Answer Key, Problem Set 7 (With explanations)

Transcription

1 Chemistry 1 Mines, Spring 014 Answer Key, Problem Set 7 (With eplanations) (Write out net ionic uations that represent what happens in each case. Represent the weak acid as HA and its conjugate base as A ); (ab) Assume 5 C!; ; & 16.59(ac); & 16.11; (abc); ; **; ; Assume ual initial volumes!; ; **; ; 14. NT1; ( all species H SO 3, HSO 3, H 3O +, OH, and SO 3. Assume T = 5 C. Also calculate the ph.); & (ecept omit H S from the problem); Suppose that a buffer contains ual amounts of a weak acid and its conjugate base. (a) What happens to the relative amounts of the weak acid and conjugate base when a small amount of strong acid is added to the buffer? (b) What happens when a small amount of strong base is added? (Write out net ionic uations in each case.) Answer: (a) The amount of conjugate base goes down a bit and the amount of weak acid goes up. (b) The amount of conjugate acid goes down and the amount of conjugate base goes up. Eplanation: (a) When a bit of strong acid (e.g., HNO 3 ) is added to a buffer (weak acid HA; weak base A ), it reacts with the best base around, which is A. The reaction that occurs can be represented by: HNO 3 + A PS HA + NO 3 % (OR H 3O + + A 100 % HA + H O [If you allow the HNO 3 to react with water first to make H 3O + ) This reaction goes until somebody runs out. I.e., it goes to completion, which is to say, it has a K value that is etremely large. This can be seen in two ways: 1) The first reaction (above) must be more product favored than the reaction of HNO 3 with H O since A is a better base than H O, and ) the second reaction (above) is the eact reverse of the acid ionization reaction uation, and thus has 1 1 K. Since HA is,by definition, weak (or else you would not have a buffer!), K a 1 1 K a Ka As such, the net result is that A is converted into HA. If the moles of strong acid added are less than the moles of A present, the strong acid is the limiting reactant, and only a fraction of the original A will be converted into HA. Thus a buffer will remain after the reaction occurs (but the ph will be a bit lower). The reaction can be visualized as follows: HA A (b) When a bit of strong base (e.g., NaOH) is added to a buffer, it reacts with the best acid around, which is HA. The reaction that occurs can be represented by: OH + HA 100 % A + H O [If you allow the NaOH to dissociate first to make OH This reaction goes until somebody runs out. + HNO 3 H 3O + + A HA + H O HA A I.e., it goes to completion, which is to say, it has a K value that is etremely large. This can be seen by recognizing that 1 the reaction (above) is the eact reverse of the base ionization reaction uation, and thus has K. Since A is, by 1 definition, weak (or else you would not have a buffer!), K b 1 1 K b As such, the net result is that HA is converted into A. If the moles of strong base added are less than the moles of HA present, the strong base is the limiting reactant, and only a fraction of the original HA will be converted into A. Thus a buffer will remain after the reaction occurs (but the ph will be a bit higher). The reaction can be visualized as follows: + OH OH + HA A + H O HA A HA A K b

2 (ab) Assume 5 C!; For each solution, calculate the initial and final ph after adding mol of NaOH. (a) 50.0 ml of pure water (b) 50.0 ml of a buffer solution that is M in HCHO and 0.75 M in KCHO Answers: (a) 7.00 before, 1.60 after; (b) 3.89 before, 4.05 after Strategy: ph s before addition of NaOH: (a) Pure water at 5 C has [H 3 O M and ph 7.00 (see PS5 and PS6) (b) As discussed in PS6, the ph of a buffer with known initial (prepared) concentrations can be determined straight away (because the small approimation applies to both HA and A ) using: [H 3 O + [HA moles HA [A moles A K a OR ph pk a + log [A moles A or [HA moles HA (or you could do a longer ICE approach with small approimations) NOTE: Find K a(hcho ) in Table ph s after addition of NaOH: Always consider dissociation of soluble ionic compounds first. Thus, the first thing to occur (in either case) is that NaOH will dissociate into Na + (a negligible acid) and OH. For (a). The only other process that occurs is a shifting in the water autoionization uilibrium. Thus, to calculate ph, find the [OH after the dissociation, and assume that is the uilibrium value. Then use the water autoionization uilibrium condition (K w [H 3 O + [OH ) to find [H 3 O + and then ph. For (b) (a buffer). The primary process that occurs after the dissociation is that OH will react with HA (see Q1 above) until somebody runs out. So first calculate the moles of HCHO in the solution and determine whether it is more or less than the mol of NaOH added. 1) If it is greater, then the resulting solution will be a buffer since there will still be some acid and base component remaining (use an ICF table to find final moles of HA and A ** ). You can determine the ph of the buffer using [H 3 O + moles HA = K a (or the HH uation) moles A **NOTE: You must let KCHO dissociate first into K + (ignore) and CHO ( A ). Then find moles CHO and use in the ICF calculation. ) If it is smaller, then find the concentration of ecess OH and proceed as in (a). Eecution of Strategy: (a) ph before addition of NaOH: 7.00 (neutral at 5 C) ph after: mol NaOH in 50.0 ml mol Na + and mol OH in 50.0 ml ( 0.50 L) mol OH [OH 0.50L 0.040M ph = log( ) = 1.60 = 1.60 (b) ph before NaOH: [H 3 O + K a [HA [A 0.195M M [H O ph log( ) Answer is reasonable. pk a here is log( ) = This buffer has more A than HA ph should be a bit higher than 3.74 and it is. M PS7

3 ph after NaOH added: (initial) moles HCHO = L 0.195molHCHO mol L (initial) moles CHO = L 0.75molCHO L Reaction that occurs: OH + HCHO CHO mol + H O (OH will react with the strongest acid around.) Decent amounts of HCHO and CHO are present after this reaction a buffer! [H 3 O + K a moles HCHO moles CHO ph log( ) NOTE: The increase in ph in pure water was units. The increase in ph in the buffer was only units. The buffer worked! Determine whether or not the miing of the two solutions will result in a buffer. (a) 75.0 ml of 0.10 M HF; 55.0 ml of 0.15 NaF (b) ml of 0.10 M HF; ml of M HCl (c) ml of 0.10 M HF; ml of M KOH (d) 15.0 ml of 0.15 M CH 3NH ; 10.0 ml of 0.5 M CH 3NH 3Cl (e) ml of 0.15 M CH 3NH ; 95.0 ml of 0.10 M HCl Answers: (a), (c), (d), and (e) will result in buffers. (b) will not Strategy: in moles OH (moles) HCHO (mol) CHO (mol) I (LR) C F NOTE: A buffer must contain some of a weak acid and some of its conjugate base. So if those two components are mied together directly, clearly a buffer will result. However, one must realize that if a strong acid or strong base is one of the substances being mied a reaction may occur to generate one of the components of the buffer that is not there initially. So you must be on the lookout for this kind of miture. As will be described below, in such an instance, a calculation needs to be done to see whether or not a buffer will result. 1) Look to see what two kinds of substances are being mied. If the following substances are mied, you can make the conclusion right away without doing a calculation: a) If two acids are mied, it won t be a buffer. (You would neither have or generate the weak base component.) b) If two bases are mied, it won t be a buffer. (You would neither have or generate the weak base component.) c) If a weak acid and a Gp I salt of its conjugate base are mied (e.g., HA and NaA or KA), it will be a buffer. d) If a weak molecular base and a salt of its conjugate acid with a negligible base anion (e.g., RNH and RNH 3 Cl, where R represents C (with other atoms bound to it) or H), it will be a buffer. ) In the following two cases, a calculation (an ICF type) will be needed: 5 M Answer is reasonable. There is more CHO and less HCHO here than in the original buffer. Thus, the ph should be a bit higher than 3.89 and it is. PS73

4 d) If a weak acid and strong base are mied, it may be a buffer because they will react with one another until one of them runs out, producing some of the conjugate base of the weak acid. The question becomes, What will be left over? So do an ICF type calculation (with moles) and assess the result: i) If there is ecess weak acid (i.e., the SB is limiting), the resulting solution is a buffer. ii) If there is ecess strong base (i.e., the WA is limiting), the resulting solution will not be a buffer (in effect, a buffer was initially created but then so much SB was around that the buffer capacity was eceeded). iii) If there is no ecess of either, the resulting solution will not be a buffer. The solution contains only the conjugate base of the weak acid. (Solution of a weak base.) e) If a weak base salt (i.e., Gp I salt of a base anion) and a strong acid are mied, it may be a buffer because they will react with one another until one of them runs out, producing some of the conjugate acid of the weak base. The question becomes, What will be left over? So do an ICF type calculation (with moles) and assess the result: Eecution of Strategy: i) If there is ecess weak base (i.e., the SA is limiting), the resulting solution is a buffer. ii) If there is ecess strong acid (i.e., the WB is limiting), the resulting solution will not be a buffer (in effect, a buffer was initially created but then so much SA was around that the buffer capacity was eceeded). iii) If there is no ecess of either, the resulting solution will not be a buffer. The solution contains only the conjugate acid of the weak base. (Solution of a weak acid.) (a) 75.0 ml of 0.10 M HF; 55.0 ml of 0.15 NaF HF and NaF are mied. HF and F will both be present. This will make a buffer. The amounts are basically irrelevant since neither one is ridiculously small. (b) ml of 0.10 M HF; ml of M HCl HF and HCl are both acids. No F will be present initially nor generated by any reaction. This will NOT make a buffer. (c) ml of 0.10 M HF; ml of M KOH HF is mied with KOH. They will react to form F : HF + OH Do an ICF type calculation to see who s limiting. moles HF: L 0.10 mol/l mol HF moles OH : L mol/l mol OH % ~100 F + H O L.R. This will make a buffer. NOTE: You need not actually complete a full ICF table to make this conclusion, but just to prove it to you (and for completeness), I will show the results of that here: in moles OH HF F I (LR) C F (d) 15.0 ml of 0.15 M CH 3 NH ; 10.0 ml of 0.5 M CH 3 NH 3 Cl CH 3 NH (a weak base) and CH 3 NH 3 + (its conjugate) will be present. This will make a buffer. (e) ml of 0.15 M CH 3 NH ; 95.0 ml of 0.10 M HCl A buffer! PS74

5 CH 3 NH is mied with HCl. They will react to form CH 3 NH 3 + : CH 3 NH + HCl % ~100 CH NH + + Cl 3 3 Do an ICF type calculation to see who s limiting. moles CH 3 NH : L 0.15 mol/l mol CH 3 NH moles HCl: L 0.10 mol/l mol HCl L.R. This will make a buffer. NOTE: You need not actually complete a full ICF table to make this conclusion, but just to prove it to you (and for completeness), I will show the results of that here: & 16.59(ac). Determine whether or not each addition eceeds the capacity of the buffer : Buffer contains 0.10 mol of a weak acid and 0.10 mol of its conjugate base in 1.0 L of solution. (a) 0.00 mol of NaOH (b) 0.00 mol of HCl (c) 0.10 mol of NaOH (d) mol of HCl 16.59: A ml buffer is M in HNO and M in KNO. (a) 50 mg NaOH (b) 350 mg KOH (c) 1.5 g HBr (d) 1.35 g HI Answers: : Only (c) eceeds the capacity; Strategy: 16.59: None eceed the capacity! Eceeding the buffer capacity means reacting away all of a buffer component. Thus, compare: moles of strong acid added to the moles of weak base component initially present OR moles of strong base added to the moles of weak acid component initially present IF: moles SA moles WB (initial) all the WB will run out (L.R.) and buffer capacity will be eceeded. OR moles SB moles WA (initial) all the WA will run out (L.R.) and buffer capacity will be eceeded. Otherwise, the buffer capacity will not be eceeded. Note the similarity of this problem to the prior one. The key idea is that whether a SA (or SB) is added to a buffer or a solution of a WB (or WA), an reaction to completion will occur, and you may end up with a buffer, or you may end up with ecess SA (or SB) [obviously not a buffer. Eecution of Strategy: in moles CH 3NH HCl CH 3NH 3 + I (LR) 0 C F A buffer! (Perhaps the authors meant the initial volume of the buffer to be 50.0 ml? This is quite silly to me. Sorry I didn t catch it earlier) : Buffer contains 0.10 mol of a weak acid and 0.10 mol of its conjugate base in 1.0 L of solution. (a) 0.00 mol of NaOH 0.00 mol OH < 0.10 mol HA OH runs out; some HA left over capacity not eceeded (b) 0.00 mol of HCl 0.00 mol HCl < 0.10 mol A HCl runs out; some A left over capacity not eceeded (c) 0.10 mol of NaOH 0.10 mol OH = 0.10 mol HA NaOH and HA will run out PS75

6 capacity is eceeded (d) mol of HCl mol HCl < 0.10 mol A HCl runs out; some A left over capacity not eceeded 16.59: A ml buffer is M in HNO and M in KNO. moles HNO : L mol/l = mol HNO moles NO : L mol/l = mol NO (a) 50 mg NaOH moles NaOH: 1g 1mol 50mgNaOH 1000mg ( )g mol OH (after dissociation) < mol HNO OH runs out; some HNO left over (b) 350 mg KOH [Removed formally from set on 3/14/1 capacity not eceeded (not even close!) (Let KNO dissociate into K + + NO ) molNaOH moles KOH: 1g 1mol 350mgNaOH mol KOH 1000mg ( )g (c) 1.5 g HBr mol OH (after dissociation) < mol HNO moles HBr: OH runs out; some HNO left over capacity not eceeded (sorry for the repetition; this is nearly identical to (a)!) 1mol 1.5gHBr ( )g mol HBr < mol NO mol HBr HBr runs out; some NO left over capacity not eceeded (still not very close!) (d) 1.35 g HI [Removed formally from set on 3/14/1 moles HBr: 1mol 1.35gHI mol HI ( )g mol HI < mol NO HI runs out; some NO left over capacity not eceeded (sorry for the repetition) PS76

7 & The graphs labeled (a) and (b) show the titration curves for two ualvolume samples of monoprotic acids, one weak and one strong. Both titrations were carried out with the same concentration of strong base (a) 1 (b) ph ph Volume of base added (ml) Volume of base added (ml) (i) What is the approimate ph at the uivalence point of each curve? Answers: (a) ~8.5; (b) ~7.0 (we cannot say eactly 7.00 because the T was not specified as 5 C) (ii) Which curve corresponds to the titration of the strong acid and which one to the titration of the weak acid? Answer: The curve in (b) is the strong acid; in (a) is the weak acid Reasons: (i) The uivalence point of a titration of a monoprotic acid is the point at which moles of added base = initial moles of acid. This point will also be the point on the curve that is the steepest point, which is also called an inflection point (it is roughly in the middle of the steep segment that appears nearly linear). These points are shown by an on each plot. (ii) At the uivalence point, by definition, all the HA has been reacted away and turned into A (because the titration reaction is: OH + HA 100 % A + H O ) Thus, the solution at the uivalence point is a solution of A (with Na + around as well): If HA is a weak acid, then A is a weak base (PS5 ideas), and the solution will be basic. If HA is a strong acid, A will be a negligible base (PS5), and the solution will be neutral. Since the uivalence point ph is well above 7.0 (basic) in (a) it must be the one with the weak acid. The one in (b), with a ph of ~7.0 (~neutral) must contain the strong acid (assume T ~5 C) The ph at the uivalence point of the titration of a strong acid with a strong base is 7.0 (AT 5 C!!). However, the ph at the uivalence point of the titration of a weak acid with a strong base is above 7.0. Why? Answer: At the e.p., the solution is the same as a solution of A. Thus if HA is weak, a solution of A, a weak base, will be basic. But if HA is strong, a solution of A, a negligible base, will be neutral. See above reason for (ii) for more details (abc) Two 5.0mL samples, one M HCl and the other M HF, were titirated with 0.00 M KOH. (a) What is the volume of added base at the uivalence point for each titration? Answer: 1.5 ml (for both) Work / Reasoning: 1) At the uivalence point (ep), moles of base added = moles of initial acid (monoprotic case) moles NaOH added moles HCl 5.0 ml 1L 1000mL 0.100molHCl L mol ) Knowing moles of NaOH added, along with the molarity of the NaOH solution, allows calculation of the volume of the NaOH solution: PS77

8 1L 1000mL molNaOH 1.5mL 0.00molNaOH L 3) No need to recalculate for HF. Answer will be the same because the molarity and volume of the HF solution is identical to that of the HCl. (b) Predict whether the ph at the uivalence point for each titration will be acidic, basic, or neutral. Answers: neutral for the HCl titration; basic for the HF titration Reasoning: At ep, HCl titration is a solution of Cl (negligible base) neutral. At ep, HF titration is a solution of F (weak base) basic. See #5 (prior problem) for details (c) Predict which titration curve will have the lower initial ph. Answer: HCl Reasoning: Initial ph refers to the solutions before any NaOH is added (i.e., the acid solutions). If initial concentrations are the same, the stronger acid will have a greater [H 3 O + and thus a lower ph Consider the curve for the titration of a weak monoprotic acid with a strong base and answer each question. (a) What is the ph and what is the volume of added base at the uivalence point? (b) At what volume of added base is the ph calculated by working an uilibrium problem based on the initial concentration and K a of the weak acid? (c) At what volume of added base does ph = pk a? (d) At what volume of added base is the ph calculated by working an uilibrium problem based on the concentration and K b of the conjugate base? (e) Beyond what volume of added base is the ph calculated by focusing on the amount of ecess strong base added? Answers: (a) ph ~ 8.7 (between 8.5 and 9.0); V ep ~ 30 ml (b) 0 ml of added base (i.e., the initial point) Because here moles HA moles (c) At ~15 ml (the half uivalence point [i.e., half of the V ep ) A ½ initial moles of HA (d) At ~30 ml (the uivalence point, because here the solution is a solution of A ) (e) Again, 30 ml. That is, after the uivalence point volume (because then there is ecess base, and that dominates the [OH and thus [H 3 O Consider the titration of a 35.0 ml sample of M HBr with 0.00 M KOH. Determine each quantity: (a) the initial ph (c) the ph at 10.0 ml of added base (b) the volume of added base ruired to reach the uivalence point (d) the ph at the uivalence point (e) the ph after adding 5.0 ml of base beyond the uivalence point Answers: (a) 0.76; (b) 30.6 ml; (c) 1.04; (d) 7.00 (assuming T = 5 C); (e) 1.15 Strategy: As noted on the PS7 sheet: In such a titration, one simply does an ICF type calculation, and then determines whether there is ecess HCl, NaOH, or neither (e.p.). There are no buffer calculations or weak acid or weak base calculations because there is never a buffer nor weak acid nor weak base! More specifically, the titration reaction is represented by (this problem involves HBr (not HCl) and KOH (not NaOH): HBr + KOH % ~100 KBr + H O (OR after dissociation : HBr + OH ~100 % H O + Br ) 1) Calculate initial moles from the volume and molarity. ph Volume of base added (ml) PS78

9 ) Calculate the moles of KOH added, when appropriate. Use stoichiometry (ICF, if you like) to determine who is limiting and who is left over (and how much, in moles). 3) Taking into account the total volume (i.e., initial volume + volume added), calculate the concentration of (left over) H 3 O + or OH. 4) If OH is left over, assume T = 5 C (because they didn t specify, and if you don t assume this, you can t find the ph), and determine [H 3 O + from [OH as usual (using K w = at 5 C). 5) For the part where you are asked to calculate V ep, do as was done in problem #6 on this set (basically a stoichiometry with molarity and volume problem). Eecution of Strategy: Initial moles of HBr: L mol/l = mol HBr (a) Initial [HBr M [H 3 O M ph log(0.175) (3 SF) (b) At ep, moles KOH added moles HBr initial mol (see above) [0.76 is OK 1L 1000mL molkoh 30.65mL 30.6mL (of 0.00 M KOH) 0.00molKOH L (c) After 10.0 ml ( L) of KOH added: mol KOH added L 0.00 mol/l mol Set up an ICF table (not necessary, but helpful to do / show stoichiometry): in moles HBr OH Br (not relevant) I (LR) C F mol Ecess HBr [HBr ecess MHBr 1L (35.0mL 10.0mL) 1000mL V HBr, initial V KOH ph log( ) (3 SF) [1.04 is OK (d) At the uivalence point, the solution must be neutral (solution of K + and Br ). Thus, assuming T = 5 C, ph = 7.00 (e) After 5.00 ml of KOH beyond the uivalence point (i.e., at 35.6 ml of KOH added), the amount of ecess KOH can be calculated directly (because they told you it was 5.00 ml (= L) past the uivalence point!): moles ecess OH = L 0.00 mol/l = mol OH [OH mol ecess MOH 1L (35.0mL 30.6mL 5.00mL) 1000mL added V HBr, initial V KOH added to reach ep V KOH added after ep [H 3 O M ph log( ) PS79

10 Consider the titration of a 0.0mL sample of M HC H 3 O with 0.15 M NaOH. Determine each quantity: (a) the initial ph (c) the ph at 5.0 ml of added base (e) the ph at the uivalence point (b) the volume of added base ruired to reach the uivalence point (d) the ph at onehalf the uivalence point (f) the ph after adding 5.0 ml of base beyond the uivalence point Answers: (a).86; (b) 16.8 ml; (c) 4.37; (d) 4.74; (e) 8.75; (f) 1.17 Strategy: a) The initial ph is determined by doing a standard weak acid problem. Find K a from Table 15.5 and use the given initial concentration to set up an ICE table and problem. Solve for, which will be ual to the [H 3 O + and then calculate ph. b) Same as in prior problem. Finding the uivalence point volume (of added base) is a stoichiometry problem with molarity and volumes. c) For points between the initial and the uivalence point, you have a buffer. Use an ICF table to find moles of HA remaining and moles of A generated. Then find the ph as discussed earlier: [H 3 O + K a [HA moles HA OR ph pk a + log [A moles A [A moles A or [HA moles HA (or you could do a longer ICE approach with small approimations) d) At half the uivalence point, ph pk a ([H 3 O + K a ) e) At the uivalence point, you must calculate the concentration of A by dividing the moles of A (= initial moles of HA) by the total volume (V initial acid solution + V base added ), in L. Then calculate K b from K a and do an ICE type of problem. Remember, however, that here is going to be ual to [OH, not [H 3 O +, so you ll need to calculate [H 3 O + before finding ph. f) After the uivalence point, the only thing that matters is how much ecess OH is left in the solution. The calculation is analogous to part (e) of the prior problem. Eecution of Strategy: (a) K a (HC H 3 O ) (at 5 C) HC H 3 O + H O C H 3 O K a [CH3O [H3O [HC H O 3 + H 3 O + ; K a 5 [CH3O [H3O [HC H O [HC H 3 O (M) [C H 3 O (M) [H 3 O + (M) I ~0 C + + E Try the small approimation (borderline, but let s hope!): Assume ( ) A negative here does not make sense so use the (+) one. PS710

11 Check assumption: %( 5%so OK! ) [H 3 O ph log( ) molHC H3O (b) At ep, moles NaOH added moles HC H 3 O initial 0.000L 1L 1L 1000mL molNaOH 16.8mL 0.15molKOH L (c) ph after 5.0 ml of 0.15 M KOH added: (initial) moles HC H 3 O (from (b)) (initial) moles OH L 0.15 mol/l mol Reaction that occurs: OH + HC H 3 O C H 3 O in moles OH HC H 3 O C H 3 O mol + H O (OH will react with the strongest acid around.) Decent amounts of HC H 3 O and C H 3 O are present after this reaction a buffer! [H 3 O + K a I (LR) C F moles HC H O 3 3 moles C H O ph log( ) (d) At the half uivalence point, ph pk a log( ) 4.74 (e) At the e.p., moles NaOH added moles HC H 3 O (part (b)) (You should not need an ICF table here because you know you re at the uivalence point, but I will include one so that you can see how this would work out if you had not calculated the e.p. first.) OH + HC H 3 O C H 3 O + H O in moles OH HC H 3 O C H 3 O I ~0 C F No HC H 3 O left not a buffer! Just a solution of C H 3 O, a weak base. ph of a weak base problem: Need K b and [C H 3 O 14 Kw K b K (HC H O ) a 3 And since the total volume is now 0.0 ml (original acid) ml (added KOH) = 36.8 ml = L, [C H 3 O mol = M L 5 Answer is reasonable. pk a here is log( ) This buffer has more HA than A ph should be a bit lower than 4.74 and it is. M PS711

12 C H 3 O + H O HC H 3 O + OH ; K b [HCH3O [OH [C H O K b [C H 3 O (M) [HC H 3 O (M) [OH (M) I ~0 C + + E ( )( ) ( ) ( ) ( ) (f) After 5.00 ml of NaOH beyond the uivalence point, the amount of ecess NaOH can be calculated directly (because they told you it was 5.00 ml ( L) past the uivalence point!): moles ecess OH = L 0.15 mol/l = mol OH [OH mol ecess M OH 1L (0.0mL 16.8mL 5.00mL) 1000mL (Assumption is clearly valid << 0.057) (Assume that ) [OH M; [H 3 O M ph log( ) V HBr, initial V KOH added to reach ep V KOH added after ep [H 3 O M ph log( ) Consider the titration curves for two weak acids, both titrated with M NaOH. Assume ual initial volumes! 14 1 (a) 14 1 (b) ph ph Volume of base added (ml) Volume of base added (ml) (i) Which acid solution is more concentrated? (ii) Which acid has the larger K a? Reasoning: Answers: (a) (b) (i) Assuming ual initial volumes, the one with greater concentration of HA has the greater number of moles of HA. Since solution (a) takes more moles of NaOH (more ml of solution~38 ml vs ~30 ml) to reach the uivalence point, it must contain more initial moles of HA. Don t be fooled by the initial ph! That value depends on both K a and [HA. Although the initial ph in (a) is higher (meaning PS71

13 the initial [H 3 O + is lower) that does not mean the initial [HA is lower it it not! The reason the [H 3 O + is lower in (a) is because the acid is considerably weaker (see part (ii)). (ii) To determine the K a of each acid, look at the half uivalence points ( s on the plots), since the p a ph there uals pk a and K a 10 K. The higher the pk a, the smaller the K a, so clearly (a) (with pk a 7.5) has the acid with lower K a and (b) (with pk a 5.5) has the acid with the greater K a. Acid (a) is weaker (which is why the ph is higher initially despite the 14 higher initial [HA (see part (i)) A 0.446g sample of an unknown monoprotic acid is titrated with M KOH. The resulting titration curve is shown here. Determine the molar mass and pk a of the acid. Answers: MM = 10 g/mol; pk a (at least) Reasoning: To determine molar mass, you need two things: grams of a sample, and moles of that sample. The grams of this sample is given. To find moles of HA here, find the moles of KOH needed to react with all of the HA (i.e., the moles of KOH needed to ph Volume of base added (ml) reach the uivalence point). This can be done using the uivalence point volume and the molarity of the KOH solution: V ep (from curve) is about 36 ml (see vertical line on plot) L L mol/l moles KOH moles HA (initial) because it s a 1:1 stoichiometry g HA Molar Mass g/mol molha The pk a is the ph at the half uivalence point ( on plot; ~18 ml). I estimate a value of about 4.3 from my plot, although it looks a bit lower than that (maybe 4.1 or 4.ish?) on the plot in the tet itself. Clearly there is uncertainly to at least 0.1 and probably a bit more Methyl red has a pk a of 5.0 and is red in its acid form and yellow in its basic form. (a) If several drops of this indicator are placed in a 5.0mL sample of M HCl, what color with the solution appear? (b) If M NaOH is slowly added (i.e., added in small increments) to the HCl sample, in what ph range will the indicator change color? Answers: (a) red; (b) from about Reasoning: The reading of the plot (V ep) limits precision here to about SF in my estimation. Also, my plot (in the key) is not eactly the same as the one in the book (it is not a copy), so there will be some variability in answers to greater than SF, to be sure. Let s call the acid form of methyl red HIn and the base form In. The ph of a M HCl solution is 1.0 ( log(0.100) = 1.00). This is 4 ph units below methyl red s pk a. Thus, the ratio of HIn / In will be : 1 (see mathematical proof of this at the end of this solution*), meaning nearly all of the indicator units are in the HIn (acid) form, which is red (see problem info). Thus the solution will appear red. Now as the ph increases, the ratio of HIn / In will get smaller and smaller, meaning that there will be fewer of the In species in the HIn (red) form and more in the In (yellow) form. When it is one unit below pk a (here, that would be 4), the ratio will be 10 : 1, so 10 out of every 11 In species will be in the HIn form. That s over 90%, so the color will still appear mostly red. But nearly 10% of the In species will now be yellow and the color will thus start to change. By the time the ph reaches pk a (here, 5), there will be ual amounts of the red form and the yellow form, so the color will be orange. And by the time the ph reaches one unit above pk a (here, 6), the ratio of HIn / In will be 1 : 10, meaning over 90% of the In species will be in the yellow form, and the solution will appear primarily yellow. In general, then, for any indicator, the ph range in which its color changes is its own buffer range : i.e., pk a 1. In this problem, that is or PS713

14 * Proof of this: From a calculational point of view, think of a solution of HIn and In as a buffer (even though it won t act like one in this solution because there is such a small amount in the system relative to the amount of HCl and/or NaOH (later)). If pk a 5.0, then K a At ph 1.0, [H 3O M and we have: 1 [H O [HIn [HIn [H3O Ka [In [In K 10 Obviously, you could see this with the HH uation as well (Tro does this), but as you already know, I prefer the modified K a epression approach to buffers (it s easier to rederive if you forget it [just one step from the K a epression uation). a Determine whether or not each compound will be more soluble in acidic solution than in pure water. Eplain. (a) Hg Br (b) Mg(OH) (c) CaCO 3 (d) AgI Answers: (b) and (c) will be more soluble in acidic solution; (a) and (d) will not be. Reasoning: The issue here is whether or not the anion of the ionic compound will react with added acid (i.e., if it is a nonnegligible base). If it does, then adding acid will have the effect of decreasing the concentration of the base anion, which will shift the solubility uilibrium to the right (i.e., toward dissolution of the solid). If, however, the anion is one of the negligible bases (conjugate to a strong acid), then there will be no such effect, and the solubility will not increase when acid is added. In this case, (a) and (d) have Br and I, both negligible bases, as the anion. Thus solubility will be unaffected. However, (b) has OH and (c) has CO 3, both nonnegligible bases. Thus these compounds solubility will increase in acidic solution. 14. NT1. You have two salts, AgX and AgY, with very similar K sp values. You know that the K a value for HX is much greater than the K a value for HY. Which salt is more soluble at low ph? Eplain. Answer: AgY is more soluble in acidic solution. Reasoning: As noted in the prior problem, solubility will increase as ph decreases (i.e, as acid is added or solution is acidified) if the anion of the salt is a nonnegligible base. This is because the added acid will react with the anion (A ) to form its conjugate acid HA thereby lowering the concentration of A (a product) and coaing more forward reaction (dissolution) to occur. This can be thought of in Le Châtelier s terms as a stress in which a product species concentration is decreased forward reaction (in this case, dissolution) will occur to help alleviate the stress. The stronger of a base the anion is, the greater will be the etent to which it reacts with a given concentration of H 3 O +, and so the greater will be the increase in solubility. In this case, since K a for HX is much greater, K b for X must be much smaller (the stronger the acid, the weaker its conjugate base; or K ak b = K w), and so AgX should be less soluble at low ph than AgY (which will be affected more by a lowering in ph). One could look at this mathematically as follows: K b [HA [A [OH [HA Kb K [H b 3O [A [OH Kw/[H3O Kw This uation shows that for a given [H 3 O + (and thus a given ph), the greater the K b, the greater is the [HA ratio, which means a greater fraction of A s are protonated and thus taken out of solution. [A K b PS714

15 Calculate the concentration of all species in a M solution of H SO 3. (By all species, let s assume they mean H SO 3, HSO 3, H 3O +, OH, and SO 3 ). Assume T = 5 C. Also calculate the ph. Answers: [H SO M; [HSO 3 [H 3 O M; ph = 1.10; Strategy: [OH M; [SO M 1) Recognize that H SO 3 is a diprotic acid, and thus there will be two acid ionization uilibria involved. Look up K a1 and K a for H SO 3 in Table ( and respectively). ) Although not necessary at this moment, it is helpful to write out the two acid ionization uations to help you see what is going on : H SO 3 (aq) + H O(l) HSO 3 (aq) + H3 O + (aq) K a1 = HSO 3 (aq) + H O(l) SO 3 (aq) + H3 O + (aq) K a = ) Now recognize that since the nd ionization is much less product favored than the first, we can consider the first uilibrium (alone) to determine [H SO 3, [HSO 3, [H 3 O + (and thus ph and [OH ). That is, the only purpose for doing any calculation involving the second ionization will be to determine [SO 3! 4) Do an weak acid ionization problem (ICE table setup) considering the first ionization only. This will result in determining [H SO 3, [HSO 3, and [H 3 O +. 5) Calculate ph and [OH from [H 3 O + using [H 3 O + [OH K w ( at 5 C) 6) Use the uilibrium values for [HSO 3 and [H 3 O + already determined as the initial concentrations in the second ionization / ICE type problem. Apply the small approimation and solve for, which here will ual [SO 3. Eecution of Strategy: H SO 3 (aq) + H O(l) HSO 3 (aq) + H3 O + (aq) K a1 3 [HSO [H3O [H SO 3 [H SO 3 (M) [HSO 3 (M) [H 3 O + (M) I ~0 C + + E K a [HSO3 [H3O ( )( ) [H SO (0.500 ) (1) ( )root (1) 3 0 (Negative root is not meaningful for this physical system.) Thus: Check (+) root : ( )( ( ) ) (matchesk [H SO M [HSO 3 [H 3 O M ph log( ) a ) PS715

16 [OH / M HSO 3 (aq) + H O(l) SO 3 (aq) + H3 O + (aq) K a [SO 3 [H3O [HSO K a [HSO 3 (M) [SO 3 (M) [H 3 O + (M) I C + + E [SO3 [H3O 8 ( )( ) ( )(0.0818) (Assuming << ) [HSO ( ) (0.0818) 3 [SO M (Clearly the assumption was good: << 0.08) & (ecept omit H S from the problem) For a binary acid, H Y, what factors affect the relative ease with which the acid ionizes? Answer: The polarity and bond strength of the HY bond both play a role. All other things being ual: The greater the polarity of the bond, with the H being more positive, the stronger the acid will be. The greater the bond strength of the HY bond, the weaker the acid will be Based on their molecular structure, arrange the binary compounds in order of increasing acid strength. Give your reasoning. H Te, HI, NaH Answer: NaH < H Te < HI Reasoning: These are binary acids. As such, the bond polarity and bond strength play a role (see above). The electronegativities of Na, Te, and I are in the order Na < Te < I. You can rationalize this as follows (i.e., you need not have the actual values of electronegativity available). Na is a metal on the far left of the periodic table. Its electronegativity will be smaller than any nonmetal. Te and I are nonmetals on the far right, but I is to the right of Te. Since EN tends to increase as you move right on the periodic table, I s EN should be greater than Te s. The greater the EN, the more an atom will pull bonding electrons to itself. Thus, the HI bond should be most polar (with H being partially positive), making HI the best acid. The HNa bond should be the most polar (with H being partially negative), making NaI the worst acid. and H Te is in the middle. NOTE: This argument ignores bond strength Based on their molecular structure, pick the stronger acid from each pair of oyacids. Give reasoning. Answers (a) H SO 4 or H SO 3 H SO 4 (b) HClO or HClO HClO (c) HClO or HBrO HClO Reasoning In both of these cases, the chosen acid has one more O bonded to the X atom (S or Cl, here). That pulls more electron density from X, which, in turn will pull more electron density from the O bonded to an acidic H, making the HO bond more polar, the H more positive, and thus the acid stronger. See picture below. The number of O s is the same, but in one case X is Cl and in the other, it is Br. Since Cl has a greater EN, it will pull more electron density from O bonded to the H, making the HO bond more polar and the acid stronger. See picture below. PS716

17 (d) CCl 3COOH or CH 3COOH CCl 3 COOH Cl has a greater EN than H. So the three Cl s will pull electron density from the C (bound to the C bound to the O) to a greater etent than the 3 H s on the other structure. As noted above, such an inductive effect will make the HO bond more poloar, the H more positive, and the acid stronger. See picture below (although imagine the picture flipped (since the H is on the right in these formulas rather than on the left). Acidic hydrogen (i.e., a H that will be donated as H + when the acid acts like a BL acid) H O X O Anything that causes electrons to move in the direction of the arrows will make the H more partially positive and thus easier to be given away as H + (i.e., makes the acid stronger). PS717