Sample Exercise 16.1 Identifying Conjugate Acids and Bases
Contents
1. Education Sample Exercise 16 18 Determining Whether Salt Solutions Are Acidic Basic or Neutral Solution Continued e This solution contains Al and ClO ions Cations such as Al that are not in groups 1A or 2A are acidic summary point 5 The ClO ion is the conjugate base of a strong acid HCIO and therefore does not affect pH summary point 1 Thus the solution of Al C1O will be acidic Practice Exercise 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 ce CH NH Cl or BaCl d NH NO or NH NO Answers a Fe NO b KBr c CH NH Cl d NH NO3 AFRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 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 acidic2. 2 18x 10 Because K is small we can neglect the small nis amount of NH that reacts with water as x 0 15 1 8 x 10 2 7 x 10 compared to the total NH concentration that 1s x NH OH ae 10 14 102 we can neglect x relative to 0 15 M Then we have JRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 Education With contributions from Patrick Woodward All rights reserved Sample Exercise 16 15 Using K to Calculate OH Solution Continued 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 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 The pH of the solution is above 7 because we are dealing with a solution of a base 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 of the two amine base
3. M b H 4 x 10 M OH 7 x 10 PM Answers a basic b neutral c acidic Havin Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 x10 M Note In this problem and all that follow we assume unless stated otherwise that the temperature is 25 C Solution Analyze We are asked to calculate the hydronium ion concentration in an aqueous solution where the hydroxide concentration is 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 OH x 10714 a Using Equation 16 16 we have AH IGH J 10 10 Gox 107 loxi a n 12 This solution is basic because H7 0H 0 010 1 0 x 10 M OH gt H b In this instance 10 x 10 14 10x 19 14 a FE OH oD EE EET a 5 6 x 10 M This solution is acidic because 18x 1 Bl gt 0H Practice Exercise Calculate the concentration of OH agq in a solution in which a H 2 x 10 M b H OH c H 100x OH Answers a 5 x 10 M
4. part a We will then use stoichiometry and the relationship between pH and H to answer parts b and c Finally we will consider acid strength in order to compare the colligative properties of the two solutions in part d a Acids have polar H X bonds From Figure 8 6 we see that the electronegativity of H is 2 1 and that of P is also 2 1 Because the two elements have the same electronegativity the H P bond is nonpolar Section 8 4 Thus this H cannot be ac dic The other two H atoms however are bonded to O which has an electronegativity of 3 5 The H O bonds are therefore polar with H having a partial positive charge These two H atoms are consequently acidic b The chemical equation for the neutralization reaction 1s H PO aq 2 NaOQH ag Na HPO3 aq 2 H Od Navies Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Integrated Exercise Putting Concepts Together Solution Continued From the definition of molarity M mol L we see that moles M x L Section 4 5 Thus the number of moles of NaOH added to the solution is 0 0233 L 0 102 mol L 2 38 x 103 mol NaOH The balanced equation indicates that 2 mol of NaOH is consumed for each mole of H PO Thus the number
5. reaction components on the right side of the equation than on the left Dilution causes the reaction to shift in the direction of the larger number of particles because this counters the effect of the decreasing concentration of particles Practice Exercise In Practice Exercise 16 11 we found that the percent ionization of niacin K 1 5 x 10 ina 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 9 b 12 AFRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 1s 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 CO aqg H2O H CO3 aq What is the pH of a 0 0037 M solution of H CO solution Analyze We are asked to determine the pH of a 0 0037 M solution of a polyprotic acid Plan H CO is a diprotic acid the two acid dissociation constants K and K Table 16 3 differ by more than a factor of 10 Consequently the pH can be dete
6. a pH of 3 26 What is the acid dissociation constant K for niacin Answers 1 5 x 10 gt Havas Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 is ionized Solution Analyze We are given the molar concentration of an aqueous solution of weak acid and the equilibrium concentration of Ht aq and asked to determine the percent ionization of the acid Plan The percent ionization is given by Equation 16 27 Solve A Jequilibrium 42x10 M Percent ionization x 100 x 100 4 2 HCOONa 0 10 M Practice Exercise A 0 020 M solution of niacin has a pH of 3 26 Calculate the percent ionization of the niacin Answer 2 7 Havin Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 12 Using K to Calculate pH Calculate the pH of a 0 20 M solution of HCN Refer to Tabl
7. equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base 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 a HPO4 aq H O HPO ag OH aq b NH4 aq OH aq NH3 aq H2O Answers a left b right oO JARA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 4 Calculating H for Pure Water Calculate the values of Ht and OH in a neutral solution at 25 C Solution Analyze We are asked to determine the concentrations of H and OH ions in a neutral solution at 25 C Plan We will use Equation 16 16 and the fact that by definition H OH in a neutral solution Solve We will represent the concentration of H and OH in neutral solution with x This gives Ee OH Gia 1 0 1a cS 1 x 1 0 x 10 M H OH In an acid solution H is greater than 1 0 x 107 M in a basic solution H is less than 1 0 x 107 M Practice Exercise Indicate whether solutions with each of the following ion concentrations are neutral acidic or basic a H 4 x 10
8. of moles of H PO in the sample is 238 X 10 mol NaOH mol n a K mol HPO 2 00 X mol NaO a LIS 10 e Hh 2 mol NaOH ees The concentration of the H PO solution therefore equals 1 19 x10 mol 0 0250 L 0 0476 M c From the pH of the solution 1 59 we can calculate at equilibrium H antilog 1 59 10 0 026 M two significant figures Because Ka1 gt gt Kaz the vast majority of the ions in solution are from the first ionization step of the acid H3PO3 aq H aq HPO aq Because one H PO ion forms for each H ion formed the equilibrium concentrations of H and H PO are equal H H PO3 0 026 M The equilibrium concentration of H PO equals the initial concentration minus the amount that ionizes to form H and H PO H PO 0 0476 M 0 026 M 0 022 M two significant figures These results can be tabulated as follows H3PO3 aq H aq HPO aq a AN M Equilibrium 0 022 M 0 026 JRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Integrated Exercise Putting Concepts Together Solution Continued The percent ionization 1s Ee ahem O26 I Percent ionization mee x 100 cane x 100 55 The first acid dissocia
9. 4 The anion CH COO is the conjugate base of the weak acid CH COOH and will hydrolyze to produce OH ions thereby making the solution basic Summary point 2 b This solution contains NH and Cl ions NH is the conjugate acid of a weak base NH and is therefore acidic Summary point 3 CI is the conjugate base of a strong acid HCl and therefore has no influence on the pH of the solution Summary point 1 Because the solution contains an ion that is acidic NH and one that has no influence on pH Cl the solution of NH Cl will be acidic c This solution contains CH NH and Br ions CH NH is the conjugate acid of a weak base CH NH an amine and is therefore acidic Summary point 3 is the conjugate base of a strong acid HBr and is therefore pH neutral summary point 1 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 Summary points and 4 making the solution neutral Navies Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved
10. Because Na is on the left side of the periodic table we know that it has a very low electronegativity As a result the hydrogen in NaH carries a negative charge Thus NaH should be the least acidic 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 We would make the same prediction by an extension of the trends shown in Figure 16 12 Further we expect that the binary hydrogen compounds of the halogens as the most electronegative element in each period will be acidic relative to water In fact HI is one of the strong acids in water Thus the order of increasing acidity is NaH lt AsH lt H O lt HI b The acids H SO and H SeO have the same number of O atoms and OH groups In such cases the acid strength increases with increasing electronegativity of the central atom Because S is more electronegative than Se we predict that H SO is more acidic than H SeO Next we can compare H SeO and H SeOs For acids with the same central atom the acidity increases as the number of oxygen atoms bonded to the central atom increases Thus H SeO should be a stronger acid than H SeO Thus we predict the order of increasing acidity to be H SeO lt H SeO lt H SO Navies Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddl
11. J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education sample Exercise 16 10 Calculating K 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 K for formic acid at this temperature solution Analyze We are given the molar concentration of an aqueous solution of weak acid and the pH of the solution and we are asked to determine the value of K for the acid 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 this problem using the method first outlined in Sample Exercise 15 9 starting with the chemical reaction and a tabulation of initial and equilibrium concentrations Solve 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 HCOOH a q H a g HC a g The equilibrium constant expression 1s F Be From the measured pH we can calculate H e H HCOO i HCOOH We can do a little accounting to determine the concentrations of the species involved in the equilibrium tee We imagine that the ba is initially 0 10 M in pH aR logit 2 38 HCOOH molecules We then consider the ionization of lo ofH 8 the acid into H
12. Sample Exercise 16 1 Identifying Conjugate Acids and Bases a What is the conjugate base of each of the following acids HC10 H S PH HCO b What is the conjugate acid of each of the following bases CN SO 7 H O HCO 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 is the parent substance plus one proton Solve 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 O and H CO3 Notice that the hydrogen carbonate ion 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 HCOt Havin Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 2 Writing Equations for Proton Transfer Reactions The hydrogen sulfite ion HSO is amphiprotic a Write an equation for the reaction of HSO w
13. This concentration is high enough that we can assume that Equation 16 37 is the only source of OH that is we can neglect any OH produced by the autoionization of H O Sep T RN a Oe ay We now assume a value of x for the initial a ee concentration of ClO and solve the equilibrium Change problem in the usual way Equilibrium Navies Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 16 Using pH to Determine the Concentration of a Salt Solution Continued i _ We now use the expression for the base K PACE Gen 2 33x 107 ee Y fi 4 dissociation constant to solve for x CO E S26 W ea Kad T eo a 3 2 X 10 0 31 M 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 is 2 00 L 0 62 mol of NaClO 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 is the molarity of the solution Answer 0 12 M Havin Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catheri
14. aOH b a 0 0011 M solution of Ca OH Solution Analyze We are asked to calculate the pH of two solutions of strong bases Plan We can calculate each pH by either of 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 OH7 to calculate pOH 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 La x0 3 57 X 10 2 4 59H log 3 57 x 1078 12 45 0 028 A X D AM pH log 3 57 X 10 12 4 H Method 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 is 2 x 0 0011 M 0 0022 M Method 1 H 1 0 x 10 455 x 10712 M gt H log 4 55 X 10712 4 3A tie Ee ee ape Method 2 Practice Exercise pOH log 0 0022 2 66 pH 14 00 pOH 11 34 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 a 7 8 x 10 M b 2 4 103M Havas Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine
15. and HCOO For each HCOOH molecule that ionizes one Ht ion and one ion HCOO A 10s 490 ke a are produced in solution Because the pH measurement indicates that H 4 2 x 10 M at equilibrium we can construct the following table HRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 10 Calculating K from Measured pH Solution Continued ayy i M fey m Notice that we have neglected the very small HCOOH aq H aq HCOO ag concentration of H aq that is due to the 0 10 M e autoionization of H O Notice also that the 42 102M ver eee amount of HCOOH that ionizes is very small po uitibrium 0 10 4 2 x 10 M 4 2 x 103M 4 2 x 103M 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 4 2 x 10 M 0 10 M We can now insert the equilibrium centrations e AII ux 2 4 into the expression for K a 0 10 1 8 x 10 Check The magnitude of our answer is reasonable because K for a weak acid is usually between 10 and 10710 Practice Exercise Niacin one of the B vitamins has the following molecular structure O cr N A 0 020 M solution of niacin has
16. b 1 0 x 107 M ce 1 0 x 10 M ARAA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 calculate pH using its defining equation Equation 16 17 Solve a In the first instance we found H to be 1 0 x10 M pH log 1 0 x 109 12 00 12 00 Because 1 0 x10 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 1 x 10 lt 5 6 X 10 Pa thee 10 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 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 Practi
17. ce Exercise a In a sample of lemon juice H is 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 10 M so pH 8 28 AFRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 calculation Solve From Equation 16 17 we have EE 376 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 H antilog 3 76 1076 1 7 x 104M or 10 that allows us 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 Check Because the pH is between 3 0 and 4 0 we know that H will be between 1 x 10 and 1 x 10 M Our calculated H falls within this estimated range Practice Exercise A
18. d solve for the unknown concentration that of Ht Solve HF aq gt H a F aq a The equilibrium reaction and equilibrium now concentrations are as follows AUF 9 The equilibrium constant expression 1s OR ys ee K HF 010 x 6 56 lt 10 When we try solving this equation using the approximation 0 10 x 0 10 that is by neglecting the concentration of acid that ionizes in comparison a x 82x10 M with the initial concentration we obtain Navies Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 Education With contributions from Patrick Woodward All rights reserved Sample Exercise 16 13 Using K to Calculate Percent lonization Solution 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 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 solut
19. e 16 2 or Appendix D for the value of K Solution Analyze We are given the molarity of a weak acid and are asked for the pH From Table 16 2 Ka for HCN is 4 9 x 10 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 HCN aq H aq CN aq the ionization reaction that forms H aq and the Teens o HJ CN IARR equilibrium constant K expression for the k a i reaction HCN HCN aq H aq CN aq Next we tabulate the concentration of the os i libr oom o 0o species involved in the equilibrium reaction letting x H at equilibrium Equilibrium 020 9M from the table into the equilibrium constant K x x 49 x 100 Substituting the equilibrium concentrations expression yields pai y JRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 Education With contributions from Patrick Woodward All rights reserved Sample Exercise 16 12 Using K to Calculate pH Solution Continued We next make the simplifying approximation 020 y 0 90 that x the amount of acid that dissociates 1s ij _ small compared with the initial c
20. e River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 20 Predicting Relative Acidities from Composition and Structure Practice Exercise In each of the following pairs choose the compound that leads to the more acidic or less basic solution a HBr HF b PH H S c HNO HNO d H SO H SeQO3 Answers a HBr b H S c HNO d H SO Copyright 2009 by Pearson Education Inc Upper Saddle River New Jersey 07458 All rights reserved AFRA Chemistry The Central Science Eleventh Edition By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy H With contributions from Patrick Woodward Sample Integrated Exercise Putting Concepts Together Phosphorous acid H PO has the following Lewis structure H O H a Explain why H PO is diprotic and not triprotic b A 25 0 mL sample of a solution of H PO is titrated with 0 102 M NaOH It requires 23 3 mL of NaOH to neutralize both acidic protons What is the molarity of the HPO solution c The original solution from part b has a pH of 1 59 Calculate the percent ionization and K for H PO assuming that K gt gt K d How does the osmotic pressure of a 0 050 M solution of HCI compare qualitatively with that of a 0 050 M solution of H PO Explain Solution We will use what we have learned about molecular structure and its impact on acidic behavior to answer
21. f oxalate ion C O 7 in this solution Answers a pH 1 80 b C O 7 6 4 x 10 M AFRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 15 Using K to Calculate OH Calculate the concentration of OH in a 0 15 M solution of NH solution Analyze We are given the concentration of a weak base and are asked to determine the concentration of OH Plan We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids that is we write the chemical equation and tabulate initial and equilibrium concentrations Solve We first write the ionization reaction NH3 aq H2O 2 NH4 aq OH aq and the corresponding equilibrium constant K INH OH7 expression K INE LB x 107 We then tabulate the equilibrium concentrations NH3 aq H200 NH4 OH aq involved in the equilibrium usm 0 0 cme en ae om We ignore the concentration of H O because it Equilibrium 0 15 x M O am o ee is not involved in the equilibrium constant expression Inserting these quantities into the NH OH7 C l equilibrium constant expression gives the Ky INEel 015 x 1 8 x 10 following
22. ion we have Solving the resultant quadratic expression we obtain The percentage of molecules ionized is PEARSON Education Chemistry The Central Science Eleventh Edition By Theodore E Brown H Eugene LeMay Bruce E With contributions from Patrick Woodward x 0 10 x 6 8 x 10 6 8 X 10 6 8 X 10 x x 6 8 X 10 x 6 8 x 10 0 b V b 4ac ax 2a 708 x 10 V 6 8 x 10 4 6 8 X 10 2 6 8 xX 10 1 6 x 10 2 e Et F 75 X10 mM concentration ionized _ Percent ionization of HF __ X 100 original concentration 7 9 x 10 M ee NAHOR SF 0 10 M 2 xX E 68 x 10 0 010 x P 1 x 107M 0 0025 M X 100 23 0 010 M Copyright 2009 by Pearson Education Inc Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 All rights reserved Sample Exercise 16 13 Using K to Calculate Percent lonization Solution Continued 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 Chatelier s principle Section 15 7 There are more particles or
23. ith water in which the ion acts as an acid b Write an equation for the reaction of HSO with water in which the ion acts as a base In both cases identify the conjugate acid base pairs Solution Analyze and Plan We are asked to write two equations representing reactions between HSO 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 3 HSO3 aq H20 1 S03 aq H30 aq The conjugate pairs in this equation are HSO acid and SO conjugate base and H O base and H O conjugate acid iq EST ii E E mee eee EE The conjugate pairs in this equation are H O acid and OH conjugate base and HSO base and H SO conjugate acid Practice Exercise When lithium oxide L1 O is dissolved in water the solution turns basic from the reaction of the oxide ion O 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 OHS is also the conjugate base of the acid H O JRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 Wi
24. l rights reserved Education Sample Exercise 16 19 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic 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 Copyright 2009 by Pearson Education Inc AFRA Chemistry The Central Science Eleventh Edition By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 All rights reserved With contributions from Patrick Woodward Sample Exercise 16 20 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 H O b H SO H SeO H SeQ Solution Analyze We are asked to arrange two sets of compounds in order from weakest acid to strongest acid In a the substances are binary compounds containing H whereas in b the substances are oxyacids Plan For the binary compounds in part a we will consider the electronegativities of As I Na and O relative to H Ahigher electronegativity will cause the H to have a higher partial positive charge causing the compound to be more acidic For the oxyacids in part b we will consider both the relative electronegativities of the central atom S and Se and the number of oxygen atoms bonded to the central atom Solve a
25. lustrate that of a Ms calculation Using the values of HCO and H BS Dee calculated above and setting CO 2 y we have the following initial and equilibrium concentration y y y values RE ES es ee _ HICO3s7 4 0 x 10 y sro pct t a ae a Assuming that y is small compared to 4 0 x 10 gt ees LS we have y 56 10 M KO 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 is negligible compared to that of H COs3 as far as production of H is concerned However it is the only source of CO 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 H CO a small fraction ionizes to form H and HCO and an even smaller fraction ionizes to give CO 7 Notice also that CO 7 is numerically equal to K JRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Sample Exercise 16 14 Calculating the pH of a Polyprotic Acid Solution 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 K b Calculate the concentration o
26. ne J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 Analyze We are asked to determine dissociation constants for F the conjugate base of HF and NH the conjugate acid of NH 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 ionization constants for each of the conjugates Solve 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 Ko 0x0 k a 1 5 x 1971 Ka 68 x 10 b K for NH is listed in Table 16 4 and in Appendix D as K 1 8 x 10 Using Equation 16 40 we can calculate K for the conjugate acid NH Kk 106 16 4 a 56 X 10 Ky 18x 10 JRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 Education With contributions from Patrick Woodward All rights reserved Sample Exercise 16 17 Calcula
27. oncentration x of acid that is _ 49x 10710 0 20 3 10 10 x 0204 9 x 10 0 98 x10 0 98 x 10 9 9 10 M H Solving for x we have A concentration of 9 9 x 10 M is 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 p log H log 9 9 x 10 5 00 Practice Exercise The Ka for niacin Practice Exercise 16 10 is 1 5 x 10 What is the pH of a 0 010 M solution of niacin Answer 3 41 ARAA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 13 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 From Appendix D we find K 6 8 x 107 Plan We approach this problem as we would previous equilibrium problems We begin by writing the chemical equation for the equilibrium and tabulating the known and unknown concentrations of all species We then substitute the equilibrium concentrations into the equilibrium constant expression an
28. or basic Because Na is a cation of group LA we know that it 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 the HPO ion We need to consider the fact that HPO can act as either an acid or a base HPO aq H q PO4 gt aq 16 45 HPO aq H2O H gt POg aq OH aq 16 46 The reaction with the larger equilibrium constant will determine whether the solution is acidic or basic Solve The value of K for Equation 16 45 as shown in Table 16 3 is 4 2 x 10 1 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 z P q Ka a Ky Ka We want to know K for the base HPO knowing the value of K for the conjugate acid HPO K HPO 4 Rolie Oe Ke 1 0 x 1071 Because K for H PO is 6 2 x 10 8 Table 16 3 we calculate K for H PO 7 to be 1 6 x 107 This is more than 10 times larger than K for H PO thus the reaction shown in Equation 16 46 predominates over that in Equation 16 45 and the solution will be basic JAFRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward Al
29. rmined by considering only K thereby treating the acid as if it were a monoprotic acid H CO3 aq H aq HCO aq Solve Proceeding as in Sample Exercises 16 12 a iibri ion Uaita 0 0087 M and 16 13 we can write the equilibrium reaction and equilibrium concentrations as follows e lt opak H HCO3 x x Ka aS The equilibrium constant expression 1s as follows aS 43 107 H2CO3 0 0037 x x 40x10 M Solving this equation using an equation solving calculator we get 0 0037 x 0 0037 x x iA X 1077 Alternatively because K is small we can 0 0037 make the simplifying approximation that x 1s a 0 008743 K 107 16 x 10 small so that x H HCO x 10 4 0 x 10 M 1 6 JRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Sample Exercise 16 14 Calculating the pH of a Polyprotic Acid Solution Solution Continued Thus Solving for x we have The small value of x indicates that our simplifying Seis a Hat a Sas e lpol 4 0 x N AY assumption was justified The pH 1s therefore pr OHE ORED H AED Comment If we were asked to solve for CO 7 HCO a H a CO3 a we would need to use K Let s il
30. s in the list Havin Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 Using the information in Equation 16 37 calculate the number of moles of NaClO that were added to the water solution Analyze We are given the pH of a 2 00 L solution of NaClO and must calculate the number of moles of NaClO needed to raise the pH to 10 50 NaClO is an ionic compound consisting of Na and ClO ions As such it is a strong electrolyte that completely dissociates in solution into Na which is a spectator ion and ClO ion which is a weak base with K 3 3 x 10 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 equilibriumconstant expression K Solve We can calculate OH by using either pOH 14 00 pH 14 00 10 50 3 50 Equation 16 16 or Equation 16 20 we will use ee eee the latter method here Se a aa
31. solution formed by dissolving an antacid tablet has a pH of 9 18 Calculate H Answer H 6 6 10 19 M Havin Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 8 Calculating the pH of a Strong Acid What is the pH of a 0 040 M solution of HCIO Solution Analyze and Plan Because HCIO is a strong acid it is completely ionized giving H C107 0 040 M Solve The pH of the solution is 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 Practice Exercise An aqueous solution of HNO has a pH of 2 34 What is the concentration of the acid Answer 0 0046 M Havin Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education Sample Exercise 16 9 Calculating the pH of a Strong Base What is the pH of a a 0 028 M solution of N
32. th contributions from Patrick Woodward All rights reserved Education 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 Solution HSO aq CO3 aq SO ag HCO3 aq 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 equilibrium 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 SO the conjugate base of HSO We can find the relative positions of these two bases in Figure 16 4 to determine which is the stronger base Solve CO appears lower in the right hand column in Figure 16 4 and is therefore a stronger base than SO 7 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 HSO aq CO3 aq SO47 aq HCO aq Ka 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
33. ting K or K for a Conjugate Acid Base Pair Practice Exercise a Which of the following anions has the largest base dissociation constant NO PO or N37 b The base quinoline 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 107 b 7 9 x 10 AFRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education 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 Al C1O Solution Analyze We are given the chemical formulas of five ionic compounds salts and asked whether their aqueous solutions will be acidic basic or neutral Plan We can determine whether a solution of a salt is acidic basic or neutral by identifying the ions in solution and by assessing how each ion will affect the pH Solve 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 summary point
34. tion constant 1s maps H H2PO3 0 026 0 026 a H3PO3 0 022 d Osmotic pressure is a colligative property and depends on the total concentration of particles in solution Section 13 5 Because HCI is a strong acid a 0 050 M solution will contain 0 050 M H ag and 0 050 M Cl aq or a total of 0 100 mol L of particles Because H PO is a weak acid it ionizes to a lesser extent than HCl and hence there are fewer particles in the H PO solution As a result the HPO solution will have the lower osmotic pressure 0 031 AFRA Chemistry The Central Science Eleventh Edition Copyright 2009 by Pearson Education Inc By Theodore E Brown H Eugene LeMay Bruce E Bursten and Catherine J Murphy Upper Saddle River New Jersey 07458 With contributions from Patrick Woodward All rights reserved Education
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