Answer Key, Problem Set 6 (With explanations)

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1 Chemistry 1 Mines, Spring 18 Answer ey, Problem Set 6 (With explanations) (a,d)*;. 16.6(a); (a,c) Assume 5 C. Also find % ionization; ; (a,b) For (a), assume 5 C, for (b), assume 4 C (at which w.9 x 1 14 ); ; Assume 5 C. Also calculate [OH ; & 16.1; *; * (Assume uimolar aqueous solutions here. Takes some thought and work; not all salts are ual!! Give detailed reasoning; if you write a short answer here, you will surely lose points); * Assume 5 C.; *; (a)*; 14. Conceptual Connection 17. (p. 786), except using the following numbers in place of 4.8, 4.5, and 4.7: pa = 5.58, ph = 6.13, and ph = 6.93; , except switch the word masses to moles and Add Part (b): Why can t an H3PO4/ NaHPO4 buffer system be used to make the buffer in this problem? Give specific reasoning. Solutions of (only) Acids or (only) Bases (strong or weak) ph of, % ionization in, [OH in, etc (a,d). Determine the ph of each solution. (a).48 M HI Answer: 1.3 Reasoning: HI is a strong acid, thus % ionization is 1%, and [I [H3O +.48 M ph log(.48) (d) A solution that is 1.9% HCl by mass (Assume a density of 1.1 g/ml for the solution.) Answer:.5 Reasoning: Assume 1 ml (.1 L) of solution (exactly). Then: mass of solution 1.1 g/ml x 1 ml 11 g mass of HCl 1.9 g HCl 11g solution x 1.11g HCl 1 g solution MM (HCl) = = g/mol moles HCl [HCl =.3 mol/.1 L =.3 M 1mol HCl 1.11g HCl x g.3 mol HCl is a strong acid, thus % ionization is 1%, and [Cl [H3O +.3 M ph log(.3) (a). What mass of HClO 4 should be present in.5 L of solution to obtain a solution with each ph value? (a) ph =.5 Answer:.159 g HClO4 Reasoning: [H3O + 1 ph 1.5 =.316 M. Since HClO4 is a strong acid (1% ionized), the original concentration of HClO4 must have been.316 M..316 M x.5 L =.158 mol HClO g MM(HClO4) = (16.) = 1.46 g/mol.158 mol x.1587 g HClO 4 mol PS61

2 (a,c) Determine the ph (and % ionization) of an HF solution of each concentration. In which cases can you not make the simplifying assumption that x is small? [Assume 5 C. (a).5 M; (c).5 M Answers: (a) ph =.3; % ionization = 3.7%; small x approximation is valid (c) ph =.55; % ionization = 11%; small x approximation is not valid 1) Write out the acid ionization uation and a expression. Find a (5 C) from Table ) Set up an ICE table, letting x = [HF that ionizes. Substitute into the a expression and solve for x. Use the small x approximation if it looks reasonable. If not, use the quadratic formula. 3) Check the small x approximation, if applicable. If good (or if you used the quadratic formula), recognize that x = [H3O +. (If the small x approximation is not good, use the quadratic formula to get x.) [H 4) ph = log[h3o + 3O ; % ionization = x1 [HA Execution of Strategy for Part (a): HF + HO F + H3O + ; a [F [H O 3 [HF 3.5 x 1 [HF (M) [F (M) [H3O + (M) I.5 ~ C x + x + x E.5 x x x a [F [H O 3 [HF 3.5 x 1 xx.5 x 3.5 x 1 Try the small x approximation: xx.5 x 3.5 x 1 Assume.5 x.5 xx x 1 x.5(3.5 x 1 ) 8.75 x 1 5 x x x 1 3 A negative x here does not make sense (can t have a negative [F or [H3O + ) so use the (+) one. Check assumption: x x %( 5% so O) [H3O + x 9.34 x 1 3 ph log(9.34 x 1 3 ) ; % ionization = 3.7% (see ) Execution of Strategy for Part (c): The setup is the same as in (a), except for using.5 M (new [HF) for.5 M: PS6

3 a [F [H O 3 [HF 3.5 x 1 xx.5 x 3.5 x 1 Try the small x approximation: xx.5 x 3.5 x 1 Assume.5 x.5 xx x 1 x.5(3.5 x 1 ) 8.75 x 1 6 x x x 1 3 A negative x here does not make sense so use the positive one. Check assumption:.958 x x %( 5% so NOT okay! Put in x.96 x 1 3 in (.5 x) (successive approximations approach): xx.5 x 3.5 x 1 Assume.5 x x.(3.5 x 1 ) xx. 7.7 x x 1 x x x 1 3 (only (+) root meaningful here) Check assumption:. (.5.77 x 1. OR, if you do the quadratic uation approach to solve for x: x)( x) 3.5 x 1 (.5 x) ( x 3.5 x 1 3 ) x 1 1.5%( 5% so O.5 x Same x x x 1 x x 3.5 x 1 x 8.75 x x 1 4(1) 8.75 x x x x 1 3 x.78..x 1 ( )root (1) 6 (Negative root is not meaningful for this physical system.) 3 3 (.8 x 1 )(.8 x 1 ) Check (+) root : 3.53 x 1 (matches ) 3 a (.5.8 x 1 ) So, [H3O x 1 3 M ph log(.78.. x 1 3 ) Makes sense. More dilute HF should be less acidic, and it is. (.56 >.3) 3.78 x 1 % ionization = x1 11.1% A.115 M solution of a weak acid (HA) has a ph of 3.9 [at some temperature. Calculate the acid ionization constant ( a ) for the acid [at this temperature. Answer: a =.3 x 1 6 Makes sense. More dilute solution should have greater % ionization (Le Châtelier) 1) Analysis: It may not look like it at first, but this is fundamentally a given the initial concentrations and one uilibrium concentration, find kind of uilibrium problem. Although [H3O + is not technically given, you are just one step away since the ph is given! Thus: PS63

4 ) Write out the acid ionization uation (the primary reaction that occurs here is acid ionization) and its expression. 3) Calculate [H3O + from ph ([H3O + = 1 ph ) 4) Use stoichoimetry (ICE table is not necessary, but helpful here) to determine the other uilibrium concentrations from [HA (given) and [H3O + (and recognizing that [A and [H3O + are both (effectively) zero, as assumed in all normal acid ionization problems). 5) Plug into a expression! Execution of [A HA + HO A + H3O + [H3O ; a [HA Since ph = 3.9, [H3O = x 1 M.518 M Since the stoichiometry is 1 : 1 : 1 and we assume there was essentially no [H3O + to begin with, this means that.518 M of H3O + formed. Thus, the [A that formed and the [HA that reacted were also.518 M. If.518 M of the initial.115 M reacted, the concentration remaining at uilibrium is just M = M. If you wanted to use an ICE table to help see these relationships, it would look like this: [HA (M) [A (M) [H3O + (M) I (initial).115 ~ C (change in) E (at uilibrium).518 [HA (M) [A (M) [H3O + (M) I (initial).115 ~ C (change in) E (at uilibrium).518 [HA (M) [A (M) [H3O + (M) I (initial).115 ~ C (change in) E (at uilibrium) x 1 M NOTE: I suppose another way to use an ICE table for this is to use an x, but then recognize that you know x here (because you know [H 3 O + from the ph): x =.518 [HA (M) [A (M) [H 3 O + (M) I (initial).115 ~ C (change in) x + x + x E (at uilibrium).115 x x x Substitute back into the uilibrium constant expression uation (Law of Mass Action): a [A [H O 3 [HA (.518)(.518).97 x 1 (.11448) x 1 PS6

5 (a,b). For each strong base solution, determine [OH, [H 3 O +, ph, and poh. For (a), assume 5 C, for (b), assume 4 C (at which w.9 x 1 14 ) (a) 8.77 x 1 3 M LiOH (at 5 C) (b).11 M Ba(OH) (at 4 C; w.9 x 1 14 ) Answers: (a) [OH = 8.77 x 1 3 M; [H3O + = 1.1 x 1 1 M; ph = 11.94; poh =.57 (.6 O) (b) [OH =.4 M; [H3O + = 1.3 x 1 1 M; ph = 11.88; poh = 1.65 (1.65 O) 1) Since these are identified as strong base solutions, you must assume that these (ionic) compounds are strong electrolytes let the ionic compounds dissociate first! ) Being careful to note stoichiometry (1 OH per formula unit in LiOH; OH per formula unit in Ba(OH)), calculate the [OH in each solution. 3) Use [H3O + [OH w to calculate [H3O +. (remembering to switch to w.9 x 1 14 in (b)!) NOTE: If you calculate poh first and then use ph + poh 14., you will get (a) correct but (b) incorrect since that uation only applies to a solution at 5 C!!). If you really want to go this route, you must use (the general) relationship that ph + poh p w 4) Calculate ph using ph = log[h3o +. Calculate poh using poh = log[oh. Execution of Strategy, Part (a): LiOH(aq) Li + (aq) + OH (aq) [OH in solution [LiOHinitial 8.77 x 1 3 M [H3O + 1. x x x 1 M ph log(1.14 x 1 1 ) poh log(8.77 x 1 3 ).57 (but.6 is fine also, since usually ph s and poh s are limited to the hundredths place) Execution of Strategy, Part (b): Ba(OH)(aq) Ba + (aq) + OH (aq) [OH in solution x[ba(oh)initial (.11).4 M [H3O x x 1 poh log(.4) (or 1.65) M ph log(1.33 x 1 1 ) NOTE: Even though this solution has a lower ph than the one in (a), it is clearly not more acidic! Remember, acidic is not defined in terms of ph Write uations showing how each weak base ionizes [in? water to form OH. Also write the corresponding expressions for b. (a) CO 3 CO3 (aq) + HO(l) HCO3 (aq) + OH (aq) ; b [HCO3 [OH [CO3 (b) C 6 H 5 NH C6H5NH(aq) + HO(l) C6H5NH3 + (aq) + OH (aq) ; (c) C H 5 NH CH5NH(aq) + HO(l) CH5NH3 + (aq) + OH (aq) ; b b 6 3 [C6H5NH [OH [C H NH [C H5NH [OH [C H NH PS65

6 Amphetamine (C 9 H 13 N) is a weak base with a p b of 4. [at 5 C. Calculate the ph of a solution containing an amphetamine concentration of 5 mg/l. Also calculate [OH. [Assume 5 C. Answer: 1.5 1) Write out the base ionization uation and b expression. ) Calculate b from pb (the same way you calculate [H3O + pb from ph: b 1 ; [because pb = log b) 3) Calculate the initial [C9H13N (in moles/l) using the mg converted into grams, and then using the (calculated) molar mass (9(1.1) + 13(1.1) = 135. g/mol) to get moles. 4) Set up an ICE table, letting x = [B that ionizes. Substitute into the b expression and solve for x. Use the small x approximation if it looks reasonable. If not, use the quadratic formula. 5) Check the small x approximation, if applicable. If good (or if you used the quadratic formula), recognize that x = [OH. To get ph, calculate [H3O + from [OH using [H3O + [OH = w = 1. x 1 14 (because we re assuming T = 5 C). 6) ph = log[h3o + Execution of C9H13N + HO C9H13NH + + OH ; b [C9H13NH [OH [C H N x 1 5 [C9H13N 5 mg 1 g 1 mol L 1 mg 135. g 3 x x x 1 M [C9H13N(M) [ C9H13NH + (M) [OH (M) I.1664 ~ C x + x + x E.1664 x x x It does not look to me like the small x approximation is even worth trying here. is not all that small, and [B is quite small. I will just go straight to the quadratic formula approach: [C9H13NH 5 [OH ( x)( x) 5 b 6.3 x 1 x 6.3 x x [C H N (.1664 x) x x x x x x x x x x 1 4(1) x x x 1 x.93..x 1 (1) Negative root is not meaningful for this physical system. [C9H13NH [OH (.93..x 1 )(.93..x 1 ) 5 Check (+) root: x 1 (close to b ) [C9H13N ( x 1 ) So, [OH.93 x 1 M and thus [H3O + 1. x 1 14 /.93 x x 1 11 M ph log( x 1 11 ) ( ) root PS66

7 Solutions of Salts (Cations & [nonoh Anions) Acidic, Basic, or Neutral? (and ph of [if only one nonnegligible) & Determine whether each anion is [basic or neutral a (nonnegligible) base or a negligible base ( ph neutral ). For those anions that are bases, write an uation that shows how the anion acts as a base. (a) C 7 H 5 O is a weak base because its CA is HC7H5O, a weak acid [not one of the six SA s (b) I (c) NO 3 (d) F C7H5O (aq) + HO(l) HC7H5O(aq) + OH (aq) is a negligible base (i.e., ph neutral) because its CA, HI, is a strong acid is a negligible base (i.e., ph neutral) because its CA, HNO3, is a strong acid is a weak base because its CA is HF, a weak acid [not one of the six SA s F (aq) + HO(l) HF (aq) + OH (aq) Determine whether each cation is [acidic or ph neutral a (nonnegligible) acid or a negligible acid ( ph neutral ). For those cations that are acids, write an uation that shows how the cation acts as an acid. (a) Sr + (b) Mn 3+ is a negligible acid Group II ions are negligible acids (don t bind waters) is an acid +3 cations bind waters (c) C 5 H 5 NH + (d) Li + Mn(HO)6 3+ (aq) + HO(l) Mn(HO)5(OH) + (aq) + H3O + (aq) is an acid because it is the CA of the weak (amine) base C5H5N C5H5NH + (aq) + HO(l) C5H5N(aq) + H3O + (aq) is a negligible Group I ions are negligible acids (don t bind waters) *. Determine whether each salt will form a solution that is acidic, basic, or phneutral. (a) Al(NO 3 ) 3 (b) C H 5 HN 3 NO 3 (c) CO 3 (d) RbI (e) NH 4 ClO Answers: (a) acidic (b) acidic (c) basic (d) neutral (e) basic 1) Let each salt ( soluble ionic compound ) dissociate first! ) Look at each ion (see outline at the bottom of the PS6 sheet) to assess whether it is an acid, base, or a negligible 3) If: a) one of the two ions is an acid and the other is a negligible, the solution will be acidic b) one of the two ions is a base and the other is a negligible, the solution will be basic c) if both of the ions are negligibles, the solution is neutral d) if the cation is an acid and the anion is a base, then the solution will be: i) basic, if b(anion) > a(cation) NOTE: You will need values in this case (not in the others). Note: ab(for conjugates) w, and you will need to use ii) acidic, if a(cation) > b(anion) this fruently to determine the a or b of ions from their respective (molecular) conjugates. Execution of (a) Al(NO 3 ) 3 Al 3+ + NO3 acid (+3 ion) Solution is acidic negligible base (HNO 3 is a strong acid) PS67

8 (b) C H 5 NH 3 NO 3 CH5NH3 + + NO3 acid (CA to C H 5 NH ) negligible base (HNO 3 is a strong acid) Solution is acidic (c) CO CO3 negligible acid (Gp I ion) nonnegligible base (HCO 3 is not a strong acid) Solution is basic (d) RbI Rb + + I negligible acid (Gp I ion) negligible base (HI is a strong acid) Solution is neutral (e) NH 4 ClO NH4 + + ClO acid (CA to NH 3 ) nonnegligible base (HClO is not a strong acid) From Table 16.8: b (NH3) =.9 x 1 8 a(nh 4+ ) 1. x x x 1 < b(clo ) 5 1. x 1.9 x x 1 7 From Table 16.5: a (HClO) =.9 x 1 8 Solution is basic (ClO ionizes more than NH 4+ ) *. Arrange the solutions in order of increasing basicity: Answer (at the level of this class): CH 3 NH 3 Br, OH, Br, CN, C 5 H 5 NHNO C5H5NHNO or CH3NH3Br < Br < CN < OH NOTE: The precise determination of the relative placement of the two acidic solutions in this problem is really beyond the scope of this course. I did not recognize this until I was writing up the key for the problem. I have sent out an regarding this and/or mentioned it in class. I apologize for this fundamental flaw in the problem (I m curious what the solution manual says about this problem. ). If you d like to know how to determine the ph of the C5H5NHNO solution (.1 M), see me in an office hour. It is completely accessible to you given what we ve done, but it is not something I would expect anyone to figure out on his / her own. Analyze each salt separately as in the prior problem. The first sorting would be by type acid, base, or (ph) neutral salt. If two salts end up in the acid or base category, then determining relative strength would be the secondary sorting. Clearly, any acid salt will make a solution that is less basic (more acidic) than a solution of any neutral salt, and any neutral salt will make a solution less basic than a solution of any base salt. If there are two salts with only a base ion, compare the b s to determine which solution will be more basic (bigger b), and if there are two salts with only an acid ion, compare the a s. At the level of this course, you are not responsible for determining the ph of a mixture containing a weak acid and a weak base (e.g., a salt with an acid cation and a base anion) that is beyond the scope of this course (although it is a direct, reasonable, and accessible extension of principles already covered). You are only responsible for determining if such a salt solution would be acidic or basic (by comparing a(cation) vs. b(anion). So, as noted above, you will not be able to completely answer this question as written because one of the two acidic solutions is from a salt with an acid cation and base anion. If I were to put a similar problem on an exam, there would be no such problematic pair of salts (the authors of this problem were just not careful enough in its construction). PS68

9 Execution of CH 3 NH 3 Br CH3NH3 + (a weak acid; CA to CH3NH) + Br (a negligible base; CB to SA HBr) (weakly) acidic OH + (a negligible acid; Gp I) + OH (considered a strong base, although technically, it is OH that is the strong base [it completely ionizes to produce OH ) (maximally) basic Br + (a negligible acid; Gp I) + Br (a negligible base; CB to SA HBr) neutral CN + (a negligible acid) + CN (a weak base; CB to WA HCN) (weakly) basic C 5 H 5 NHNO C5H5NH + (a weak acid; CA to C5H5N) + NO (a weak base; CB to WA HNO) From Table 16.8: b (C5H5N) = 1.7 x 1 9 a(c 5 H 5 NH + ) 1. x x x 1 > 9 b(no ) (weakly) acidic (C 5 H 5 NH + ionizes more than NO ) x 1.17 x x 1 11 From Table 16.5: a (HNO) = 4.6 x 1 Thus, at this point, you could certainly say: Least basic: C5H5NHNO or CH3NH3Br < Br < CN < OH Most basic Regarding the two acidic solutions, you need to look at more data: Namely, the a of CH3NH3 +. If the a of this acid were greater than the a for C5H5NH +, then you could clearly state that the x 1 11 solution of CH3NH3Br would be more acidic. However, its a.7 x 1, which is 4.4 x 1 considerably smaller. However, you cannot state based on this information alone that the solution of CH3NH3Br must be less acidic (more basic). Although it is a poorer acid than C5H5NH +, the solution of C5H5NH + also has an ual concentration of a weak base present in the solution (NO ) which will make the solution more basic than if it were only C5H5NH +! As such, it is not technically possible to (correctly) conclude which solution is more basic (i.e., further calculation would be needed). This is why the problem, as written, is not solvable at the level of this course. As it turns out, the C5H5NHNO does end up more acidic, but not by all that much (less than ph units)! Again, come by my office if you want to see the proof of this *. Determine the ph of each solution. [Assume 5 C. (a). M CHO (b). M CH 3 NH 3 I (c). M I Answers: (a) 8.5 (b) 5.67 (c) 7. 1) First let each salt dissociate to see what ions are present. ) Analyze each ion as in the previous problems. If you are asked to calculate the ph, then at most, one of the ions will be an acid or a base (not both). 3) If the cation and anion are both negligibles, the solution will be neutral (ph = 7. if you assume 5 C). PS69

10 4) If the cation is an acid (with the base negligible), determine the a of the acid* and treat the problem as an acid ionization problem (because it is exactly that!). I.e., use the strategy outlined in Problem 3 (16.66) of this problem set. 5) If the anion is a base (with the cation negligible), determine the b of the base and treat the problem as a base ionization problem (because it is exactly that!). I.e., use the strategy outlined in Problem 7 (16.9) [except that you d get b likely by other means*. 6) *Note that you will likely get a or b by using ab = w as in the prior couple of problems. Execution of Strategy, Part (a) CHO + (a negligible acid) + CHO (a weak base; CB to WA HCHO) Thus, the solution is uivalent to being a solution of. M CHO (weakly basic) a(hcho) = 1.8 x 1 (Table 16.5) b(cho ) x x x 1 11 CHO + HO HCHO + OH ; b [HCHO [CHO [OH x 1 11 [CHO (M) [HCHO (M) [OH (M) I. ~ C x + x + x E. x x x Try the small x approximation: x x x x x 5.56 x x Assume. x. 11 x.(5.56 x 1 ) 1.11 x x 1.11 x x A negative x here does not make sense so use the positive one x 1 Check assumption: x 1.16% ( << 5% so O). So, [OH 3.33 x 1 6 M and thus [H3O + 1. x 1 14 /3.33 x x 1 9 M ph log(3... x 1 9 ) (weakly basic, as expected) Execution of Strategy, Part (b) CH3NH3I CH3NH3 + (a weak acid; CA to CH3NH) + I (a negligible base; CB to SA HI) Thus, the solution is uivalent to being a solution of. M CH3NH3 + (weakly acidic) b(ch3nh) = 4.4 x 1 (Table 16.8) a(ch3nh3 + ) x 1.7 x x 1 11 CH3NH3 + + HO CH3NH + H3O + ; a [CH NH [H O x 1 + [CH3NH 3 PS61

11 [CH3NH3 + (M) [ CH3NH (M) [H3O + (M) I. ~ C x + x + x E. x x x Try the small x approximation: a xx. x [CH NH [H O x 1.7 x 1 + [CH3NH 3 x x x x x.7 x 1.7 x Assume. x. 11 x.(.7 x 1 ) 4.54 x x 4.54 x 1.13 x A negative x here does not make sense (can t have a negative [F or [H3O + ) so use the (+) one x 1 Check assumption: x 1.11% ( 5% so O). [H3O + x.13 x 1 6 ph log(.13 x 1 6 ) (weakly acidic, as expected) Execution of Strategy, Part (c) I + (a negligible acid; Gp I) + I (a negligible base; CB to SA HI) neutral at 5C, ph = 7. Solutions Containing Both a Weak Acid and its (Weak) Conjugate Base (i.e., Buffer Solutions) *. Solve an uilibrium problem (using an ICE table) to calculate the ph of each solution [at 5 C: (a) a solution that is.195 M in HC H 3 O and.15 M in C H 3 O Answers: (a) 4.55 (b) a solution that is.55 M in CH 3 NH and.135 M in CH 3 NH 3 Br (b) 1.9 Strategy and Execution: Set up the problem as an acid ionization problem (regardless of whether or not the acid is of the form HA or of the form BH +!) with some of the product ion present initially ( common ion ) (a) HCH3O + HO CH3O + H3O + ; a [CH3O [H3O x 1 (Table 16.5) [HC H O 3 [HCH3O (M) [CH3O (M) [H3O + (M) I ~ C x + x + x E.195 x.15 + x x PS611

12 a x x [CHO [HO.15 + x 1.8 x x 1 [HC H O Use the small x approximation: x.15 + x x.15 x 1.8 x x Assume.197 x Assume.15 x x [H3O 1.8 x 1 x.8..x x 1 Check assumption: x 1...% ( 5% so O).15 ph log(.8..x 1 5 ) NOTE: Once you get this, you can go straight to this uation for buffer in the future if you wish: [HA Note how this uation matches this one [H3O a x buffer [A (b) The acid component of this buffer is CH3NH3 +. Look up b for CH3NH in Table 16.8 and use that to calculate a [.7 x 1 11 (see prior problem, (b)). Answer is reasonable. p a = 4.74 here, and there is more HA than A. Thus ph should be a bit more acidic (lower) than 4.74 & it is! CH3NH3 + + HO CH3NH + H3O + ; a [CH NH [H O x 1 + [CH3NH 3 Shortcut once you get familiar with buffer situations. Buffers are WYSIWYG! a x x [CH NH [H O.55 + x x 1.7 x 1 + [CH3NH With a a of ~1 11 the small x approximation is a slam dunk : [CH3NH3 + (M) [CH3NH (M) [H3O + (M) I ~ C x + x + x E.135 x.55 + x x x.55 + x x.55 x.7 x 1.7 x Assume.55 x Assume.135 x x [H3O.7 x 1 x 1...x x [HA [H3O a x buffer [A Check assumption (for kicks): x 1 ~ 9 x 1 % (!)( 5%!) ph log(1...x 1 11 ) Answer is reasonable. p a = 1.64 here, and there is more B than BH +. Thus ph should be a bit more basic (higher) than 1.64 & it is! PS61

13 (a)*. Calculate the ph of the solution that results from each mixture: (a) 15. ml of.5 M HF with 5. ml of.3 M NaF Answer: ) BE CAREFUL! Although this is a buffer problem, it involves the mixing of two solutions. Thus, you cannot use the reagent bottle concentrations (given in the problem). Recognize that each reagent s concentration will change upon mixing (because of dilution). **Also remember the earlier guideline to Let any soluble ionic compounds dissociate first! That is how you can see that NaF is effectively a solution of F in this problem. ) You can either: a) Calculate the initial concentrations after mixing (i.e., do a dilution calculation) OR b) Calculate initial moles and use those in place of initial concentrations. NOTE: Why is it okay to use moles here (when moles is obviously not the same as concentration)? Because in this specific instance, it is the ratio of concentrations that matters, not each individual value, and since the volume is the same for both species in the buffer solution, the ratio of the moles uals the ratio of the concentrations (see p. 71 in Tro). 3) Find the a for HF from Table 16.5 (a 3.5 x 1 ) (Recognize that you need the a for the acid component of a buffer in order to determine buffer ph.) 4) Once () is completed, you can (if you are comfortable with the prior problem!!) substitute directly into: [HA [H O x buffer OR [H O x buffer 3 a [A moles HA 3 a moles A 5) Calculate ph from [H3O + NOTE: You may use the HendersonHasselbalch in place of (4) and (5) if you like, but I have chosen not to do that in my key since I have found that students tend to lose touch [and make more sign errors when trying to use the HH uation. Execution of 15. ml [HF.5 M x.1 M (15. ml + 5. ml) dilution calculation: x V Mafter Mbefore V 5. ml [F.3 M x.18 M (15. ml + 5. ml) before after.1 [H3O 3.5 x 1 x 1.94 x 1.18 ph log(1.94 x 1 ) = 3.71 Check answer. It makes sense! p a = log(3.5 x 1 ) = [F > [HF ph should be a bit higher (more basic). 14. Conceptual Connection 17. (p. 786), except using the following numbers in place of 4.8, 4.5, and 4.7: p a = 5.58, ph = 6.13, and ph = A buffer contains the weak acid HA and its conjugate base A. The weak acid has a p a of 5.58 and the buffer has a ph of (i) Which statement is true of the relative concentrations of the weak acid and conjugate base in the buffer? (a) [HA > [A (b) [HA < [A (c) [HA = [A (ii) Which buffer component would you add to change the ph of the buffer to 6.93? PS613

14 1Answer ey, Problem Set 6 Answer: (i) (b) is correct; (ii) Add more A Reasoning: The ph of the buffer is greater (more basic) than the pa. Thus, the buffer must contain more of the base component than the acid component. (This is because the ph = pa when the HA / A ratio is precisely 1. That s why I call a ph of pa home for a buffer. This is the ph around which the buffer system will act as a buffer.) Getting the ph from 6.13 to 6.93 means making the buffer more basic (higher ph). Thus, you must add the base component of the buffer, which is A (a)(i) Which buffer system is the best choice to create a buffer with ph = 9.? HF/NaF NH 3 /NH 4 Cl HNO /NO HClO/ClO (a)(ii) For the best system, calculate the ratio of the moles of the buffer components ruired to make the buffer. (b) Why can t an H 3 PO 4 /NaH PO 4 buffer system be used to make the buffer in this problem? Give specific reasoning. Answers: (a)(i) The NH3/NH4Cl system; (ii) [HA [A 8OR.[A [HA (b) a1 for H3PO4 is 7.5 x 1 3 (Sorry, this was from Table 16.1). This is way too strong of an acid to make a buffer at ph 9.! Note that pa =.1, and thus buffers with H3PO4 and NaHPO4 would only be possible within a ph unit (or two) around.1 (i.e., the maximum ph would be around 4.1, which is well below 9.). (Note: At ph 9., the HA / A ratio would be.133 : 1! It will not act like a buffer with that tiny ratio!) 1) Recognize that a buffer system tends to only work as a buffer at ph s near to the pa of its acid component. This is because of the relationship: [HA [H3O a x buffer [A If HA / A is 1, [H3O + = a and ph uals pa. If the ratio is 1 or 1/1, the ph will be one unit below or above pa. If the ratio is 1 or 1/1, the ph will be two units below or above pa. Once the ratio gets this high (Tro says even lower than this), the system won t act much like a buffer since there will be too little of one component. ) Find the a s for all the acids of the potential buffer systems from either Table 16.5, or (in the case of NH4 +, from w/b, where b is from Table 16.8). 3) Calculate pa for each (pa = loga). 4) Pick the one that has the pa closest to the desired ph (here 9.). 5) For (a)(ii), use the relationship in (1) above to solve for the HA / A ratio (in moles). You will need to calculate [H3O + from ph first. Execution of Buffer System HF/NaF NH 3 /NH 4 Cl HNO /NO HClO/ClO Acid Component HF NH4 + HNO HClO a 3.5 x /1.8 x x x 1.9 x 1 8 pa PS614

15 Thus, the NH3/NH4Cl buffer system will work best at ph 9. (9.5 is the closest to 9.). Part (a)(ii): [H3O + = 1 9. M 9. [H 3O [HA [HA 1 3 a 1 1 [A [A a 5.56 x 1 [H O x [A a 5.56 x 1 (You could also have solved for 9. [HA [H3O 1 not specify which ratio it wanted.).8.56since the problem did PS615