Problem 1 C 6 H 5 [ COOH C 6 H[H 5 COO + ] - + H [ I C - x + x + x E x x x
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1 Problem 1 What is the ph of a 291mL sample of 2.993M benzoic acid (C 6 H 5 COOH) (K a =6.4x10 5 )? Write out acid dissociation reaction: C 6 H 5 COOH C 6 H 5 COO H Make an ICE chart since this is a weak acid equilibrium: C 6 H 5 [ COOH C 6 H[H 5 COO ] H [ I C x x x E x x x Write out K a and solve: K a [C6H5COO ][H ] [C H COOH] 2 5 x x M [H 6 ph log( m) ph ]
2 Problem 2 A 489mL sample of M HNO 3 is mixed with 427mL sample of NaOH (which has a ph of 14.06). What is the ph of the resulting solution? Write out the reaction: H NO 3 Na OH Na NO 3 H 2 O Or after eliminating spectator ions: H OH H 2 O Now, make an ICE chart using moles to find out what s left after reaction: What s [NaOH]? H [ OH[ H 2 [ O I C E Now solve for poh using moles of OH, and then find ph: moles [OH ] ( )L [OH ] M poh ph13.38 poh poh 0.06 [OH ] [OH ] 1.148
3 The K a values at 25 o C for a series of acids is given: 1.8 x x x x x 10 2 Which of the following acids has a K a = 1.4 x 10 3? First realize all the answers are forms of ethanoic acid. Then, realize they want you to pick the 2 nd strongest acid of the bunch since it has the 2 nd largest K a. So you want to analyze each acid s conjugate base stability: Problem 3 c Notice that the electronegative halogens act to stabilize the conjugate base, which come from the strongest acids. Cl is most electronegative, and most stabilizing.
4 Problem 4 Consider a solution containing 0.45M HCN and 0.69M NaCN for the next two questions. The K a for HCN = 6.2 x What is the ph of this solution? Write out the reaction of the acid dissociation: HCN H CN Write out the ions that the salt produces: NaCN Na CN Now, make a chart using molarity to find out what s in this buffer solution: HCN [ H [ CN [ I Now solve for ph using HendersonHasselbalch equation: [CN ] ph pka log [HCN] log( ph ) log 0.69M 0.45M
5 Problem 5 Consider a solution containing 0.45M HCN and 0.69M NaCN for the next two questions. The K a for HCN = 6.2 x Now calculate the ph of this solution after 0.25 mol of NaOH is added to 1.00L of the solution. Assume that the volume does not change. The added NaOH (base) will react with the HCN (acid), so we should form an ICE chart to see what remains after reaction: Now solve for ph using HendersonHasselbalch equation: [CN ] ph pka log [HCN] log( ph 9.88 HCN [ OH [H H] 2 O CN [ I C E ) log 0.94moles 0.20moles
6 Problem 6 Consider a 7.3 x M solution of HCl at 285 K. What is the ph of the solution? Since this is a strong acid, we can simply take [HCl]=[H ] since it fully dissociates, and find ph easily: ph log (7.310 ph M) Does this make any sense? What s wrong with it? Think about it. We re calculating ph of an acid yet ph>7. What does this mean? Well, look at the concentration of HCl; it is extremely tiny, so it is negligible. In other words, when compared to water, HCl would lose in this case since it s so dilute. Simply ph is simply that of neutral water (ph=7)!
7 Problem 7 For the following reaction, K < 1 at room temperature. HOCl (aq) HCO 3 (aq) H 2 CO 3 (aq) OCl (aq) Which of the following statements is true? A) H 2 CO 3 is a stronger acid than HOCl. True. Since K<1, we know that the stronger acid is on the product side (H 2 CO 3 ). B) HOCl is a stronger acid than H 2 CO 3. False. See A. C) OCl is a stronger acid than HOCl. False. OCl is a base. D) HCO 3 is stronger base than OCl. False. Since K<1, we know the stronger base is on the product side (OCl ). E) More information is needed.
8 Problem 8 What is the ph of a solution that results from adding mol Ca(OH) 2 (s) to 1.37L of a buffer comprised of 0.21M HF and 0.21M NaF? First, write out the buffer solution s reaction: HF H F Now, if we add Ca(OH) 2 (strong base), it will react with the HF (acid). So let s construct an ICE table in moles to find out what s left after reaction: HF [ OH [H H] 2 O F[ Notice how Ca(OH) I (0.081) C E Now solve for ph using HendersonHasselbalch equation: ph pk a log( ph 3.70 [F ] log [HF] 4 ) log moles moles will dissociate to form twice as many (OH ).
9 Problem 9 A 0.251L solution of 1.89M sulfurous acid (K a1 = 1.5x10 2 and K a2 = 1.0x10 7 ) is titrated with 1.25M NaOH. What will the ph of the solution be when L of the NaOH has been added? First, write out the first acid dissociation: (notice how you don t need to know what sulfurous acid is, just keep in mind it s diprotic) H 2 A HA H Now, if we add NaOH (strong base), it will react with the H 2 A (acid). So let s construct an ICE table in moles to find out what s left after reaction: Are we done? No. Remember, this is diprotic and so we have excess OH that will react with HA. continued on next slide. H 2 [ A OH [H H] 2 O HA [ I C E
10 Problem 9 A 0.251L solution of 1.89M sulfurous acid (K a1 = 1.5x10 2 and K a2 = 1.0x10 7 ) is titrated with 1.25M NaOH. What will the ph of the solution be when L of the NaOH has been added? continued from previous slide Now, write out the second acid dissociation: HA A 2 H Now, if we have extra OH (strong base), it will react with the HA (acid). So let s construct an ICE table in moles to find out what s left after reaction: HA [ OH [H H] 2 O [ A 2 I C E Now use HendersonHasselbalch to get the ph: 2 [A ] ph pka2 log [HA ] log( ph ) log moles moles
11 Problem 10 The titration of 1.00L of a 1.00M solution of the triprotic acid, H 3 A, with 1.00M NaOH is shown below. It is not drawn to scale. What is the ph at D? At point D, we are at the 2 nd halfway point. So we should use the following formula: ph pk a2 log( ph ) H 3 A H 2 O H 3 O H 2 A K a1 =3.29x10 3 H 2 A H 2 O H 3 O HA 2 K a2 =6.43x10 7 HA 2 H 2 O H 3 O A 3 K a3 =4.25x10 10
12 Problem 11 The titration of 1.00L of a 1.00M solution of the triprotic acid, H 3 A, with 1.00M NaOH is shown below. It is not drawn to scale. What is the ph at C? At point C, we are at the 1 st equivalence point. So we should use the following formula: pka1 pka2 ph 2 log( ph ) log( ) H 3 A H 2 O H 3 O H 2 A K a1 =3.29x10 3 H 2 A H 2 O H 3 O HA 2 K a2 =6.43x10 7 HA 2 H 2 O H 3 O A 3 K a3 =4.25x10 10
13 Problem 12 Consider a 0.35M solution of each of the following salts in distilled water. Will the solution be acidic, neutral, or basic? You may need to reference a table of K a and K b Tables values.. NH 4 OBr: NH 4 H 2 O NH 3 H 3 O K a =5.6x10 10 OBr H 2 O HOBr OH K b =5.0x10 6 Kb > Ka Basic BaClO 4 : Ba 2 H 2 O no reaction (from strong base) ClO 4 H 2 O no reaction (from strong acid) RbCH 3 COO: Rb H 2 O no reaction (from strong base) CH 3 COO H 2 O CH 3 COOH OH Neutral Basic (CH 3 CH 2 ) 3 NHNO 2 : (CH 3 CH 2 ) 3 NH H 2 O (CH 3 CH 2 ) 3 N H 3 O K a =2.5x10 11 NO 2 H 2 O HNO 2 OH K b =2.5x10 11 Kb = Ka Neutral
14 Problem 13 How many of the following are true? 1. The buffer with the greatest buffering capacity for a phenol/sodium phenolate buffer system will have a ph of False, phenol: K a =1.6x10 10, so ph=pk a = A buffer resists ph change when base is added by converting hydroxide ions to hydronium ions.. False, the hydroxide (OH ) would react with the acid (HA) to form H 2 O, not H 3 O. 3. A buffered system can be formed by combining a weak acid/conjugate base pair in solution. True, this is a weak acidsalt combination. 4. A buffer is formed when you add HCl to NH 3. True, but only near the halfway point. 5. A buffer is formed when you add HCl to CH 3 COOH. False, this is a weak acid and strong acid addition, ph would just increase a lot.
15 Problem 14 Which of the following solutions will produce a buffer with ph near 10.50? L of 0.50M HSO 4 (K a = 1.0x10 7 ) 1.00L of 0.25M KOH. No, ph=pk a = L 0.50M H 2 NNH 2 (K b = 3.0x10 6 ) 2.00L 0.25M HI. No, K a =3.33x10 9, so ph=pk a = L 1.00M H 3 BO 3 (K a = 5.8x10 10 ) 0.020L 2.50M HBr. No, ph=pk a = L of 0.50M HONH 2 (K b = 1.1x10 8 ) 0.50L of 0.050M HNO 3. No, K a =9.1x10 7, so ph=pk a = L 1.00M CH 3 NH 2 (K b = 4.4x10 4 ) 1.00L 0.25M HI. Yes, K a =2.27x10 11, so ph=pk a =10.64.
16 Problem 15 The next three problems deal with the titration of 145mL of 1.35M methylamine CH 3 NH 2 (K b = 4.4 x 10 4 ) with 0.25M HCl. Water will be a major species throughout the titration. The chemical species, in addition to water, that can be found in this reaction mixture during the titration are: I. H II. OH III. Cl IV. CH 3 NH 2 V. CH 3 NH 3 What is the ph at equivalence point? At equivalence point, we want equal moles of CH 3 NH 2 and HCl: CH[ 3 NH 2 H CH 3 NH 3 I C E Now, we want to calculate the ph of CH 3 NH 3 in water: Remember to use molarities, so divide by total volume 0.145L from methylamine 0.784L from HCl (you get this from 0.196moles/0.25M=0.784L) 0.929L Total Volume continued on next slide
17 Problem 15 The next three problems deal with the titration of 145mL of 1.35M methylamine CH 3 NH 2 (K b = 4.4 x 10 4 ) with 0.25M HCl. Water will be a major species throughout the titration. The chemical species, in addition to water, that can be found in this reaction mixture during the titration are: I. H II. OH III. Cl IV. CH 3 NH 2 V. CH 3 NH 3 What is the ph at equivalence point? continued from previous slide So, we should write out the ion s (it s an acid) reaction in water: [use molarities!] First, find Ka: CH 3 NH [ 3 H 2 O CH 3 NH 2 H 3 O I C x x x E x x x K K a a K a [CH3NH3 ][OH [CH NH ] 2 11 x x ] M [H O 3 ] ph log( ph M)
18 Problem 16 The next three problems deal with the titration of 145mL of 1.35M methylamine CH 3 NH 2 (K b = 4.4 x 10 4 ) with 0.25M HCl. Water will be a major species throughout the titration. The chemical species, in addition to water, that can be found in this reaction mixture during the titration are: I. H II. OH III. Cl IV. CH 3 NH 2 V. CH 3 NH 3 How many ml of HCl will need to be added to reach ph=10.64? First, realize from 18 that you are before equivalence point, so use Henderson Hasselbalch equation: This is the halfway point! [CH NH ] 3 3 poh pkb log So we know the number of [CH3NH2] moles of HCl is half that of 4 [CH3NH3 ] base CH 3.36 log( ) log 3 NH 2! 1 n 1.35M HCl n CH3NH2 2 [CH3NH3 ] log 0.196moles 1.35M 2 n 0.098moles [CH3NH 0 log 1.35M [CH3NH M 3 ] 3 ] V V HCl HCl HCl moles 0.25 M 0.392L
19 Problem 17 The next three problems deal with the titration of 145mL of 1.35M methylamine CH 3 NH 2 (K b = 4.4 x 10 4 ) with 0.25M HCl. Water will be a major species throughout the titration. The chemical species, in addition to water, that can be found in this reaction mixture during the titration are: I. H II. OH III. Cl IV. CH 3 NH 2 V. CH 3 NH 3 What are the major species when 783mL of HCl has been added? Let s write out the ICE chart in moles to see what s left after reaction: CH[ 3 NH 2 H CH 3 NH 3 I C E So this is equivalence point! We only have CH 3 NH 3 and Cl (unreacted) in solution
20 Problem 18 A 0.613M solution of a weak acid has a ph of at 309K. What is the K a of the weak acid? Write out acid dissociation reaction (generic is fine): HA A H Make an ICE chart since this is a weak acid equilibrium: HA [ [H A ] H [ I C x x x E x x x First, find x from ph: x [H 10 ] x M Write out K a (plug in x) and solve: K a K a [A ][H [HA] ] 2 x ( )
21 Consider the following titration curve: Which is based on the following reactions: H 3 A H H 2 A H 2 A H HA 2 HA 2 H A 3 Problem 19 How many of the following are true? 1. The ph at point B is pk a1. True, this is the 1 st halfway. 2. The ph at point E will be the average of pk a1 and pk a2. False, you want pk a2 and pk a3 since it s the 2 nd eq. pt. 3. To the right of point F [A 3 ] > [HA 2 ]. True, it s after the 3 rd halfway point. 4. HA 2 will be a major species at point E. True, you have lost all H 2 A. 5. The second dissociation of H will play a large role in the ph level at point E. False, only the first dissociation matters.
Problem 1. What is the ph of a 291mL sample of 2.993M benzoic acid (C 6 H 5 COOH) (K a =6.4x10-5 )?
Problem 1 What is the ph of a 291mL sample of 2.993M benzoic acid (C 6 H 5 COOH) (K a =6.4x10-5 )? Problem 2 A 489mL sample of 0.5542M HNO 3 is mixed with 427mL sample of NaOH (which has a ph of 14.06).
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