Solutions for Chapter 9 End-of-Chapter Problems

Solutions for Chapter 9 End-of-Chapter Problems Problem 9.1. (a) Balanced chemical equations for the formation of pale blue Cu(OH) 2 (s) and deep blue-violet [Cu(NH 3 ) 4 ] 2+ (aq) when NH 3 (aq) is slowly added to a light blue solution of Cu 2+ (aq) are: 2NH 3 (aq) + 2H 2 O(l) + Cu 2+ (aq) Cu(OH) 2 (s) + 2NH + 4 (aq) light blue pale blue Cu(OH) 2 (s) + 4NH 3 (aq) [Cu(NH 3 ) 4 ] 2+ (aq) + 2OH (aq) pale blue deep blue-violet (b) To understand how the formation of Cu(OH) 2 (s) from Cu 2+ (aq) will be affected by basic conditions, we need to consider the reactions that make up the net reaction written in part (a): 2NH 3 (aq) + 2H 2 O(l) 2NH + 4 (aq) + 2OH (aq) Cu 2+ (aq) + 2OH (aq) Cu(OH) 2 (s) Additional OH (aq) is a disturbance to these equilibria (Le Chatelier s principle) and the system will respond by using up some of the additional OH (aq). Thus, the first reaction will shift toward reactants, because OH (aq) is a product, and the second reaction will shift toward products because OH (aq) is a reactant. The net result will be to favor formation of Cu(OH) 2 (s) from Cu 2+ (aq). (c) The equation written in part (a) suggests that the formation of [Cu(NH 3 ) 4 ] 2+ (aq) from Cu(OH) 2 (s) will be favored under mildly acidic conditions, because H 3 O + (aq) will react with OH (aq) ions to form water. This reaction is a disturbance to the equilibrium, because it removes a product of the reaction, The system should respond by trying to form more OH (aq), which also would mean the formation of more [Cu(NH 3 ) 4 ] 2+ (aq). This analysis is flawed because it neglects the basicity of NH 3 (aq), which will also react with added H 3 O + (aq): NH 3 (aq) + H 3 O + (aq) NH + 4 (aq) This reaction removes a reactant, NH 3 (aq), necessary for formation of [Cu(NH 3 ) 4 ] 2+ (aq), so the presence of H 3 O + (aq) in the solution will not favor formation of [Cu(NH 3 ) 4 ] 2+ (aq). Note that the reaction of H 3 O + (aq) with OH (aq) also removes a reactant in the second reaction in part (b), so Cu(OH) 2 (s) will tend to dissolve or not form under acidic conditions. (d) None of the reactions involves oxidation or reduction. Copper has an oxidation number of +2 in all cases. Problem 9.2. If Fe 3+ (aq) and SCN (aq) formed a one-to-two metal ion complex, the complex in Figure 9.1 would have to show the new ratio of two SCN (aq) ions to every Fe 3+ (aq) ion. Twice as many SCN (aq) ions would be used in forming the complex and there would be four more unreacted Fe 3+ (aq) ions in the right-hand part of the figure, as shown in this alternative figure: March 2005 ACS Chemistry Cahpter 9 suggested solutions 1

Chemical Equilibria Chapter 9 Problem 9.3. (a) Addition of CN (aq) to the equilibrium reaction, Cd 2+ (aq) + 6CN (aq) Cd(CN) 6 2 (aq), will cause the reaction to shift toward formation of more of the product, Cd(CN) 6 2 (aq). Addition of more of a reactant is a disturbance to the equilibrium system (Le Chatelier s principle) and the system responds by minimizing the disturbance, using up some of the added CN (aq) to form more of the product. (b) Identification of change in the CoCl 2 aqueous equilibrium is possible because of color differences between the reactants and the products. Both species in this equilibrium are colorless, so there is no visual clue to the position of equilibrium. Problem 9.4. The statements in this problem refer to the equilibrium reaction: H 2 (g) + Cl 2 (g) 2HCl(g). (a) When this reaction has reached a state of equilibrium, no further reaction occurs. This is not a true statement. No further net reaction occurs, but there is a dynamic equilibrium in which molecules of HCl(g) continuously form, even as other HCl(g) molecules are decomposing into H 2 and Cl 2. While the macroscopic concentrations are not changing, there is constant change at the microscopic level. (b) When equilibrium is established, the number of moles of reactants equals the number of moles of products for this reaction. This is not a true statement. The stoichiometric coefficients that balance the equation do not give the relative number of moles present at equilibrium. It is necessary to know the position of the equilibrium and to use the equilibrium constant at a given temperature to find relative numbers of moles of reactants and products. (c) The concentration of each substance in the system will be constant at equilibrium. This is a true statement. The observable concentrations are constant if dynamic equilibrium has been established, but there still is constant change at the microscopic level. Problem 9.5. The reaction of interest is: CO(g) + 2O 2 (g) 2CO 2 (g). In this reaction, a C O triple bond (BH = 1071 kj mol 1 ) and two O O double bonds (BH = 498.7 kj mol 1 ) are broken, and four C O double bonds in carbon dioxide (BH = 799 kj mol 1 ) are made. The overall change in enthalpy for the reaction is 1127 kj mol 1, that is, the reaction is exothermic. If the reaction is at equilibrium and is disturbed by adding energy to raise its temperature, the system will respond (Le Chatelier s principle) by using up some of the added energy which it can do by proceeding in reverse (the endothermic direction) toward increase in the concentration of the reactants, CO(g) and O 2 (g), at the expense of some of the reactant, CO 2 (g). 2 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria Problem 9.6. The reaction of interest is: CO(g) + 2H 2 (g) CH 3 OH(g). (a) If some CH 3 OH(g) is removed while the volume of the system is held constant, the concentration of the product decreases. The system will respond to this disturbance (Le Chatelier s principle) by producing more product at the expense of a decrease in the concentrations (amounts) of the reactants. (b) Adding CO(g) to the system while the volume is held constant increases the concentration of one of the reactants. The system will respond to this disturbance by reacting to reduce the concentration of this reactant with the consequence that the concentration of the other reactant will be reduced and the concentration of the product will increase. (c) If the system is compressed to a smaller volume, the concentrations of all the species will be increased. In response to this disturbance, the system will react to reduce the number of moles of gas, so as to try to bring the concentrations as near as possible to where they had been. The number of moles of gas will be reduced if reactants combine to form the product, so the result will be an increase in the concentration of the product and reductions in the concentrations of the reactants. (d) N 2 (g) has no role in the reaction and adding it at constant volume does not affect the concentrations of the reactants and products. There will be no change in the equilibrium system when this addition is made. Problem 9.7. The reaction of interest is this Lewis acid-base equilibrium reaction between boric acid and water: B(OH) 3 (s) + 2H 2 O(aq) [B(OH) 4 ] (aq) + H 3 O + (aq) (a) Addition of B(OH) 3 (s) will have no effect on the equilibrium. Addition of a solid reactant does not change its concentration. (b) Addition of H 3 O + (aq) will cause this equilibrium reaction to shift toward the reactants. Increasing the H 3 O + (aq) concentration is a disturbance to the equilibrium system (Le Chatelier s principle) and the system will respond by using up some of the added H 3 O + (aq). It can do this by proceeding in reverse, using up some [B(OH) 4 ] (aq) and forming more B(OH) 3 (s). (c) Addition of OH (aq) will cause this equilibrium reaction to shift toward the products. Added OH (aq) will react with H 3 O + (aq), which will disturb the equilibrium system by decreasing the concentration of a product. The system will respond by producing more H 3 O + (aq) to replace part of what is lost, thus forming more [B(OH) 4 ] (aq) and using up some B(OH) 3 (s). Problem 9.8. The reaction of interest is: HA(aq) + H 2 O(aq) H 3 O + (aq) + A (aq). (a) This reaction equation tells us that, in the forward direction, a molecule of HA, HA(aq), (solvated by the water) interacts with a molecule of water, H 2 O, to transfer a proton to the water to form a solvated hydronium ion, H 3 O + (aq), and a solvated anion, A (aq), HA without its acidic proton. In the reverse direction a hydronium ion interacts with a solvated anion to transfer a proton to the anion to form a water molecule and a solvated molecule of the acid, HA(aq). The way the equation is written, it appears that the same hydronium and anion that form in the forward reaction react in the reverse reaction. This is an incorrect interpretation. The Web Companion animation illustrates the correct interpretation. ACS Chemistry Chapter 9 suggested solutions 3

Chemical Equilibria Chapter 9 (b) The beginning (proton transfer from HA(aq) to H 2 O(aq)) and end (proton transfer from H 3 O + (aq) to A (aq)) of the reaction shown in the Web Companion, Chapter 9, Section 9.4, page 1, animation are exactly consistent with the reaction written and described above. What we see in the middle of the animation is that protons can be transferred between hydronium ions and water molecules, that it is not the same proton transferred in each case, and that the anion to which a proton is transferred from a hydronium is not the same one from which a proton originally came. What is missing in the equation is a way to show: (1) that any one of the acid molecules in the mixture might transfer its proton to a water molecule, (2) that the hydronium ion formed is just one of many hydronium ions in the solution, (3) that the hydronium ions are continuously transferring protons to water molecules to form different hydronium ions, and (4) that any one of the acid anions and any one of the hydronium ions might be the ones that react to transfer a proton from hydronium ion to anion. Although reaction equations are a very compact way to represent the stoichiometry of reactions, they do not very well represent the details of what is occurring at the molecular level. As you use such equations, always try to use what you are learning about reactions to visualize what is going on at the molecular level as well. Problem 9.9. The equilibrium constant expression for ammonia synthesis is: K eq = ( NH 3 (g)) 2 ( N 2 (g))h 2 (g) ( ) 3 Since all species in this equilibrium constant expression are gases, the dimensionless concentration ratios should be expressed in bar (standard state = 1 bar). Problem 9.10. It requires 50 drops of a 1.0 M weak acid, HA(aq), solution added to 10.0 ml of distilled water to produce enough ions to light the bulb in a conductivity detector as brightly as 1 drop of 1 M HCl(aq) added to 10.0 ml of distilled water. We can conclude that there are only about 2% = ( 1 drop 50 drops) 100% as many ions formed by the weak acid as by an equivalent number of moles of HCl(aq), hydrochloric acid. We know that HCl(aq) reacts completely to form H 3 O + (aq) and Cl (aq) ions, so we estimate that only about 2% of the weak acid reacts to transfer its proton to water. Thus, in a 1 M solution of HA(aq), the concentrations are [HA(aq)] = 1 M, [H 3 O + (aq)] = 0.02 M, and [A (aq)] = 0.02 M. The proton transfer reaction and equilibrium constant (dissociation constant) for the weak acid are: HA(aq) + H 2 O(aq) H 3 O + (aq) + A (aq) ( H 3 O + (aq))( A (aq)) K eq = ( HA(aq) )( H 2 O(aq) ) eq = (0.02)(0.02) (1)(1) = 4 10 4 Problem 9.11. (a) The reaction equation and equilibrium constant expression for the decomposition of calcium carbonate are: CaCO 3 (s) CaO(s) + CO 2 (g) 4 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria K eq = ( ) ( CaCO 3 (s)) ( CaO(s) ) CO 2 (g) For pure solids the dimensionless concentration ratio is 1, unity. For the gas we need to use bar as the unit (standard state = 1 bar). (b) If the equilibrium pressure of CO 2 (g) is 0.37 bar, then: (0.37 bar) K eq ( (1 bar) )1 () = = 0.37 () 1 Here, we have substituted the values for the concentration ratios to get K eq. Problem 9.12. (a) Addition of four more thiocyanate anions, SCN(aq), to the equilibrium solution depicted in the center of Figure 9.2 leads to the formation of a total of three iron-thiocyanate complexes at equilibrium in the solution. As in Worked Example 9.14, we count the number of each species represented in our solutions and assume the numbers are proportional to their molarities in the solution. After adding four more thiocyanate ions to the middle panel of Figure 9.2 and forming three iron-cyanate complexes in the resulting mixture, there will be nine thiocyanate ions unreacted (the original eight plus the four added minus the three in the complexes) and 17 iron ions (the original 20 minus the three in the complexes). The equilibrium constant for this system is then: [ Fe(SCN) 2+ (aq)] K = [ Fe 3+ (aq)]scn [ (aq)] = 3c (17c)(9c) = 1 51c (b) The equilibrium constants in Worked Example 9.14 and Check This 9.15 are 1/54c and 1/55c, respectively. In part (a), we get 1/51c. Within the limitations of these simple pictures with countable numbers of molecules, the agreement among these three values is quite good. Problem 9.13. (a) The pressure of each gas in the original mixture in the 1.00 L container at 500 K is given by P = nrt/v, where the number of moles of each gas is given in the problem statement: P(HI) = (2.21 10 3 mol)(8.314 10 2 L bar K 1 mol 1 )(500K) = 9.19 10 2 bar 1.00 L P(I 2 ) = (1.46 10 3 mol)(8.314 10 2 L bar K 1 mol 1 )(500K) = 6.07 10 2 bar 1.00 L P(H 2 ) = (2.09 10 5 mol)(8.314 10 2 L bar K 1 mol 1 )(500K) = 8.69 10 4 bar 1.00 L (b) The equilibrium constant expression for the reaction, H 2 (g) + I 2 (g) 2HI(g), is: ( HI(g) ) 2 K eq = H 2 (g) ( )( I 2 (g)) (c) Each of the dimensionless concentration terms in the equilibrium constant expression is a ratio of the pressure (in bar) of the reactant to the standard pressure, 1 bar. Therefore, we can ACS Chemistry Chapter 9 suggested solutions 5

Chemical Equilibria Chapter 9 substitute the numeric values for the pressures in part (a) into the expression in part (b) to get the numeric value of the equilibrium constant: ( 9.19 10 2 ) 2 K eq = = 1.60 ( 8.69 10 4 )6.07 ( 10 2 102 ) Problem 9.14. {NOTE: There is a typographical error in this problem. The concentrations should be [HIn(aq)] = 6.3 10 5 M and [In (aq)] = 8.7 10 5 M. Both are labeled [HIn(aq)].} (a) For the reaction, HIn(aq) + H 2 O(aq) H 3 O + (aq) + In (aq), the equilibrium constant expression is: ( K eq = H 3O + (aq))in ( (aq)) HIn(aq) ( )( H 2 O(aq) ) (b) The molar concentrations of the acidic and basic forms of the indicator are given and we can use the ph to get the molar concentration of the hydronium ion, [H 3 O + (aq)] = 10 7.41 M = 3.9 10 8 M. The ratio of these molar concentrations to the standard concentration, 1 M, gives the dimensionless concentration terms for these species. Since the solution is dilute, the water concentration is essentially the same as in pure water, 55.5 M, so the dimensionless concentration term for water is unity. The numeric value for the equilibrium constant is: 8 ( 3.9 10 )8.7 ( 10 5 ) K eq = = 5.4 10 ( 8 6.3 10 5 )1 () Problem 9.15. The information in this table is used in this problem and the next. Acid name Formula K a acetic acid CH 3 C(O)OH 1.8 10 5 chloroacetic acid ClCH 2 C(O)OH 1.4 10 3 (a) The acid equilibrium (proton transfer to water) equations for the acids are: CH 3 C(O)OH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 C(O)O (aq) ClCH 2 C(O)OH(aq) + H 2 O(l) H 3 O + (aq) + ClCH 2 C(O)O (aq) (b) The equilibrium constant expressions for K a for the reactions in part (a) are: K a ( H 3 O + (aq))( CH 3 C(O)O (aq)) (acetic acid) = (CH 3 C(O)OH(aq)) ( H 2 O(aq) ) H 3 O + (aq) K a (chloroacetic acid) = (ClCH 2 C(O)OH(aq)) H 2 O(aq) eq ( )( ClCH 2 C(O)O (aq)) ( ) eq 6 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria (c) The equilibrium constant for chloroacetic acid is nearly two orders of magnitude greater than that for acetic acid, that is, chloroacetic acid is a stronger acid than acetic acid (but still does not transfer all its protons to water). For solutions of equal concentration, chloroacetic acid will release more hydronium ions into solution, resulting in a lower ph than for acetic acid. Problem 9.16. (a) Use the equilibrium constant expressions and equilibrium constants from the previous problem to find the ph of a 0.050 M solution of acetic acid: ( )( CH 3 C(O)O (aq)) H K a (acetic acid) = 1.8 10 5 3 O + (aq) = (CH 3 C(O)OH(aq)) ( H 2 O(aq) ) species H 3 O + (aq) CH 3 C(O)O (aq) CH 3 C(O)OH(aq) initial mol 1.0 10 7-0 - 0.050 change in mol x formed x formed x reacts final mol x x 0.050 x Because K a (acetic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.050: K a = 1.8 10 5 x x = 0.050 x x 2 = 0.050 (1.8 10 5 ) = 9.0 10 7 x x 0.050 = x 2 0.050 x = 9.5 10 4 Check the assumption that the value of x is small and can be neglected relative to 0.050: 9.5 10 4 0.050 100% = 1.9% Our rule of thumb is that a 5% error is acceptable and this is well within the limit. ph = log (H 3 O + (aq)) = log (9.5 10 4 ) = 3.02 (b) Use the equilibrium constant expressions and equilibrium constants from the previous problem to find the ph of a 0.050 M solution of chloroacetic acid: H K a (chloroacetic acid) = 1.4 10 3 3 O + (aq) = (ClCH 2 C(O)OH(aq)) H 2 O(aq) eq ( )( ClCH 2 C(O)O (aq)) ( ) species H 3 O + (aq) ClCH 2 C(O)O (aq) ClCH 2 C(O)OH(aq) initial mol 1.0 10 7-0 - 0.050 change in mol x formed x formed x reacts final mol x x 0.050 x Because K a (chloroacetic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.050: eq ACS Chemistry Chapter 9 suggested solutions 7

Chemical Equilibria Chapter 9 K a = 1.4 10 3 x x x x x 2 = 0.050 x 0.050 = 0.050 x 2 = 0.050 (1.4 10 3 ) = 7.0 10 5 x = 8.4 10 3 Check the assumption that the value of x is small and can be neglected relative to 0.050: 8.4 10 3 0.050 100% = 17% Our rule of thumb is that a 5% error is acceptable and this value is far outside the limit, so we will have to discard the assumption and solve the problem more exactly. One approach is to use successive approximations. We start with the result just obtained for x, and use it to give us an approximation for the concentration ratio of the unreacted acid: 0.050 x = 0.050 0.0084 0.042 Then we repeat the above calculation with this new value for the acid concentration: K a = 1.4 10 3 x x x 2 0.042 = 0.042 x 2 = 0.042 (1.4 10 3 ) = 5.9 10 5 x = 7.7 10 3 If we consider a third approximation we would use this new value of x to give us another approximation for the concentration of the unreacted acid: 0.050 x = 0.050 0.0077 0.042 This value is identical (within the limits of uncertainty of the data given) to the one used in the previous calculation, so it will give the same result, x = 7.7 10 3. Thus, we take this result as the correct value for x = (H 3 O + (aq)). Another approach would have been to begin with the exact equation, rearrange it to quadratic form, ax 2 +bx + c = 0, and then solve using the quadratic formula: x = b ± b 2 4ac. The result is x = 7.7 10 3, so both methods give the same value. 2a The ph of the 0.050 M chloroacetic acid solution is: ph = log (H 3 O + (aq)) = log (7.7 10 3 ) = 2.11 (c) The results calculated in parts (a) and (b) confirm the prediction made in our solution to Problem 9.15(c). For acetic acid and chloroacetic acid at the same concentrations, the stronger acid, chloroacetic, transfers more protons to water and forms more hydronium ions. Thus, the ph is lower for chloroacetic acid. Problem 9.17. (a) For the autoionization reaction, H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq), K w = (H 3 O + (aq))(oh (aq)) and pk a = log{(h 3 O + (aq))(oh (aq))}. If the value of pk a decreases as the temperature increases, this means that the value of K w increases with increasing temperature: at 25 C: K w = 10 pk w = 10 14.0 = 1 10 14 at 60 C: K w = 10 pk w = 10 11.5 = 3 10 12 8 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria The product concentrations increase at the higher temperature. Le Chatelier s principle predicts that equilibria will shift toward formation of more product if energy is added (the temperature is increased) in endothermic reaction. The autoionization of water is an endothermic reaction. (b) An aqueous solution with H 3 O + (aq) concentration equal to 1.0 10 7 M at 25 C will have a ph of 7.0 [= log(h 3 O + (aq)) = log(1.0 10 7 )]. To find out whether the solution is acidic, basic, or neutral, we need to know how the concentrations of hydronium ion, [H 3 O + (aq)], and hydroxide ion, [OH (aq)], compare. Use K w to find [OH (aq)]: K w = 1 10 14 = (H 3 O + (aq))(oh (aq)) = (1.0 10 7 )(OH (aq)) (OH (aq)) = 1 10 7 [OH (aq)] = 1 10 7 M Because [H 3 O + (aq)] = [OH (aq)] (within the uncertainty of the data given), a solution in which [H 3 O + (aq)] = 1.0 10 7 M at 25 C is neutral. (c) An aqueous solution with H 3 O + (aq) concentration equal to 1.0 10 7 M at 60 C will have a ph of 7 [= log(h 3 O + (aq)) = log(1.0 10 7 )]. To find out whether the solution is acidic, basic, or neutral, we need to know how the concentrations of hydronium ion, [H 3 O + (aq)], and hydroxide ion, [OH (aq)], compare. Use K w to find [OH (aq)]: K w = 3 10 12 = (H 3 O + (aq))(oh (aq)) = (1.0 10 7 )(OH (aq)) (OH (aq)) = 3 10 5 [OH (aq)] = 3 10 5 M Because [H 3 O + (aq)] < [OH (aq)], a solution in which [H 3 O + (aq)] = 1.0 10 7 M at 60 C is basic. Note that a ph of 7.0 at 60 C means the solution is basic, but a ph of 7.0 at 25 C indicates a neutral solution. The relationship between [H 3 O + (aq)] and [OH (aq)] determines whether a solution is acidic, neutral, or basic. The neutral ph at a particular temperature depends upon the equilibrium constant at that temperature. Neutral ph at 60 C is about 5.8. Problem 9.18. We are asked to determine the order of acid strength for species acting as acids in these reactions, given that the position of equilibrium lies to the right (products are favored over reactants) in each reaction: (i) N 2 H + 5 (aq) + NH 3 (aq) NH + 4 (aq) + N 2 H 4 (aq) (ii) NH 3 (aq) + HBr(aq) NH + 4 (aq) + Br (aq) (iii) N 2 H 4 (aq) + HBr(aq) N 2 H + 5 (aq) + Br (aq) It is convenient to use the Brønsted-Lowry approach to identify the acids, proton donors, in these reactions. The acids are: N 2 H + 5 (aq), NH + 4 (aq), and HBr(aq). Recall that the position of equilibrium in an acid-base reaction favors the weaker acid (and base). Since these equilibria all favor products, the weaker acid in each reaction is on the product side. Note that HBr(aq) is never on the product side in these reactions, so we can conclude that it is probably the strongest of the three acids. Conversely, NH + 4 (aq) is always found on product side, so we can conclude that it is probably the weakest of the three acids. This leaves N 2 H + 5 (aq) as the intermediate strength acid and you see in reaction (i) that it is stronger than NH + 4 (aq). In reaction (iii), you see that N 2 H + 5 (aq) is weaker than HBr(aq). Thus, the relative acid strengths (strongest to weakest) are: HBr(aq) > N 2 H + 5 (aq) > NH + 4 (aq). ACS Chemistry Chapter 9 suggested solutions 9

Chemical Equilibria Chapter 9 Problem 9.19. (a) The ph of a 0.1 M solution of lactic acid can be calculated from its equilibrium constant expression and equilibrium constant (let LacH = lactic acid and Lac = lactate ion): LacH(aq) + H 2 O(aq) Lac (aq) + H 3 O + (aq) K a (lactic acid) = 1.4 10 4 (H = 3 O + (aq))(lac (aq)) (LacH(aq))(H 2 O(aq)) species H 3 O + (aq) Lac (aq) LacH(aq) initial mol 1.0 10 7-0 - 0.1 change in mol x formed x formed x reacts final mol x x 0.1 x Because K a (lactic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1: K a = 1.4 10 4 = x x 0.1 x x 2 = 0.1 (1.4 10 4 ) = 1.4 10 5 x x 0.1 = x2 0.1 x = 3.7 10 3 (uncertainty is larger than the ±4% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1: 3.7 10 3 0.1 100% = 3.7% Our rule of thumb is that a 5% error is acceptable and this result is within the limit. ph = log (H 3 O + (aq)) = log (3.7 10 3 ) = 2.4 (b) The ph of a 0.1 M solution of benzoic acid can be calculated from its equilibrium constant expression and equilibrium constant (let BzOH = benzoic acid and BzO = benzoate ion): BzOH(aq) + H 2 O(aq) BzO (aq) + H 3 O + (aq) K a (benzoic acid) = 6.5 10 5 (H = 3 O + (aq))(bzo (aq)) (BzOH(aq))(H 2 O(aq)) species H 3 O + (aq) BzO (aq) BzOH(aq) initial mol 1.0 10 7-0 - 0.1 change in mol x formed x formed x reacts final mol x x 0.1 x Because K a (benzoic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1: K a = 6.5 10 5 = x x 0.1 x x 2 = 0.1 (6.5 10 5 ) = 6.5 10 6 x x 0.1 = x2 0.1 eq eq 10 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria x = 2.5 10 3 (uncertainty is larger than the ±4% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1: 2.5 10 3 0.1 100% = 2.5% Our rule of thumb is that a 5% error is acceptable and this result is within the limit. ph = log (H 3 O + (aq)) = log (2.5 10 3 ) = 2.6 (c) The ph of a 0.1 M solution of aniline can be calculated from its equilibrium constant expression and equilibrium constant (let An = aniline and AnH + = protonated aniline): An(aq) + H 2 O(aq) AnH + (aq) + OH (aq) K b (aniline) = 4.3 10 10 (OH (aq))(anh + (aq)) = (An(aq))(H 2 O(aq)) species OH (aq) An(aq) AnH + (aq) initial mol 1.0 10 7 0.1-0- change in mol x formed x reacts x formed final mol x 0.1 x x Because K b (aniline) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1: K b = 4.3 10-10 = x x 0.1 x x 2 = 0.1 (4.3 10-10 ) = 4.3 10-11 x x 0.1 = x2 0.1 x = 6.6 10 6 (uncertainty is larger than the ±2% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1: 6.6 10 6 100% = 0.007% 0.1 Our rule of thumb is that a 5% error is acceptable and this result is well within the limit. poh = log (OH - (aq)) = log (6.6 10 6 ) = 5.2 ph = 14.00 poh = 14.0 5.2 = 8.8 (d) The ph of a 0.1 M solution of hydrazine can be calculated from its equilibrium constant expression and equilibrium constant: N 2 H 4 (aq) + H 2 O(aq) N 2 H + 5 (aq) + OH (aq) K b (hydrazine) = 8.9 10 7 (OH (aq))(n = 2 H + 5 (aq)) (N 2 H 4 (aq))(h 2 O(aq)) eq eq ACS Chemistry Chapter 9 suggested solutions 11

Chemical Equilibria Chapter 9 species OH (aq) N 2 H 4 (aq) N 2 H + 5 (aq) initial mol 1.0 10 7 0.1-0- change in mol x formed x reacts x formed final mol x 0.1 x x Because K b (hydrazine) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1: K b = 8.9 10 7 = x x 0.1 x x 2 = 0.1 (8.9 10 7 ) = 8.9 10 8 x x 0.1 = x2 0.1 x = 3.0 10 4 (uncertainty is larger than the ±3% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1: 3.0 10 4 100% = 0.3% 0.1 Our rule of thumb is that a 5% error is acceptable and this result is well within the limit. poh = log (OH - (aq)) = log (3.0 10 4 ) = 3.5 ph = 14.00 poh = 14.0 3.5 = 10.5 Problem 9.20. From Table 9.2, pk a = 9.24 for NH + 4 (aq) + H 2 O NH 3 (aq) + H 3 O + (aq). We are interested in the basic reaction of ammonia, NH 3 (aq) + H 2 O(aq) NH + 4 (aq) + OH (aq), and its pk b, which is 4.76 (= 14.00 pka). Using this value, we can calculate the ph of a 0.250 M aqueous ammonia solution. K b (NH 3 ) = 10 pk b = 10 4.76 = 1.7 10 5 (OH (aq))(nh + = 4 (aq)) (NH 3 (aq))(h 2 O(aq)) species OH (aq) NH 3 (aq) NH + 4 (aq) initial mol 1.0 10 7 0.250-0- change in mol x formed x reacts x formed final mol x 0.250 x x Because K b (NH 3 ) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.250: K b = 1.7 10 5 = x x 0.250 x x 2 = 0.250 (1.7 10 5 ) = 4.3 10 6 x x 0.250 = x 2 0.250 x = 2.1 10 3 Check the assumption that the value of x is small and can be neglected relative to 0.250: eq 12 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria 2.1 10 3 0.250 100% = 0.84% Our rule of thumb is that a 5% error is acceptable and this result is within the limit. poh = log (OH - (aq)) = log (2.1 10 3 ) = 2.68 ph = 14.00 poh = 14.00 2.68 = 11.32 Problem 9.21. The [H 3 O + (aq)] in a 0.350 M solution of aqueous hydrogen fluoride, HF, solution can be calculated from its equilibrium constant expression and equilibrium constant: HF(aq) + H 2 O(aq) F (aq) + H 3 O + (aq) K a (HF) = 6.46 10 4 (H = 3 O + (aq))(f (aq)) (HF(aq))(H 2 O(aq)) species H 3 O + (aq) F (aq) HF(aq) initial mol 1.0 10 7-0 - 0.350 change in mol x formed x formed x reacts final mol x x 0.350 x Because K a (HF) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.350: K a = 6.46 10 4 x x = 0.350 x x 2 = 0.350 (6.46 10 4 ) = 2.26 10 4 x x = 0.350 eq x 2 0.350 x = 1.50 10 2 Check the assumption that the value of x is small and can be neglected relative to 0.1: 1.50 10 2 100% = 4.30% 0.350 Our rule of thumb is that a 5% error is acceptable and this result is just within the limit. This percentage also represents the percentage of the HF(aq) that transfers its protons to water. Problem 9.22. Assume that the only source of hydronium ions in ph 3.4 lemon juice is transfer of protons from citric acid to water, H 3 Cit(aq) + H 2 O(aq) H 3 O + (aq) + H 2 Cit (aq), with K a = 7.4 10-4. From the reaction stoichiometry, (H 3 O + (aq)) = (H 2 Cit (aq)) = 10 ph = 10 3.4 = 4 10 4. We can use the equilibrium constant expression and solve to find (H 3 Cit(aq)) in the solution: K = 7.4 10 4 = (H 3 O+ (aq))(h 2 Cit (aq)) (H 3 Cit(aq)) (H 3 Cit(aq)) = (4 10 4 ) 2 = (4 10 4 ) 2 (H 3 Cit(aq)) 7.4 10 4 = 2 10 4 [H 3 Cit(aq)] = 2 10 4 M ACS Chemistry Chapter 9 suggested solutions 13

Chemical Equilibria Chapter 9 This is the amount of citric acid left undissociated. The total concentration of citric acid present in the solution is the concentration left unreacted, [H 3 Cit(aq)], plus the amount present as the anion, [H 2 Cit (aq)], that is, 2 10 4 M + 4 10 4 M = 6 10 4 M. Another way to do this problem is to let the original concentration (before proton transfer) of citric acid be c M. Then the equilibrium concentration is c [H 3 O + (aq)] M, so we have: 7.4 10 4 (4 10 4 ) 2 = (c (4 10 4 )) c = (7.4 10 4 )(4 10 4 ) + (4 10 4 ) 2 (7.4 10 4 = 6 10 4 M ) We see that both methods give the same result for the total concentration of citric acid in the solution (unreacted and monoanionic base form). Problem 9.23. (a) The aluminum cation, Al 3+, is relatively small (it has three fewer electrons than an aluminum atom) and has a high charge. The electric field produced by the ion is quite strong; water molecules can get fairly close to the ion and interact strongly with the field. When a water molecule is part of the hydration layer of the cation, the positive charge on the cation attracts electron density from the oxygen atoms toward the cation and makes the bonds between the oxygen atom and hydrogen atoms in the water much more polar than in the water molecule alone. Thus, it is easier for the hydrogen atoms on the waters in the hydration layer to be transferred to adjacent water molecules, than for proton transfer between water molecules in pure water. The result is that there is more proton transfer and higher hydronium ion concentration in the cation solutions than in pure water. The sodium ion (with only one fewer electrons than its atom) is a larger ion with lower charge than the aluminum ion. Water molecules are not nearly so strongly held by this larger ion with lower charge, so there is negligible effect on the hydronium ion concentration of the solution. (b) For the reaction, Al(H 2 O) 3+ 6 (aq) + H 2 O(aq) H 3 O + (aq) + Al(OH)(H 2 O) 2+ 5 (aq), the acid equilibrium constant expression is: K a ( H 3 O + (aq))( Al(OH)(H 2 O) 2+ 5 (aq)) = ( Al(H 2 O) 3+ 6 (aq))( H 2 O(aq) ) To find the numeric value for the equilibrium constant, we need to find the molar concentrations of the reactants and products in this equation. The ph of the solution, 2.9, gives us: [H 3 O + (aq)] = 10 -ph =10 2.9 M = 1 10 3 M The stoichiometry of the reaction tells us that: [H 3 O + (aq)] = [Al(OH)(H 2 O) 2+ 5 (aq)] = 1 10 3 M The solution contains 0.1 M aluminum ion, so, assuming all of this ion gets hydrated, the stoichiometry gives: [Al(H 2 O) 3+ 6 (aq)] = (0.1 M) [Al(OH)(H 2 O) 2+ 5 (aq)] = (0.1 M (1 10 3 M) = 0.1 M At these low solute concentrations, the water concentration is essentially the same as in pure water, 55.5 M, so the dimensionless concentration term for water is unity. The numeric value for the equilibrium constant is: 14 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria ( K a = 1 10 3 )1 ( 10 3 ) = 1 10 5 ; pk a = 5.0 ( 0.1)1 () (c) The smaller the value of pk a, the larger the value of K a. Since the pk a for the Fe(III) ion, 2.5, is much smaller than that for Al(III), 5.0 [from part (b)] the acid equilibrium reaction for Fe(III) goes further toward products and produces more hydronium ions. Fe(III) solutions are more acidic than Al(III) solutions. Problem 9.24. (a) For the reaction, HA(aq) + H 2 O(aq) H 3 O + (aq) + A (aq), the acid equilibrium constant expression is: K a ( H 3 O + (aq))( A (aq)) ( = ( HA(aq) )( H 2 O(aq) ) = H 3 O + (aq))( A (aq)) ( HA(aq) ) For a 0.215 M HA solution, with a ph of 4.66, we have: (H 3 O + (aq)]) = 10 ph = 10 4.66 = 2.2 10 5 and the stoichiometry of the reaction tells us that: (H 3 O + (aq)) = (A (aq)) = 2.2 10 5 It is easy to observe that much less than 5% of the original acid, [HA(aq)] = 0.215 M, has reacted. ( 2.2 10 5 )2.2 ( 10 5 ) K a = = 2.3 10 9 0.215 ( ) (b) For the reaction, A (aq) + H 2 O(aq) OH (aq) + HA(aq), the base equilibrium constant expression is: K b ( OH (aq))( HA(aq) ) = A (aq) ( ) We know that K a K b = K w = 1.00 10 14, so: K b = K w = (1.00 10 14 ) = 4.4 Ka (2.3 10 9 10 6 ) The stoichiometry of the base reaction shows us that [OH (aq)] = [HA(aq)], so, assuming that a negligible amount of A (aq) reacts, we can rearrange the equilibrium constant expression to find the hydroxide concentration and ph of the solution: (OH (aq))(ha(aq)) = (OH (aq)) 2 = K b (A (aq)) = (4.4 10 6 )(0.175) = 7.7 10 7 (OH (aq)) = 8.8 10 4 ; poh = 3.06 ph = 14.00 poh = 14.00 3.06 = 10.94 Check the assumption that the value of (OH (aq))x is small and can be neglected relative to 0.175: 8.8 10 4 100% = 0.50%.175 ACS Chemistry Chapter 9 suggested solutions 15

Chemical Equilibria Chapter 9 Our rule of thumb is that a 5% error is acceptable and this result is well within the limit. Problem 9.25. (a) For the reaction, HO(O)CC(O)O (aq) + H 2 O(aq) O(O)CC(O)O (aq) + H 3 O + (aq), pk a2 = 4.27 ( a2 because it is the second proton transferred from oxalic acid). Since pk a + pk b = 14.00 for a conjugate acid-base pair, the pk b for O(O)CC(O)O (aq) is 9.73. (b) The molar mass of disodium oxalate is 134 g mol 1, so a solution containing 35 g of this salt in a liter of water (assumed to yield a liter of solution) gives [ O(O)CC(O)O (aq)] = 0.26 M. The basic reaction in this solution is: O(O)CC(O)O (aq) + H 2 O(aq) HO(O)CC(O)O (aq) + OH (aq) The stoichiometry gives [OH (aq)] = [HO(O)CC(O)O (aq)] and we can use the equilibrium constant value from part (a), with the assumption that a negligible amount of the oxalate ion reacts, to get the (OH (aq)) and the ph of the solution: K b ( OH = 10 9.73 = 1.9 10 10 (aq))( HO(O)CC(O)O (aq) ( OH (aq)) 2 = = O(O)CC(O)O (aq ( 0.26) ( ) (OH (aq)) = (1.9 10 10 )(0.26) = 7.0 10 6 ; poh = 5.15 ph = 14.00 poh = 14.00 5.15 = 8.85 Checking the assumption, we see that 7.0 10 6 M (concentration of hydrogen oxalate ion) is much less than 5% of the original, 0.26 M, oxalate ion, so a negligible amount of the oxalate has reacted. Problem 9.26. (a) For ease in writing, let HA = HC(O)OH (methanoic acid) and A = HC(O)O (methanoate anion). The proton transfer reaction, equilibrium reaction (for a dilute solution), and equilibrium constant are: HA(aq) + H 2 O(aq) H 3 O + (aq) + A (aq) K a ( H 3 O + (aq))( A (aq)) ( = ( HA(aq) )( H 2 O(aq) ) = H 3 O + (aq))( A (aq)) = 1.8 10-4 ( HA(aq) ) From the reaction stoichiometry we see that (H 3 O + (aq)) = (A (aq)), as in several of the previous problems. Substitute this equality, rearrange the equilibrium constant equation, and solve for (H 3 O + (aq)): (H 3 O + (aq)) 2 = K a ( HA(aq)) (H 3 O + (aq)) = K a (HA(aq)) Assuming that a negligible amount of HA(aq) transfers its protons to water, we can begin with the approximation that (HA(aq)) = c = initial concentration of HA(aq) in the solution, so: (H 3 O + (aq)) = K a c In the present example, we have: (H 3 O + (aq)) = K a c = (1.8 10 4 )(0.100) = 4.2 10 3 16 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria ph = log(h 3 O + (aq)) = log(4.2 10 3 ) = 2.38 We can check the assumption (approximation) and get the percent ionization of the acid in one operation: 4.2 10 3 % ionization = 100% = 4% 0.100 Although this result is within the limit we have set (5%) for accepting the approximation that a negligible amount of acid has transferred its protons, we should check to see whether this much ionization affects the result we get for (H 3 O + (aq)). One way to do this is to assume that 4% of the acid reacts and use the lower concentration in the equation to see what (H 3 O + (aq)) is: (H 3 O + (aq)) = (1.8 10 4 )(0.096) = 4.1 10 3 Within the precision of the data, this result is the same as the previous one, so the 4% ionization result is acceptable. (b) Le Chatelier s principle predicts that disturbing an equilibrium system by decreasing the concentration of a reactant results in the system responding to form more of the reactant, thereby decreasing the concentration of the product(s). Le Chatelier s principle is, however, powerless to make a quantitative prediction of the result, so the best we can say is that the [H 3 O + (aq)] and [A (aq)] will be lower when [HA(aq)] is lower. (c) Begin by using the simplest approach and then checking to see whether the results make sense or whether further work is necessary. For a 0.050 M solution of HA(aq), we find: (H 3 O + (aq)) = (1.8 10 4 )(0.050) = 3.0 10 3 Our prediction of a lower concentration of H 3 O + (aq) [and A - (aq)] is borne out. The percent ionization is: 3.0 10 3 % ionization = 100% = 6% 0.050 This result, 6% ionization, is larger than for the more concentrated acid, a result that Le Chatelier s principle could not predict and is also larger than we usually are willing to accept without trying a less approximate calculation. We should again try what we did in part (b), substitute this tentative result back into the equation to see how it affects the outcome: (H 3 O + (aq)) = (1.8 10 4 )(0.047) = 2.9 10 3 Again we find that our result, within the precision of the data, is not changed, so we can have reasonable confidence that our conclusion is valid: as the concentration of a weak acid is decreased, the percent that ionizes increases. Problem 9.27. In essentially all the problems we have presented, the concentration of hydronium ion produced by the transfer of protons from an acid is a good deal larger than 10 7 M, the maximum concentration of hydronium ions that water autoionization can produce. In addition, the presence of the hydronium ion from the acid decreases the amount of hydronium ion produced by water autoionization because the added hydronium ion is a disturbance to the autoionization equilibrium and the response of the system is to react to try to use up the added product which reduces the amount of hydroxide ion and hydronium ion formed by autoionization. If the concentration of hydronium ion from the acid we add to the system is very low, comparable to ACS Chemistry Chapter 9 suggested solutions 17

Chemical Equilibria Chapter 9 10 7 M, then we have to account for the autoionization, because it will be adding a significant amount of hydronium ion to the solution. The conditions under which we have to be account for the autoionization of water can be determined by looking at the equation developed in the solution for Problem 9.26(a): (H 3 O + (aq)) = K a c. When the added concentration of acid is very low and/or when the acid dissociation constant for the added acid is very small, the amount of hydronium ion from the added acid will be tiny and could be comparable to 10 7 M. Problem 9.28. A solution contains a buffer system, if significant concentrations of the protonated and unprotonated forms of a weak acid-base conjugate pair are present. In case (i), a solution of HClO 4 and NaClO 4, no significant concentration of the protonated form, HClO 4 (aq) exists in the solution, because HClO 4 (aq) is a strong acid and transfers essentially all its protons to water. In case (ii), a solution of Na 2 CO 3 and NaHCO 3, the HCO 3 (aq) anion is a weak acid (also a weak base that is not relevant to this discussion) that is in equilibrium with its conjugate base: HOCO 2 (aq) + H 2 O(aq) H 3 O+(aq) + CO 2 3 (aq) Since the solution contains both the conjugate acid and conjugate base, this equilibrium reaction provides buffering action. Problem 9.29. The reaction that is relevant to this problem is: HOAc(aq) + H 2 O(aq) H 3 O + (aq) + OAc (aq) If NaOAc(s) is added to an aqueous solution of acetic acid, the concentration of OAc (aq) will increase as the solid dissolves. Adding a product is a disturbance to the equilibrium system, so the system will react by trying to decrease the effect of the disturbance, that is, by shifting toward reactants to use up some of the added product. The result will also be to decrease the concentration of H 3 O + (aq), because it is required to react with the added acetate ion. The result of addition of the conjugate base to a weak acid-base equilibrium system is to lower the (H 3 O + (aq)), that is, raise the ph of the solution. Problem 9.30. The reaction that is relevant to this problem is: NH 3 (aq) + H 2 O(aq) NH + 4 (aq) + OH (aq) If NH 4 Br(s) is added to an aqueous solution of ammonia, the concentration of NH + 4 (aq) will increase as the solid dissolves. Adding a product is a disturbance to the equilibrium system, so the system will react by trying to decrease the effect of the disturbance, that is, by shifting toward reactants to use up some of the added product. The result will also be to decrease the concentration of OH (aq), because it is required to react with the added ammonium ion. The result of addition of the conjugate acid to a weak acid-base equilibrium system is to lower the (OH (aq)), that is, lower the ph of the solution. Problem 9.31. The reaction that is relevant to this problem is: HOAc(aq) + H 2 O(aq) H 3 O + (aq) + OAc (aq) 18 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria Because HOAc(aq) is a weak acid, we know that only a tiny amount of it will transfer its protons to water in a 0.250 M solution, so (H 3 O + (aq)) = (OAc (aq)) will be rather small in this solution. Using the equation in the solution for Problem 9.26 we have: (H 3 O + (aq)) = K a c = (1.7 10 5 )(0.250) = 2.1 10 3 Addition of sufficient solid NaOAc to make the final sodium ion concentration, [Na + (aq)], 1.50 M will enormously increase (OAc (aq)) and decrease the amount of HOAc(aq) that reacts by the above equation. We can start with the approximation that (HOAc(aq)) and (OAc (aq)) have the values, 0.250 and 1.50, respectively, they would have, if no proton transfers took place. Thus, we have: K a ( H = 1.7 10-5 3 O + (aq))oac ( (aq)) ( H = 2 O(aq) )( 1.50) ( HOAc(aq) ) 0.250 (H 3 O + (aq)) = 2.8 10-6 ; ph = 5.55 As we said, addition of the OAc (aq reduces the (H 3 O + (aq)) (by a factor of about 1000 in this case), which is also the direction of the change predicted in the solution for Problem 9.29. The hydronium ion that is no longer present reacted with the added OAc (aq), the amount of OAc (aq) reacted is about 0.14% = (2.1 10 3 ) (1.50) 100%, so a negligible amount has been used up and our assumptions about the concentrations of the conjugate acid and conjugate base are valid. Problem 9.32. If x = (OAc (aq)) in a 0.25 M acetic acid-acetate buffer solution, then (HOAc(aq)) = (0.25 x) in this solution. If the ph = 5.36, then (H 3 O + (aq)) = 4.4 10-6. Substitute these values into the equilibrium constant expression and solve for x to find the concentrations of acetate and acetic acid that will produce this buffer solution: ()4.4 x K a ( 10-6 ) = 1.7 10-5 = 0.25 x x = (OAc (aq)) = 0.20; (HOAc(aq)) = (0.25 x) = 0.05 [OAc (aq)] = 0.20 M [HOAc(aq)] = 0.05 M Problem 9.33. (a) The equilibrium reaction, equilibrium constant expression, and equilibrium constant for lactic acid transferring a proton to water are: HC 3 H 5 O 3 (aq) + H 2 O(aq) H 3 O + (aq) + C 3 H 5 O 3 (aq) K a ( H = 1.4 10 4 3 O + (aq))c ( 3 H 5 O 3 (aq)) ( + (aq))c ( 3 H 5 O 3 (aq)) = HOC 3 H 5 O 2 (aq) HOC 3 H 5 O 2 (aq) ( )( H 2 O(aq) ) = H 3 O ( ) For a buffer solution composed of 0.15 M lactic acid, HOC 3 H 5 O 2, and 0.10 M sodium lactate, NaOC 3 H 5 O 2, we begin by making the approximation that these concentrations are negligibly affected by the proton transfers, and solve the equilibrium constant expression to find (H 3 O + (aq)): ACS Chemistry Chapter 9 suggested solutions 19

Chemical Equilibria Chapter 9 1.4 10 4 = ( H 3 O + (aq)) 0.10 ( ) ( 0.15) (H 3 O + (aq)) = 2.1 10 4 ; ph = log(2.1 10 4 ) = 3.68 Because (H 3 O + (aq)) is small compared to the initial concentrations of the conjugate acid-base pair, the approximation that their concentrations are not appreciably affected is valid. An alternative (equivalent) solution is to use the Henderson-Hasselbalch equation [with pk a = -logk a = log(1.4 10 4 ) = 3.85]: ( conjugate base) 0.10 ph = pk a + log ( conjugate acid = 3.85 + log = 3.85 + ( 0.18) = 3.67 ) 0.15 Within the round-off uncertainty of the calculations, the ph calculated by each route is the same. (b) Half way to the equivalence point in the titration between 0.15 M lactic acid and 0.15 M NaOH, the concentration of the lactic acid and its conjugate base will be equal, and the ph will equal the pk a, 3.85, as we see from the Henderson-Hasselbalch equation when (conjugate base) = (conjugate acid): ( conjugate base) ( conjugate base) ph = pk a + log ( conjugate acid = 3.85 + log ) ( conjugate base = 3.85 + log(1) ) = 3.85 + 0 = 3.85 In part (a), where there was more conjugate acid than conjugate base, the ph of the buffer solution was lower, more acidic. The solution had a higher concentration of hydronium ion than in the solution here. Problem 9.34. (a) This is a buffer solution containing a weak acid, HA = phenol, pk a = 9.98, and its conjugate base, A = phenolate ion. The ph of the solution is: ph = pk a + log (A (aq)) = 9.98 + log 0.377 (HA(aq)) 0.551 = 9.81 Note that this solution with a higher concentration of the conjugate acid than the conjugate base has a lower numeric value of ph (more acidic) than the numeric pk a value. (b) Assume that all the hydronium ion produced by the added HCl (0.100 M in the 1 L of solution) reacts with the phenolate anion to produce phenol. The new concentration of the phenolate anion will be 0.227 M (= 0.377 M 0.100 M) and the concentration of phenol will be 0.651 M (= 0.551 M + 0.100 M). The new ph will be: ph = 9.98 + log 0.277 0.651 = 9.61 As we would expect, the addition of hydronium ion (from the HCl) lowers the ph, but by only 0.20 ph units in this buffered solution. (c) Assume that all the added hydroxide ion (0.125 M in 1 L of solution) from the NaOH reacts with phenol to produce phenolate anion. The new concentration of phenolate anion will be 0.502 20 ACS Chemistry Chapter 9 suggested solutions

Chapter 9 Chemical Equilibria M (= 0.377 M + 0.125 M) and the concentration of phenol will be 0.426 M (= 0.551 M 0.125 M). The new ph will be: ph = 9.98 + log 0.502 0.426 = 10.05 As we would expect, the addition of hydroxide ion (from the NaOH) raises the ph, but by only 0.24 ph units in this buffered solution. Problem 9.35. (a) If the density of bleach and water are the same, then 1 L of bleach has a mass of 1.00 kg and 5% of this, about 50 g, is NaOCl. The molar mass of NaOCl is 74.5 g mol 1, so the molarity of NaOCl in the solution and, hence, the concentration of the aqueous anion is: [OCl 50 g L 1 (aq)] = = 0.7 M 1 74.5 g mol Only one significant figure is justified by the data (5% solution). (b) The reaction of the hypochlorite ion (a weak base) with water is: OCl (aq) + H 2 O(aq) HOCl(aq) + OH (aq) The equilibrium constant expression for this reaction is: K b ( OH (aq))( HOCl(aq) ) = ( OCl (aq))( H 2 O(aq) ) The symbol, K b, we have used is the usual designation for the equilibrium constant that refers to the reaction of a base with water to form hydroxide ion. The dimensionless concentration ratio for water is included in the expression, but often (usually) it is left out, since, in dilute solutions, it is unity. The numeric value for K b is easily obtained from the pk a = 7.75 value for hypochlorous acid, the conjugate acid of the hypochlorite anion: pk b = 14.00 pk a = 6.25; K b = 10 6.25 = 5.6 10 7 (c) Assuming that a negligible amount of the hypochlorite anion reacts and that (HOCl(aq)) = (OH (aq)) in the resulting solution, we can write: (OH (aq)) 2 = K b (OCl (aq)) = (5.6 10 7 )(0.7) = 4 10 7 (OH (aq)) = 6 10 4 (H 3 O + 1.00 10 14 (aq)) = = 2 10 11 ; ph = log(2 10 11 ) = 10.8 6 10 4 (d) To lower the ph of the bleach solution to 6.5, you would have to add hydrochloric acid to react with the hypochlorite ion to convert much of it to hypochlorous acid. This target ph is a lower numeric value than the pk a for hypochlorous acid, so the ratio of conjugate base to conjugate acid (assuming this is a buffer solution) is less than unity. (e) If we assume (as in part (d)) that the solution of hypochlorous acid and hypochlorite ion is a buffer, we can use the equation for the ph of a buffer solution (the Henderson-Hasselbalch equation) rearranged to find the ratio of the base to acid: ACS Chemistry Chapter 9 suggested solutions 21