one of the strong acids in water Thus the order of increasing acidity is NaH lt AsH HI b The acidity of oxyacids increases as the number of oxygen atoms bonded to the central atom increases Thus 56 will be a stronger acid than H SeQ in fact the Se atom in HJSeQ Is in its maximum positive oxidation state and so we expect it to be a comparatively strong acid much like H5SeO is an oxyacid of a nonmetal that is similar to H SO As such we expect that 56 is able to donate a proton to H O indicating that H SeO is a stronger acid than H O Thus the order of increasing acidity is lt H SeO lt H5 SeO PRACTICE EXERCISE In each of the following pairs choose the compound that leads to the more acidic or less basic solution HBr HF b PH H S c HNO HNO d H SO H SeO Answers a b 5 c HNO d 50 ewis Acids H F H m 7 gt H F Lewis Lewis base acid Lewis acids are defined as electron pair acceptors e Atoms with an empty valence orbital can be Lewis acids ewis Bases H H F d dli Lewis Lewis base acid Lewis bases are defined as electron pair donors e Anything that could be a Brensted Lowry base 15 a Lewis base Lewis bases can in
30. t b 1s 2 x 0 0011M 0 0022 M Method 1 a 101079 in 10 M pH log 4 55 x 10 11 34 Method 2 log 0 0022 2 66 pH 14 00 pOH 11 34 SAMPLE EXERCISE 16 9 continued PRACTICE EXERCISE What is the concentration of a solution of a KOH for which the pH is 11 89 b Ca OH for which the pH is 11 68 Answers 7 8 x 103 M b 2 4 x 10 M Methyl violet Thymol blue Methyl orange Methyl red Bromthymol blue Phenolphthalein Alizarin yellow R Yellow How Do We Measure pH 0 Red olet pH range for color change 4 6 8 10 12 Yellow Yellow Blue ed Yello Red Yellow Yellow Blue Colorless E Pink Yellow Red For less accurate measurements one Can use Litmus paper e paper turns blue above 8 e paper turns red below pH 5 gt An Indicator How Do We Measure pH For more accurate measurements one uses a pH meter which measures the voltage in the solution Strong Acids You will recall that the seven strong acids are HCl Hl H 50 HCIO and HCIO These are by definition strong electrolytes and exist totally as ions in aqueous solution For the monoprotic strong acids Strong Bases Strong bases are the soluble hydroxides which are the alkali metal and heavier alkaline earth metal hydroxides Ca 2 and Ba Agai
31. teract with things other than protons however
32. therwise that the temperature is 25 Solution Analyze We are asked to calculate the hydronium ion concentration in an aqueous solution where the hydroxide concentration 1s known Plan We can use the equilibrium constant expression for the autoionization of water and the value of K to solve for each unknown concentration Solve a Using Equation 16 16 we have H IOH 1 0 x 107 0 1077 H 1 0 x 1071 M EF OH 0 010 This solution 1s basic because OH gt b In this instance 10x10 165010 H 5 56 109 OH 1 8 10 This solution 1s acidic because H gt OH SAMPLE EXERCISE 16 5 continued PRACTICE EXERCISE Calculate the concentration of OH aq in a solution in which H 2 x 10 M b H OH 100 x OH Answers a 5 x 10 M b 1 0 x 107 M c 1 0 x 105 M OH OH is defin ed b then ntration ium ION O 3 pH in pure water 1 0 x 10714 Because in pure water H4O OH H4O 1 Ox 101412 1 0 x 107 pH e Therefore in pure water OH log 1 0 x 107 7 00 e An acid has a higher H30 than pure water SO Its pH IS lt A base has a lower H4O than pure water SO Its pH IS gt 7 Solution Type H M OH M pH Value Acidic dx if lt 1 0 x 107 lt 7 00 Neutral 1 0 X 107 1
33. tions with Water Weak electrostatic interaction Weak shift of electron density W Q 0 Strong electron interaction Strong Strong shift of electron density Attraction between nonbonding electrons on oxygen and the metal causes shift of the electron density in water e This makes the bond more polar and the water more acidic e Greater charge and smaller size make a cation more acidic Effect of Cations and Anions 1 Ananion that is the conjugate base of a strong acid will not affect the pH 2 Ananlion that is the conjugate base of a weak E acid will increase the pH 3 Acation that is the conjugate acid of a weak base will decrease the pH Salt NaNO Ca NO3 Zn NO3 2 Al NO3 3 Indicator Bromthymol Bromthymol Methyl Methyl blue blue red Estimated pH Effect of Cations and Anions 4 Cations of the strong Arrhenius bases will not affect the pH 5 Other metal ions will cause a decrease in pH 6 When a solution contains both the conjugate base sem Lr NN of weak acid Estimated 7 0 6 9 5 5 3 5 conjugate acid of weak base the affect on pH depends on the and values SAMPLE EXERCISE 16 17 Predicting the Relative Acidity of Salt Solutions List the following solutions in order of increasing pH 1 0 1 Ba C H 0 11 0 1 M NH4CI 0 1 M NH4CH Br iv 0 1 M KNO Solution Analyze We are
34. um concentrations as follows H CO3 a 8 a T HCO q q 0007 0 0 The equilibrium constant expression is as follows H HCOj x x 7 Ka 43 x 10 H CO3 0 0037 x SAMPLE EXERCISE 16 13 continued solving this equation using an equation solving calculator we get 00 4 109 M Alternatively because K is small we can make simplifying approximation that 1s small so that 0 0037 x 0 0037 Thus x x 7 43x1 00037 22210 Solving for x we have x 0003 X 1077 18 1077 HCO VIE X 510 The small value of indicates that our simplifying assumption was justified The pH 185 therefore pH log H log 4 0 x 107 4 40 Comment If we were asked to solve for CO32 we would need to use K Let s illustrate that calculation Using the values of HCO and calculated above and setting 7 1 y we have the following initial and equilibrium concentration values SAMPLE EXERCISE 16 13 continued H ag COs aq 40 x 103 M 40 x 103 M Equilibrium 40 x 10 y M 4 0 x 10 y M Assuming that is small compared to 4 0 x 107 we have H COS 4 0 10 y 7 4 0 10 y 5 6 lt 10 COs 5 6 107 The value calculated for is indeed very small compared to 4 0 107 showing that our assumption was justified It also shows
35. ve more than one acidic proton If the difference between the for the first dissociation and subsequent values Is 10 or more the pH generally depends only on the first dissociation Name Ascorbic Carbonic Citric Oxalic Phosphoric Sulfurous Sulfuric Tartaric Formula HCO H3C6H507 H3PO HSO HSO H5C4H40g 8 0 107 43 107 74 10 5 9 1072 75 10 1 7 x 107 Large 1 0 x 10 1 6 X 10712 5 6 x 1071 1 7 103 6 4 X 103 62 103 6 4 103 1 2 x 107 4 6 107 4 0 107 42 x 1079 SAMPLE EXERCISE 16 13 Calculating the pH of a Polyprotic Acid Solution The solubility of CO in pure water at 25 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 H5CO which is produced by reaction between the CO and H O CO aq H gt O 1 H2CO3 aq What is the pH of a 0 0037 M solution of H CO Solution Analyze We are asked to determine the pH of 0 0037 solution of polyprotic acid Plan H CO is a diprotic acid the two acid dissociation constants K and Table 16 3 differ by more than a factor of 10 Consequently the pH can be determined by considering only K thereby treating the acid as if it were monoprotic acid Solve Proceeding as in Sample Exercises 16 11 and 16 12 we can write the equilibrium reaction and equilibri
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