HALFWAY to EQUIVALENCE POINT: ph = pk a of the acid being titrated.

CHEMISTRY 109 Help Sheet #33 Titrations Chapter 15 (Part II); Section 15.2 ** Cover topics appropriate for your lecture** Prepared by Dr. Tony Jacob http://www.chem.wisc.edu/areas/clc (Resource page) Nuggets: Titration Curves and ph Calculations - Equivalence Point, Endpoint, Halfway to Equivalence Point; Past Equivalence Point; Indicators; Acid rain TITRATION CURVES (See figures 15.6, 15.7, 15.9, 15.10) ph s are at the equivalence point; be familiar with the graphs and which corresponds to what type of titration (i.e., what does SA + SB or SB + SA look like); underlined chemical in graphs below is the titrant SA + SB ph = 7 SA + SB ph = 7 WA + SB ph > 7 WB + SA ph < 7 Polyprotic acid titration EQUIVALENCE POINT: point at which moles of base added (titrant) = moles of acid being titrated. At equivalence point: SA + SB ph = 7; WA + SB ph > 7; WB + SA ph < 7 HALFWAY to EQUIVALENCE POINT: ph = pk a of the acid being titrated. INDICATORS - Indicators are weak acids that change color depending on if they are protonated (has the H + attached) or deprotonated (H + not attached); that is, HIn has one color while In - has a different color. HIn H + + In - K ind = [H + ][In - ]/[HIn]; ph range of an indicator: ph = pk ind ± 1; note that K ind is just the K a of the indicator; on each end of the ph range, the indicator has a different color; choose the indicator such that pk a indicator ph at the equivalence point just blue Example 1: An indicator, HIn, (K a = 1.0 x 10-9 ) is blue when protonated (HIn) and red when deprotonated (In - ). What is the color of this indicator at ph = 10.5? Answer 1: pk ind = pk a = 9; ph range = pk a ± 1; ph range: 8 to 10; ph = 8 blue; ph = 10 red; at ph = 10.5 color = mainly red; see diagram at right Endpoint: Point at which the indicator changes color; if the correct indicator has been chosen for the experiment, the endpoint occurs at or near the equivalence point. TYPES of TITRATIONS: I. Strong acid + strong base neutral salt + water; ph = 7 at equivalence point HCl + NaOH H 2 O + NaCl (at equivalence point: Na +, Cl -, and H 2 O exist) Initial ph depends on concentration of strong acid II. Strong acid + weak base weak acid + salt/water; ph < 7 at equivalence point HCl + NaF HF + NaCl Titration: at equivalence point: Na +, Cl -, and HF exist, ph < 7 since HF is an acid; initial ph depends on K b of weak base NaF (before addition of HCl); final ph at equivalence point depends on [HF] III. Strong base + weak acid weak base + salt/water; ph > 7 at equivalence point HCN + NaOH NaCN + H 2 O Titration: at equivalence point: Na +, CN -, and H 2 O exist, ph > 7 since CN - is a base; initial ph depends on K a of weak acid (before NaOH is added); final ph at equivalence point depends on [CN- ] ph = pk a just red blue red blue/red ph 7 8 9 10 11 effective range of indicator

ph TITRATIONS CALCULATIONS A. ph initially (see below, right graph, where the letters occur on the graph) regular WA or WB calc B. ph before the half-way to equivalence point (buffer; [WA] [WB]) I. SA + WB or WA + SB reaction in mol to completion; use SCF table II. buffer exists; use H-H equation to find ph C. ph at the halfway to equivalence point (buffer with [WA] = [WB]) use ph = pk a since log([b]/[a]) = 0 in ph = pk a + log([b]/[a]) D. ph at the equivalence point ([WA] = 0 and replaced by WB or vice-versa) I. SA + WB or WA + SB reaction in mol to completion; use SCF table; all WA or WB consumed II. only WA or WB exists; determine new concentrations (add volumes); regular WA or WB calc E. ph past the equivalence point (SA or SB added to excess) (not always covered) I. SA + WB or WA + SB reaction in mol to completion; all WA or WB consumed and excess SA or SB remains along with WA or WB (i.e., WA/SA or WB/SB exists) II. regular SA or SB calculation (assume WA or WB is insignificant as compared to SA or SB) Example 2: A 150ml solution, 0.20M HCOOH (K a = 1.8 x 10-4 ) is titrated with 1.5M NaOH. a. What volume of NaOH is required to reach the equivalence point? b. What is the initial ph of the solution before the titration has started (Pt. A)? c. What is the ph after 5.0ml NaOH is added (a point between A and C)? d. What is the ph after 10.0ml NaOH is added (Pt. C)? e. What is the ph after 20.0ml NaOH is added (Pt. D)? f. What is the ph after 22.0ml NaOH is added (Pt. E)? Answer 2: a. At equivalence point: mol HCOOH = mol NaOH; mol HCOOH = M x L = (0.2)(0.15) = 0.030mol HCOOH 0.030mol NaOH needed; V = mol/m = 0.03/1.5 = 0.020L = 20. ml NaOH b. HCOOH HCOO - + H + K a = [H + ][HCOO - ]/[HCOOH]; 1.8 x 10-4 = x 2 /0.2-x; make approximation: I C 0.20 0 0 -x +x E 0.20-x x x c. OH - + HCOOH HCOO - + H 2 O S C F 0 0.0225 0.0075 d. OH - + HCOOH HCOO - + H 2 O S +x 0.0075 0.030 0-0.0075-0.0075 +0.0075 C -0.015-0.015 +0.015 F 0 0.015 0.015 (SCF Table is not needed if it is recognized that you are at the halfway to eq. pt.) 0.015 0.030 0 e. OH - + HCOOH HCOO - + H 2 O S C -0.030-0.030 +0.030 F I C E 0.030 0.030 0 0 0 0.030 HCOO - + H 2 O HCOOH + OH- 0.176 0 0 -x +x 0.176-x f. OH - + HCOOH HCOO - + H 2 O S C -0.030-0.030 +0.033 F x +x 0.033 0.030 0 0.003 0 0.033 x 1.8 x 10-4 = x 2 /0.2; x = 0.0060 = [H + ]; ph = -log(0.0060) = 2.22 A buffer exists! Fastest way is to use H-H equation: ph = pk a + log([base]/[acid]) = -log(1.8 x 10-4 ) + log(0.0075/0.0225) = 3.26 mol NaOH = M x L = (1.5)(0.010) = 0.015mol NaOH; do SCF Table (note: half of HCOOH is consumed!); a buffer is created use H H eqn; ph = pk a + log([base]/[acid]) = -log(1.8 x 10-4 ) + log(0.015/0.015) = 3.74 A simpler way: Recognize 20ml needed to reach eq. pt. so 10ml is at the halfway to eq. pt. (1/2 HCOOH consumed) ph = pk a = -log(1.8 x 10-4 ) = 3.74 At eq. pt.; mol NaOH = M x L = (1.5)(0.020) = 0.030mol NaOH; do SCF Table (note: all HCOOH is consumed eq. pt!); no buffer exits; just a WB problem do an ICE Table; find [HCOO - ] = mol HCOO - /total volume = 0.030mol/(0.15+0.02) = 0.176M; K b = x 2 /0.176-x; K b = 1 x 10-14 /1.8 x 10-4 = 5.56 x 10-11 ; make approximation: 5.56 x 10-11 = x 2 /0.176; x = 3.13 x 10-6 = [OH - ]; poh = -log(3.13 x 10-6 ) = 5.50; ph = 14-5.50 = 8.50 Past the equivalence point! There exists a strong base and a weak base. The strong base dominates and the weak base can be ignored. Find [OH - ]: [OH - ] = mol OH - /L solution = 0.003/((150+22)/1000) = 0.0174M poh = -log(0.0174) = 1.76; ph = 14.00 1.76 = 12.24

ACID RAIN (if covered) any rain that has a ph < 5.6; ph of natural rain is slightly acidic due to carbonic acid: CO 2 (g) + H 2 O(l) H 2 CO 3 (aq) H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) Precursor chemical sources of acid rain: NO 2 (g) and SO 2 (g) 2NO 2 (g) + H 2 O(l) HNO 3 (aq) + HNO 2 (aq) SO 2 (aq) + H 2 O(l) 2SO 2 (aq) + O 2 (g) SO 3 (aq) + H 2 O(l) H 2 SO 3 (aq) 2SO 3 (g) H 2 SO 4 (aq) 1. I. What is the ph at the equivalence point for the titration of benzoic acid, C 6 H 5 COOH, with NaOH? a. less than 7 b. equal to 7 c. greater than 7 d. depends on the amount of NaOH added II. What is the ph at the equivalence point for the titration of trimethylamine, (CH 3 ) 3 N, with HCl? a. less than 7 b. equal to 7 c. greater than 7 d. depends on the amount of NaOH added 2. A 50.0ml solution of benzoic acid, C 6 H 5 COOH, is titrated to the equivalence point with 42.55ml of 0.215M NaOH. What is the concentration of the original benzoic acid solution? 3. A 35.0ml solution of 0.155M methylamine, (CH 3 )NH 2, is titrated to the equivalence point with 0.185M HCl. What volume of HCl is needed to reach the equivalence point? 4. In the graph to the right, the titrant is NaOH and the solution being titrated is propanoic acid, CH 3 COOH. The concentration of CH 3 COOH = 0.075M and there is 100. ml of solution (K a CH 3 COOH = 1.3 x 10-5 ). a. At the equivalence point (A), approximately how much propanoic acid remains? b. What chemical specie(s) dominate at point B? (Ignore water, spectator ions.) c. What is the ph at point B? d. What chemical specie(s) dominate at point C? (Ignore water, spectator ions.) e. What is the ph at point C? f. Will the ph of the solution be 7, greater than 7, or less than 7 at point A? ph C* B * * A ml of titrant 5. A 50. ml solution of 0.20M KOH is titrated with a 0.40M HI solution. a. What volume of HI is needed to reach the equivalence point? b. What is the final ph of the solution at the equivalence point? 6. A 100ml solution of 0.15M HNO 3 is titrated with 0.30M KOH. What volume is needed to reach the equivalence point and what is the ph at the equivalence point? 7. For the titration of 0.10M CH 3 COOH with 2.0M NaOH it required 30.0ml of NaOH to reach the equivalence point. What is the ph after 15.0ml of NaOH was added? (K a CH 3 COOH = 1.8 x 10-5 ) a. 9.0 x 10-6 b. 14.3 c. 5.05 d. 4.74 e. 2.87 8. When does the pk a equal the ph of a weak acid solution, HA, titrated by a strong base? a. when ph = 1 b. when K a = ph c. when [H + ] = [A - ] d. when [HA] = [A - ] e. none of these

9. a. What is the ph at the equivalence point when 0.50M HCl is used to titrate 35.0ml 0.25M (CH 3 ) 2 NH (K b = 5.8 x 10-4 )? b. What is the K eq for the reaction of HCl with (CH 3 ) 2 NH? 10. a. What is the ph at the equivalence point when 1.00M KOH is used to titrate 75.0ml 0.15M HOCN (K a = 3.5 x 10-4 )? b. What is the K eq for the reaction of KOH with HOCN? 11. A 100. ml solution of 0.25M C 6 H 5 NH 2 solution (K b = 4.0 x 10-10 ) was titrated with 25.0ml HCl to reach the equivalence point. a. What was the original concentration of the HCl solution? b. What was the ph of the solution before the titration began? c. What was the ph of the solution after 5.00ml of HCl was added? d. What was the ph of the solution after a total of 12.5ml of HCl was added? e. What was the ph of the solution after a total of 25.0ml of HCl was added? f. What was the ph of the solution after a total of 30.0ml of HCl was added? 12. A 100. ml solution of 0.10M HCOOH solution (K a HCOOH = 1.8 x 10-4 ) was titrated with 5.0ml of 2.0M NaOH to reach the equivalence point. a. What was the ph of the solution before the titration began? b. What was the ph of the solution after 2.5ml of NaOH were added? c. What was the ph of the solution after a total of 5.0ml of NaOH was added? d. What was the ph of the solution after a total of 7.5ml of NaOH was added? 13. Methyl orange is an indicator with a K a of 1.0 x 10-4. Its acid form HIn is red while its base form In- is yellow. At ph = 6.0, the indicator will be a. red b. orange c. yellow d. blue 14. What is the effective ph range for an indicator with a K a = 1.7 x 10-5? 15. Which indicator would be a good candidate for a titration with an equivalence point at ph of 4.15? phenolphthalein, pk a = 9.10; thymol blue, pk a = 1.65; methyl yellow, pk a = 3.25; bromothymol blue, pk a = 7.30

16. I. The titration curve to the right is a titration of a. a weak base titrated with a strong acid b. a weak acid titrated with a strong base c. a strong base titrated with a strong acid d. a strong acid titrated with a strong base e. a polyprotic acid titrated with a strong base II. What is the ph at the equivalence point? a. 2.8 b. 4.8 c. 9.0 d. 13.0 e. need K a III. The chemical being titrated is: a. NH 3 (K b = 1.8 x 10-5 ) b. HClO 4 c. H 2 CO 3 (K a1 = 4.5 x 10-7 ) d. HC 2 H 3 O 2 (K a = 1.8 x 10-5 ) e. HCN (K a = 4.0 x 10-10 ) IV. Which indicator would be a good choice for this titration? a. methyl orange (K ind = 2.0 x 10-4 ) b. bromophenol blue (K ind = 1.0 x 10-4 ) c. methyl red (K ind = 7.9 x 10-6 ) d. bromothymol blue (K ind = 1.0 x 10-7 ) e. phenolphthalein (K ind = 5.0 x 10-10 ) 17. The graph shows a triprotic acid titrated with NaOH. a. On the graph, mark equivalence point 1. (Hint: It s around the 10ml point of the titration!) b. On the graph, mark the halfway to equivalence point 1, and determine the ph at that point. c. From part b, determine pk a1 and K a1. d. On the graph, mark equivalence point 2. e. On the graph, mark the halfway to equivalence point 2, and determine the ph at that point. f. From part e, determine pk a2 and K a2. g. On the graph, mark equivalence point 3. (Hint: It is just past the end of the graph!) h. On the graph, mark the halfway to equivalence point 3, and determine the ph at that point. i. From part h, determine pk a3 and K a3.

ANSWERS 1. I. c. greater than 7 {at equivalence point only the conjugate base of benzoic acid, C 6 H 5 COO -, exists; solution is basic} II. a. less than 7 {at equivalence point only the conjugate acid of trimethylamine, (CH 3 ) 3 NH +, exists; solution is acidic} 2. 0.183M {[C 6 H 5 COOH] = mol/l; L C 6 H 5 COOH = 0.0500L; need to find mol C 6 H 5 COOH; mol NaOH = (0.04255)(0.215) = 9.148 x 10-3 mol NaOH x (1mol C 6 H 5 COOH/1mol NaOH) = 9.148 x 10-3 mol C 6 H 5 COOH; [C 6 H 5 COOH] = 9.148 x 10-3 mol/0.0500l = 0.1830M} 3. 29.3ml HCl {find mol (CH 3 )NH 2 = M x L = 0.155M x 0.0350L = 0.005425mol (CH 3 )NH 2 ; at the equivalence point mol HCl = mol (CH 3 )NH 2 ; L HCl = mol/m = (0.005425mol)/(0.185M) = 0.0293L = 29.3ml HCl} 4. a. 0 remains {all the acid has been reacted with NaOH at eq. pt.} b. CH 3 COOH, CH 3 COO - {CH 3 COOH + NaOH CH 3 COO - + H 2 O; as the NaOH is added, it is completely consumed leaving the WA and its conjugate WB} c. ph = 4.89 (halfway to eq. pt) {ph = pk a at halfway pt; ph = -log(1.3 x 10-5 ) = 4.886} d. CH 3 COOH {Point C is before the titration begins so only the WA is present} e. ph = 3.01 {WA problem: {CH 3 COOH H + + CH 3 COO - ; from ICE Table, I row: 0.075, 0, 0; C row: -x, +x, +x; E row: 0.075-x, x, x; K a = [H + ][CH 3 COO - ]/[CH 3 COOH]; x 2 /0.075-x = 1.3 x 10-5 ; make approximation: x 2 /0.075 = 1.3 x 10-5 ; x = 0.0009874 = [H + ]; ph = -log(0.0009874) = 3.005} f. greater than 7 {Point A is eq. pt.; at eq. pt. all the WA is consumed and only the WB remains ph > 7} 5. a. 25ml HI {H + + OH - H 2 O; 0.05L x 0.2M = 0.010mol OH - ; at equivalence point: mol KOH = mol HI = 0.010mol; L HI = mol/m = 0.010mol/0.40M = 0.025L = 25ml} b. 7.00 {since strong acid + strong base yields only water ph = 7 at eq. pt.} 6. ph = 7; volume = 50ml {mol HNO 3 = (0.1L)(0.15M) = 0.015mol HNO 3 ; at equivalence point: mol HNO 3 = mol KOH; mol KOH = 0.015mol; M = mol/l; L = 0.015mol/0.30M = 0.050L = 50ml; ph = 7 at equivalence point since it is a strong acid being titrated with a strong base} 7. 4.74 {15.0ml is halfway to the equivalence pt; at ½ way to the equivalence pt: ph = pk a ; ph = -log(1.8 x 10-5 ) = 4.74} 8. d {from the H-H equation} 9. a. ph = 5.77 {find vol HCl needed to reach equivalence pt: mol (CH 3 ) 2 NH = M x L = (0.25)(0.0350) = 0.00875mol (CH 3 ) 2 NH; mol HCl at equivalence point = 0.00875; volume HCl = mol HCl/M HCl = 0.00875/0.50 = 0.0175L; at equivalence point: HCl + (CH 3 ) 2 NH (CH 3 ) 2 NH 2 + + Cl - ; using SCF Table: S row: 0.00875, 0.00875, 0, ; C row: -0.00875, -0.00875, +0.00875, +0.00875; F row: 0, 0, 0.00875, ; only a WA is left: (CH 3 ) 2 NH 2 + ; [(CH 3 ) 2 NH 2 + ] = 0.00875mol/(0.035L + 0.0175L) = 0.1667M; write dissociation reaction: (CH3 ) 2 NH 2 + (CH 3 ) 2 NH + H + ; ICE Table: I row: 0.1667, 0, 0; C row: -x, +x, +x; E row: 0.1667-x, x, x; K a = K w /K b = 1.0 x 10-14 /5.8 x 10-4 = 1.72 x 10-11 ; K a = [H + ][(CH 3 ) 2 NH 2 + ]/[(CH3 ) 2 NH]; 1.72 x 10-11 = x 2 /(0.1667-x); since (100)K a < [ ] o make approximation; 1.72 x 10-11 = x 2 /(0.1667); solve for x: x = 1.69 x 10-6 = [H + ]; ph = -log(1.69 x 10-6 ) = 5.77} b. K eq = 5.8 x 10 10 {reaction: H + + (CH 3 ) 2 NH (CH 3 ) 2 NH 2 + ; reverse of the Ka of (CH 3 ) 2 NH 2 + ; K a = K w /K b = 1 x 10-14 /5.8 x 10-4 = 1.72 x 10-11 ; for reverse invert: 1/K a = 1/1.72 x 10-11 = 5.8 x 10 10 }

10. a. ph = 8.29 {find vol KOH needed to reach equivalence point: mol HOCN = M x L = (0.15)(0.0750) = 0.00115mol HOCN; mol KOH at equivalence point = 0.01125; volume KOH = mol KOH/M KOH = 0.01125/1.00 = 0.01125L; at equivalence point: OH - + HOCN OCN - + H 2 O; using SCF Table: S row: 0.0115, 0.0115, 0, ; C row: -0.0115, -0.0115, +0.0115, +0.0115; F row: 0, 0, 0.0115, ; only a WB is left: OCN - ; [OCN - ] = 0.0115mol/(0.075L + 0.01125L) = 0.1333M; write hydrolysis reaction: OCN - + H 2 O HOCN + OH - ; ICE Table: I row: 0.1333,, 0, 0; C row: -x, -x, +x, +x; E row: 0.1333-x,, x, x; K b = K w /K a = 1.0 x 10-14 /3.5 x 10-4 = 2.86 x 10-11 ; K b = [OH - ][HOCN]/[OCN - ]; 2.86 x 10-11 = x 2 /(0.1333-x); since (100)K b < [ ] o make approximation; 2.86 x 10-11 = x 2 /(0.1333); solve for x: x = 1.95 x 10-6 = [OH - ]; poh = -log(1.95 x 10-6 ) = 5.71; ph = 14.00 5.71 = 8.29} b. K eq = 3.5 x 10 10 {reaction: OH - + HOCN OCN - + H 2 O; this reaction is a combination of two reactions: Rxn 1: HOCN H + + OCN - and Rxn 2: H + + OH + H 2 O; Rxn 1 = K a ; Rxn 2 = reverse of K w = 1/K w ; when you add reactions multiply the K values; K eq = K a x (1/K w ) = (3.5 x 10-4 )(1/1 x 10-14 ) = 3.5 x 10 10 } 11. a. 1.0M HCl {At eq. pt.: mol C 6 H 5 NH 2 = mol HCl; mol C 6 H 5 NH 2 = M x L = (0.25)(0.1) = 0.025mol C 6 H 5 NH 2 0.025mol HCl needed; M = mol/l = 0.025/0.025 = 1.0M HCl} b. ph = 9.00 {WB problem; C 6 H 5 NH 2 + H 2 O C 6 H 5 NH 3 + + OH - ; set up ICE Table: I row: 0.25,, 0, 0; C row: -x,, +x, +x; E row: 0.25-x,, x, x; plug into K b = [C 6 H 5 NH 3 + ][OH - ]/[C6 H 5 NH 2 ]; 4.0 x 10-10 = x 2 /0.25-x; make approximation: 4.0 x 10-10 = x 2 /0.25; x = 1.0 x 10-5 M = [OH - ]; poh = -log(1.0 x 10-5 ) = 5.00; ph = 14-pOH = 9.00} c. ph = 5.20 {mol C 6 H 5 NH 2 = M x L = (0.25)(0.1) = 0.025mol C 6 H 5 NH 2 ; mol HCl = M x L = (1)(0.005) = 0.005mol HCl; C 6 H 5 NH 2 + H + C 6 H 5 NH 3 + ; do SCF Table: S row: 0.025, 0.005, 0; C row: -0.005, -0.005, +0.005; F row: 0.020, 0, 0.005; both WA/WB present buffer use H H eqn; K a = K w /K b = 1 x 10-14 /4 x 10-10 = 2.5 x 10-5 ; ph = pk a + log([base]/[acid]) = -log(2.5 x 10-5 ) + log(0.020/0.005) = 5.20} d. ph = 4.60 {mol C 6 H 5 NH 2 = M x L = (0.25)(0.1) = 0.025mol C 6 H 5 NH 2 ; mol HCl = M x L = (1)(0.0125) = 0.0125mol HCl; C 6 H 5 NH 2 + H + C 6 H 5 NH 3 + ; halfway to equivalence point; ph = pka = -log(2.5 x 10-5 ) = 4.60; if you do not see that this is the halfway to the eq. pt., you can do it the long way: do SCF Table: S row: 0.025, 0.0125, 0; C row: -0.0125, -0.0125, +0.0125; F row: 0.0125, 0, 0.0125; both WA/WB present buffer use H H eqn; K a = K w /K b = 1 x 10-14 /4 x 10-10 = 2.5 x 10-5 ; ph = pk a + log([base]/[acid]) = -log(2.5 x 10-5 ) + log(0.0125/0.0125) = 4.60} e. ph = 2.65 {mol C 6 H 5 NH 2 = M x L = (0.25)(0.1) = 0.025mol C 6 H 5 NH 2 ; mol HCl = M x L = (1)(0.025) = 0.025mol HCl; C 6 H 5 NH 2 + H + C 6 H 5 NH 3 + ; do SCF Table: S row: 0.025, 0.025, 0; C row: -0.025, -0.025, +0.025; F row: 0, 0, 0.025; only a WA is present; C 6 H 5 NH 3 + C 6 H 5 NH 2 + H + ; find [C 6 H 5 NH 3 + ] = mol C6 H 5 NH 3 + /total volume = 0.025mol/(0.025+0.100) = 0.200M; use ICE Table: I row: 0.200, 0, 0; C row: -x, +x, +x; E row: 0.200-x, x, x; plug into K a = [C 6 H 5 NH 2 ][H + ]/[C 6 H 5 NH 3 + ]; 2.50 x 10-5 = x 2 /0.200-x; make approximation: 2.50 x 10-5 = x 2 /0.200; x = 0.00224 = [H + ]; ph = -log(0.00224) = 2.65} f. ph = 1.41 {mol C 6 H 5 NH 2 = M x L = (0.25)(0.1) = 0.025mol C 6 H 5 NH 2 ; mol HCl = M x L = (1)(0.030) = 0.030mol HCl; C 6 H 5 NH 2 + H + C 6 H 5 NH 3 + ; do SCF Table: S row: 0.025, 0.030, 0; C row: -0.025, -0.025, +0.025; F row: 0, 0.005, 0.025; past eq. pt.; no buffer exits; contains SA and WA; SA dominates ignore WA; find [H + ] = mol H + /total volume = 0.005mol/(0.030+0.100) = 0.0385M = [H + ]; ph = -log(0.0385) = 1.41} 12. a. 2.37 {K a = 1.8 x 10-4 = x 2 /0.1-x; x = 0.004243; ph = 2.37} b. 3.74 {mol HCOOH = (0.1)(0.1) = 0.01mol HCOOH; mol NaOH = (0.0025)(2) = 0.005mol NaOH; halfway to equivalence point; ph = pk a = -log(1.8 x 10-4 ) = 3.74} c. 8.37 {mol HCOOH = (0.1)(0.1) = 0.01mol HCOOH; mol NaOH = (0.005)(2) = 0.01mol NaOH; at equivalence point; all that remains is 0.01mol HCOO - ; K b = [OH - ][HCOOH]/[HCOO - ]; [HCOO - ] = 0.01mol/0.105L = 0.0952M; K b = 1 x 10-14 /1.8 x 10-4 = 5.56 x 10-11 = x 2 /(0.0952-x); x = 2.30 x 10-6 ; poh = 5.63; ph = 8.37} d. 12.67 {excess NaOH; 0.0025L x 2M = 0.005mol NaOH and 0.01mol HCOO - ; [HCOO - ] = 0.01mol/0.1075L = 0.0930M; [OH - ] from excess NaOH = 0.005mol/0.1075L = 0.0465M; the OH - from the excess strong base dominates and you don t have to consider the weak base; [OH - ] = 0.0465; poh = 1.33; ph = 12.67}

just red ph = pk a just yellow red yellow red/yellow ph 2 3 4 5 6 effective range of indicator 13. c { } 14. 3.77 < ph < 5.77 {ph = pk a ± 1; pk a = -log(1.7 x 10-5 ) ± 1 = 4.77 ± 1} 15. methyl yellow {ph indicators range is pk ind ± 1; for a reaction with an equivalence point of 4.15, the indicator must have a pk a between 3.15 and 5.15; methyl yellow does} 16. I. b {not multiple equivalence points so e is incorrect; ph starts low (acid) so a and c are incorrect; equivalence point is II. c III. d not 7 so d is incorrect} {the eq pt occurs in the middle of the vertical region of the curve; once that point is noted, read across to the y-axis for the ph} {at the halfway to the equivalence point, the ph = pk a ; the equivalence point is at 0.1mol of base added so the halfway point is at 0.05mol added; the ph at that point is 4.7 so the pk a = 4.7; the K a = 10 4.7 = 2.0 x 10-5 ; therefore it is acetic acid; note: pk b for NH 3 = 4.74 which means its pk a = 9.26} IV. e {indicator ph range = pk ind ± 1; the indicator ph range should encompass the ph at which the equivalence point lands; phenolphthalein: pk ind = -log(5.0 x 10-10 ) = 9.3; phenolphthalein ph range: 8.3 10.3 includes the equivalence pt at ph = 9} 17. a. See graph. b. ph 2.3 c. pk a1 2.3; K a1 5 x 10-3 {at the halfway to the eq pt: pk a = ph; K a = 10 -pk a} d. See graph. e. ph 7.2 f. pk a2 7.2; K a2 6 x 10-8 {at the halfway to the eq pt: pk a = ph; K a = 10 -pk a} g. See graph. h. ph 12.5 c. pk a3 12.5; K a3 3 x 10-13 {at the halfway to the eq pt: pk a = ph; K a = 10 -pk a}