CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS

Transcription

1 CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS 19.1 The purpose of n cid-bse buffer is to mintin reltively constnt ph in solution The wek cid component neutrlizes dded bse nd the wek bse component neutrlizes dded cid so tht the ph of the buffer solution remins reltively constnt. The components of buffer do not neutrlize one nother when they re conjugte cid/bse pir. 19. The presence of n ion in common between two solutes will cuse ny equilibrium involving either of them to shift in ccordnce with Le Châtelier s principle. For exmple, ddition of NF to solution of HF will cuse the equilibrium HF(q) H 2 O(l) H O (q) F (q) to shift to the left, wy from the excess of F, the common ion ) Buffer hs equl, high concentrtions of both HA nd A. It hs the highest buffering cpcity. b) All of the buffers hve the sme ph rnge. The prcticl buffer rnge of ph pk ± 1, nd is independent of concentrtion. c) Buffer 2 hs the gretest mount of wek bse nd cn therefore neutrlize the gretest mount of dded cid A buffer with high cpcity hs gret resistnce to ph chnge. A high buffer cpcity results when the wek cid nd wek bse re both present t high concentrtion. Addition of 0.01 mol of HCl to high-cpcity buffer will cuse smller chnge in ph thn with low-cpcity buffer, since the rtio [HA] / [A ] will chnge less Only (c) hs n ffect on the buffer cpcity. In theory, ny conjugte pir (of ny pk ) cn be used to mke high cpcity buffer. With proper choice of components, it cn be t ny ph. The buffer rnge chnges long with the buffer cpcity, but does not determine it. A high-cpcity buffer will result when comprble quntities (i.e., buffer-component rtio < 10:1) of wek cid nd wek bse re dissolved so tht their concentrtions re reltively high The buffer component rtio refers to the rtio of concentrtions of the cid nd bse tht mke up the buffer. When this rtio is equl to 1, the buffer resists chnges in ph with dded cid to the sme extent tht it resists chnges in ph with dded bse. The buffer rnge extends eqully in both the cidic nd bsic direction. When the rtio shifts with higher [bse] thn [cid], the buffer is more effective t neutrlizing dded cid thn bse so the rnge extends further in the cidic thn bsic direction. The opposite is true for buffer where [cid] > [bse]. Buffers with rtio equl to 1 hve the gretest buffer rnge. The more the buffer component rtio devites from 1, the smller the buffer rnge pk (formic).74; pk (cetic) Formic cid would be the buffer choice, since its pk is closer to the desired ph of.5. If cetic cid were used, the buffer component rtio would be fr from 1:1 nd the buffer s effectiveness would be lower. The NOH serves to prtilly neutrlize the cid nd produce its conjugte bse ) The buffer component rtio nd ph increse with dded bse. The OH rects with HA to decrese its concentrtion nd increse [NA]. The rtio [NA] / [HA] thus increses. The ph of the buffer will be more bsic becuse the concentrtion of bse, A, hs incresed nd the concentrtion of cid, HA, decresed. b) Buffer component rtio nd ph decrese with dded cid. The H O rects with A to decrese its concentrtion nd increse [HA]. The rtio [NA] / [HA] thus decreses. The ph of the buffer will be more cidic becuse the concentrtion of bse, A, hs decresed nd the concentrtion of cid, HA, incresed. 19-1

2 c) Buffer component rtio nd ph increse with the dded sodium slt. The dditionl NA increses the concentrtion of both NA nd HA, but the reltive increse in [NA] is greter. Thus, the rtio increses nd the solution becomes more bsic. Whenever bse is dded to buffer, the ph lwys increses, but only slightly if the mount of bse is not too lrge. d) Buffer component rtio nd ph decrese. The concentrtion of HA increses more thn the concentrtion of NA, so the rtio is less nd the solution is more cidic ) ph would increse by smll mount. b) ph would decrese by smll mount. c) ph would increse by very smll mount. d) ph would increse by lrge mount The buffer components re propnoic cid nd propnote ion. The sodium ions re specttor ions nd re ignored becuse they re not involved in the buffer. The rection tble tht describes this buffer is: Concentrtion (M) CH CH 2 COOH(q) H 2 O(l) CH CH 2 COO (q) H O (q) Initil Chnge x x x Equilibrium 0.15 x 0.5 x x Assume tht x is negligible with respect to both 0.15 nd 0.5 since both concentrtions re much lrger thn K. HO CHCH2COO K [ x][ 0.5 x] x ][ 0.5 ] CHCH2COOH [ 0.15 x] [ 0.15] 1. x 10 5 [H O 2 ] K CH CH COOH ( 1. x 10 ) CHCH2COO x 10 6 M Check ssumption: percent error (5.6 x 10 6 / 0.15)100% 0.007%. The ssumption is vlid. ph log [H O ] log ( x 10 6 ) Another solution pth to find ph is using the Henderson Hsselblch eqution: [bse] ph pk log [cid] pk log(1. x 10 5 ) [CHCH2COO ] ph log [CH CH 2 COOH] ph [0.5] log [0.15] CA C 6 H 5 COOH CB C 6 H 5 COO Neglect N [ ][ ]. Assume x nd x re negligible. K H O CB [ x ][ 0.28 x ] [ x ][ 0.28 ] 6. x 10 5 CA 0. x 0. [ ] [ CA ] [H O ] K [ CB ] [ ] [ ] (6. x 10 5 ) (0. / 0.28) x x 10 5 M Check ssumption: percent error (7.425 x 10 5 / 0.28)100% 0.026%. The ssumption is vlid. ph log [H O ] log (7.425 x 10 5 ) The buffer components re nitrous cid, HNO 2, nd nitrite ion, NO 2. The potssium ions re ignored becuse they re not involved in the buffer. Set up the problem with rection tble. Concentrtion (M) HNO 2 (q) H 2 O(l) NO 2 (q) H O (q) Initil Chnge x x x Equilibrium 0.55 x 0.75 x x Assume tht x is negligible with respect to both 0.55 nd 0.75 since both concentrtions re much lrger thn K. 19-2

3 HO NO 2 K HNO [ x][ 0.75 x] [ x ][ 0.75 ] [ 2 ] [ 0.55 x] [ 0.55] [ HNO ] [H O 2 ] K ( ) [ ] x 10 [ ] 7.1 x 10-4 NO x x 10-4 M 2 Check ssumption: percent error ( x 10-4 / 0.55)100% 0.095%. The ssumption is vlid. ph [H O ] log ( x 10-4 ) Verify the ph using the Henderson-Hsselblch eqution. [bse] pk log(7.1 x 10 4 ) [NO 2 ] [0.75] ph.149 log.149 log [HNO 2 ] [0.55] ph ph pk log [cid] The cid component is HF nd the bse component is F. Neglect K. Assume x nd x re negligible. HO F K [ x][ 0.25 x] [ x ][ 0.25 ] 6.8 x 10 4 HF 0.20 x 0.20 [ ] [ [H O ] K HF] [ ] [ ] (6.8 x 10 4 ) (0.20 / 0.25) 5.44 x x 10 4 M F Check ssumption: percent error (5.44 x 10 4 / 0.20)100% 0.27%. The ssumption is vlid. ph log [H O ] log (5.44 x 10 4 ) Verify the ph using the Henderson-Hsselblch eqution The buffer components re formic cid, HCOOH, nd formte ion, HCOO. The sodium ions re ignored becuse they re not involved in the buffer. Clculte K from pk nd write rection tble for the dissocition of formic cid. K 10 pk x 10 4 (unrounded) Concentrtion (M) HCOOH(q) H 2 O(l) HCOO (q) H O (q) Initil Chnge x x x Equilibrium 0.45 x 0.6 x x Assume tht x is negligible becuse both concentrtions re much lrger thn K. HO HCOO K [ x][ 0.6 x] x ][ 0.6 ] HCOOH 0.45 x x 10 4 (unrounded) [ ] [ [H O ] K HCOOH] [ ] ( ) [ ] x 10 [ ] HCOO [ 0.6 ] x x 10 4 M Check ssumption: percent error ( x 10 4 / 0.45)100% 0.029%. The ssumption is vlid. ph [H O ] log ( x 10 4 ) Verify the ph using the Henderson-Hsselblch eqution. [bse] ph pk log [cid] [HCOO ] ph.74 log [HCOOH] ph [0.6].74 log [0.45] 19-

4 19.16 K 10 pk x 10 9 (unrounded) The cid component is HBr nd the bse component is BrO. Neglect K. Assume x nd x re negligible. HO BrO K [ x][ 0.68 x] [ x ][ 0.68 ] x 10 9 (unrounded) HBrO 0.95 x 0.95 [ ] [ [H O ] K HBrO] [ ] [ ] ( x 10 9 ) (0.95 / 0.68) x x 10 9 M BrO Check ssumption: percent error ( x 10 9 / 0.68)100% %. The ssumption is vlid. ph log [H O ] log ( x 10 9 ) Verify the ph using the Henderson-Hsselblch eqution The buffer components re phenol, C 6 H 5 OH, nd phenolte ion, C 6 H 5 O. The sodium ions re ignored becuse they re not involved in the buffer. Clculte K from pk nd set up the problem with rection tble. K 10 pk x Concentrtion (M) C 6 H 5 OH(q) H 2 O(l) C 6 H 5 O (q) H O (q) Initil Chnge x x x Equilibrium 1.2 x 1. x x Assume tht x is negligible with respect to both 1.0 nd 1.2 becuse both concentrtions re much lrger thn K. HO CHO 6 5 K [ x][ 1. x] x ][ 1. ] CHOH 6 5 [ 1.2 x] [ 1.2] 1.0 x [H O 6 5 ] K CHOH ( 1.0 x 10 ) CHO M Check ssumption: percent error ( x / 1.2)100% 7.7 x 10 9 %. The ssumption is vlid. ph log ( x ) Verify the ph using the Henderson-Hsselblch eqution: [bse] ph pk log [cid] - [C6H5O ] [1.] ph log [C 6 H 5 OH] log [1.2] ph K 10 pk x (unrounded) The cid component is H BO nd the bse component is H 2 BO. Neglect N. Assume x nd x re negligible. HO HBO 2 K [ x][ 0.82 x] [ x ][ 0.82 ] x 10 HBO [ 0.12 x] [ 0.12] 10 (unrounded) [H O ] K HBO HBO ( x ) (0.12 / 0.82) x M (unrounded) 2 Check ssumption: percent error ( x / 0.12)100% 7.0 x 10 8 %. The ssumption is vlid. ph log [H O ] log ( x ) Verify the ph using the Henderson-Hsselblch eqution. 19-4

5 19.19 The buffer components re phenol, NH, nd mmonium ion, NH 4. The chloride ions re ignored becuse they re not involved in the buffer. Clculte K b from pk b nd set up the problem with rection tble. K b 10 pkb x 10 5 (unrounded) Concentrtion (M) NH (q) H 2 O(l) NH 4 (q) OH (q) Initil Chnge x x x Equilibrium 0.25 x 0.15 x x Assume tht x is negligible with respect to both 0.25 nd 0.15 becuse both concentrtions re much lrger thn K b. K b 10 pkb x 10 5 (unrounded) NH 4 OH K b [ 0.15 x ] OH [ 0.15 ] OH x 10 NH [ 0.25 x] [ 0.25] 5 (unrounded) [OH ] K NH b ( x 10 ) NH x 10 5 M (unrounded) Check ssumption: percent error ( x 10 5 / 0.25)100% 0.012%. The ssumption is vlid. poh log [OH ] log ( x 10 5 ) (unrounded) ph poh ph poh Verify the ph using the Henderson-Hsselblch eqution. To do this, you must find the pk of the cid NH 4 : 14 pk pk b pk 14 pk b [bse] ph pk log [cid] [NH ] [0.25] ph 9.25 log [NH 9.25 log 4 ] [0.15] ph K b 10 pkb x 10 4 (unrounded) The bse component is CH NH 2 nd the cid component is CH NH. Neglect Cl. Assume x nd x re negligible. CHNH OH K b [ 0.60 x ] OH [ 0.60 ] OH x 10 4 CHNH2 [ 0.50 x] [ 0.50] [OH 2 ] K CH NH b ( x 10 4 ) (0.50 / 0.60) x 10 4 M (unrounded) CHNH Check ssumption: percent error ( x 10 4 / 0.50)100% 0.074%. The ssumption is vlid. poh log [OH ] log ( x 10 4 ) (unrounded) ph poh Verify the ph using the Henderson-Hsselblch eqution ) The buffer components re HCO from the slt KHCO nd CO 2 from the slt K 2 CO. Choose the K vlue tht corresponds to the equilibrium with these two components. K 1 refers to crbonic cid, H 2 CO losing one proton to produce HCO. This is not the correct K becuse H 2 CO is not involved in the buffer. K 2 is the correct K to choose becuse it is the equilibrium constnt for the loss of the second proton to produce CO 2 from HCO. b) Set up the rection tble nd use K 2 to clculte ph. Concentrtion (M) HCO (q) H 2 O(l) CO 2 (q) H O (q) Initil Chnge x x x Equilibrium 0.22 x 0.7 x x 19-5

6 Assume tht x is negligible with respect to both 0.22 nd 0.7 becuse both concentrtions re much lrger thn K. 2 HO CO K [ x][ 0.7 x] [ x ][ 0.7 ] 4.7 x 10 HCO [ 0.22 x] [ 0.22] 11 HCO [H O ] K ( 4.7 x 10 ) 2 CO x 1011 M (unrounded) Check ssumption: percent error ( x / 0.22)100% 1. x 10-8 %. The ssumption is vlid. ph [H O ] log ( x ) Verify the ph using the Henderson-Hsselblch eqution. [bse] ph pk log pk log(4.7 x ) [cid] 2 ph log [CO ] [0.7] [HCO ] log [0.22] ph ) The conjugte cid-bse pir re relted by K 2 (6. x 10 8 ). b) Assume tht x is negligible with respect to both 0.50 nd 0.40 becuse both concentrtions re much lrger thn K. The cid component is H 2 PO 4 nd the bse component is HPO 2 4. Neglect N. Assume x nd x re negligible. 2 HO HPO 4 K [ x][ 0.40 x] [ x ][ 0.40 ] 6. x 10 HPO [ 0.50 x] [ 0.50] HPO [H O 2 4 ] K (6. x 10 8 ) (0.50 / 0.40) x 10 8 M (unrounded) 2 HPO 4 Check ssumption: percent error (7.875 x 10 8 / 0.50)100% 1.6 x 10 5 %. The ssumption is vlid. ph log [H O ] log (7.875 x 10 8 ) Verify the ph using the Henderson-Hsselblch eqution Given the ph nd K of n cid, the buffer-component rtio cn be clculted from the Henderson-Hsselblch eqution. pk log K log (1. x 10 5 ) (unrounded) [bse] ph pk log [cid] [Pr ] log [HPr] [Pr ] log Rise ech side to 10 x. [HPr] [Pr ] [HPr]

7 19.24 Given the ph nd K of n cid, the buffer-component rtio cn be clculted from the Henderson-Hsselblch eqution. pk log K log (7.1 x 10 4 ) (unrounded) [bse] ph pk log [cid] [NO 2 ] log [HNO 2 ] [NO 2 ] log Rise ech side to 10 x. [HNO 2 ] [NO 2 ] [HNO ] Given the ph nd K of n cid, the buffer-component rtio cn be clculted from the Henderson-Hsselblch eqution. pk log K log (2. x 10 9 ) [bse] ph pk log [cid] [BrO ] log [HBrO] [BrO ] log Rise ech side to 10 x. [HBrO] [BrO ] [HBrO] Given the ph nd K of n cid, the buffer-component rtio cn be clculted from the Henderson-Hsselblch eqution. pk log K log (1.8 x 10 5 ) (unrounded) [bse] ph pk log [cid] [CH CO O ] log [CH COOH] [CHCOO ] log [CH COOH] Rise ech side to 10 x. [CHCOO ] [CH COOH] Determine the pk of the cid from the concentrtions of the conjugte cid nd bse, nd the ph of the solution. This requires the Henderson-Hsselblch eqution. [bse] ph pk log [cid] [A ].5 pk log [HA].5 pk pk [0.1500] pk log [0.2000] 19-7

8 Determine the moles of conjugte cid (HA) nd conjugte bse (A ) using (M)(V) moles. Moles HA ( L) ( mol HA / L) mol HA Moles A ( L) ( mol A / L) mol A The rection is: HA(q) NOH(q) N (q) A (q) H 2 O(l) Initil: mol mol mol Chnge: mol mol mol Finl: mol 0 mol mol NOH is the limiting regent. The ddition of mol NOH produces n dditionl mol A nd consumes mol of HA. Then: [A mol A ] 0.15 M A L mol HA [HA] M HA L [A ] ph pk log [HA] [0.15] ph log [0.197] Determine the pk of the cid from the concentrtions of the conjugte cid nd bse nd the ph of the solution. This requires the Henderson-Hsselblch eqution. [bse] ph pk log [cid] [B] [0.40] 8.88 pk log [BH pk log ] [0.25] 8.88 pk pk Determine the moles of conjugte cid (BH ) nd conjugte bse (B) using (M)(V) moles Moles BH (0.25 L) (0.25 mol BH / L) mol BH Moles B (0.25 L) (0.40 mol B / L) 0.10 mol B The rection is: B(q) HCl(q) BH (q) Cl (q) H 2 O(l) Initil: 0.10 mol mol mol Chnge: mol mol mol Finl: mol 0 mol mol HCl is the limiting regent. The ddition of mol HCl produces n dditionl mol BH nd consumes mol of B. Then: mol B [B] 0.92 M B 0.25 L [BH mol BH ] M BH 0.25 L [B] ph pk log [BH ] [0.92] ph log [0.258] 19-8

9 19.29 Determine the pk of the cid from the concentrtions of the conjugte cid nd bse nd the ph of the solution. This requires the Henderson-Hsselblch eqution. [bse] ph pk log [cid] - [Y ] [0.220] 8.77 pk log pk log [HY] [0.110] 8.77 pk pk Determine the moles of conjugte cid (HY) nd conjugte bse (Y ) using (M)(V) moles. Moles HY (0.50 L) (0.110 mol HY / L) mol HY Moles Y (0.50 L) (0.220 mol Y / L) mol Y ) The rection is: 2 HY(q) B(OH) 2 (q) B 2 (q) 2 Y (q) 2 H 2 O(l) Initil: mol mol mol Chnge: mol mol mol Finl: mol 0 mol mol B(OH) 2 is the limiting regent. The ddition of mol B(OH) 2 will produce 2 x mol Y nd consume 2 x mol of HY. Then: [Y mol Y ] M Y 0.50 L mol HY [HY] M HY 0.50 L [Y ] ph pk log [HY] [ ] ph log [ ] Determine the pk of the cid from the concentrtions of the conjugte cid nd bse nd the ph of the solution. This requires the Henderson-Hsselblch eqution. [bse] ph pk log [cid] [B] [1.05] 9.50 pk log [BH pk log ] [0.750] 9.50 pk pk Determine the moles of conjugte cid (BH ) nd conjugte bse (B). Moles BH (0.500 L) (0.750 mol BH / L) 0.75 mol BH (unrounded) Moles B (0.500 L) (1.05 mol B / L) mol B The rection is: B(q) HCl(q) BH (q) Cl (q) H 2 O(l) Initil: mol mol 0.75 mol Chnge: mol mol mol Finl: mol 0 mol 0.80 mol HCl is the limiting regent. The ddition of mol HCl will produce mol BH nd consume mol of B. 19-9

10 Then [B] [BH ] mol B L 0.80 mol BH L ph pk log [BH ] 1.04 M B [B] M BH [1.04] ph log [ 0.760] 19.1 ) The hydrochloric cid will rect with the sodium cette, NC 2 H O 2, to form cetic cid, HC 2 H O 2 : HCl NC 2 H O 2 HC 2 H O 2 NCl Clculte the number of moles of HCl nd NC 2 H O 2. All of the HCl will be consumed to form HC 2 H O 2, nd the number of moles of C 2 H O 2 will decrese mol HCl 10 L Initil moles HCl ( 204 ml L ) mol HCl (unrounded) 1mL mol NC 2H O 2 Initil moles NC 2 H O 2 ( L) mol NC 2 H O 2 L HCl NC 2 H O 2 HC 2 H O 2 NCl Initil: mol mol 0 mol Chnge: mol mol mol Finl: 0 mol mol mol Totl volume L (204 ml) (10 L / 1 ml) L mol [HC 2 H O 2 ] M (unrounded) L [C 2 H O mol 2 ] M (unrounded) L pk log K log (1.8 x 10 5 ) [C2HO 2 ] ph pk log [HC 2 H O 2 ] [ ] ph log [ ] b) The ddition of bse would increse the ph, so the new ph is ( ) The new [C 2 H O 2 ] / [ HC 2 H O 2 ] rtio is clculted using the Henderson-Hsselblch eqution. - [C2HO 2 ] ph pk log [HC 2 H O 2 ] - [C2HO 2 ] log [HC 2 H O 2 ] - [C2HO 2 ] log [HC 2 H O 2 ] [C2HO 2 ] (unrounded) [HC H O ]

11 From prt (), we know tht [HC 2 H O 2 ] [C 2 H O 2 ] ( M M) M. Although the rtio of [C 2 H O 2 ] to [HC 2 H O 2 ] cn chnge when cid or bse is dded, the bsolute mount does not chnge unless cetic cid or n cette slt is dded. Given tht [C 2 H O 2 ] / [ HC 2 H O 2 ] nd [HC 2 H O 2 ] [C 2 H O 2 ] M, solve for [C 2 H O 2 ] nd substitute into the second eqution. [C 2 H O 2 ] [HC 2 H O 2 ] nd [HC 2 H O 2 ] [HC 2 H O 2 ] M [HC 2 H O 2 ] M nd [C 2 H O 2 ] M. Moles of C 2 H O 2 needed ( mol C 2 H O 2 / L) (0.500 L) mol (unrounded) Moles of C 2 H O 2 initilly ( mol C 2 H O 2 / L) (0.500 L) mol (unrounded) This would require the ddition of ( mol mol) mol C 2 H O 2 (unrounded) The KOH dded rects with HC 2 H O 2 to produce dditionl C 2 H O 2 : HC 2 H O 2 KOH C 2 H O 2 K H 2 O(l) To produce mol C 2 H O 2 would require the ddition of mol KOH. Mss KOH ( mol KOH g KOH ) 1 mol KOH g KOH 19.2 ) The sodium hydroxide will rect with the sodium bicrbonte, NHCO, to form crbonte ion, CO 2 : NOH NHCO 2 N CO 2 H 2 O Clculte the number of moles of NOH nd NHCO. All of the NOH will be consumed to form CO 2, nd the number of moles of NHCO will decrese. The HCO is the importnt prt of NHCO mol NOH 10 L Initil moles NOH ( 10.7 ml L 1mL ) mol NOH (unrounded) Initil moles HCO mol NHCO 1 mol HCO 10 L ( 50.0 ml ) L 1mol NHCO 1mL mol HCO NOH NHCO 2 N 2 CO H 2 O Initil: mol mol 0 mol Chnge: mol mol mol Finl: 0 mol mol mol Totl volume (50.0 ml 10.7 ml) (10 L / 1 ml) L [HCO mol ] L M (unrounded) [CO mol ] L M (unrounded) pk log K log (4.7 x ) [CO ] ph pk log [HCO - ] [ ] ph log [ ]

12 b) The ddition of cid would decrese the ph, so the new ph is ( ) The new [CO 2 ] / [HCO ] rtio is clculted using the Henderson-Hsselblch eqution. 2 ph pk log [CO ] [HCO ] log [CO ] [HCO ] 2 [CO ] log [HCO ] 2 [CO ] (unrounded) [HCO ] From prt (), we know tht [HCO ] [CO 2 ] ( M M) M. Although the rtio of [CO 2 ] to [HCO ] cn chnge when cid or bse is dded, the bsolute mount does not chnge unless cetic cid or n cette slt is dded. Given tht [CO 2 ] / [ HCO ] nd [HCO ] [CO 2 ] M, solve for [CO 2 ] nd substitute into the second eqution. [CO 2 ] [HCO ] nd [HCO ] [HCO ] M [HCO ] M nd [CO 2 ] M (unrounded) Moles of HCO needed ( mol HCO / L) (10 L / 1 ml) (25.0 ml) mol (unrounded) Moles of HCO initilly ( mol HCO / L) (10 L / 1 ml) (25.0 ml) mol (unrounded) This would require the ddition of ( mol mol) mol HCO (unrounded) The HCl dded rects with CO 2 to produce dditionl HCO : 2 CO HCl HCO Cl To produce mol HCO would require the ddition of mol HCl g HCl Mss HCl ( mol HCl) 1 mol HCl g HCl 19. Select conjugte pirs with K vlues close to the desired [H O ]. ) For ph 4.5, [H O ] x 10 5 M. Some good selections re the HOOC(CH 2 ) 4 COOH/ HOOC(CH 2 ) 4 COOH conjugte pir with K equl to.8 x 10 5 or C 6 H 5 CH 2 COOH/C 6 H 5 CH 2 COO conjugte pir with K equl to 4.9 x From the bse list, the C 6 H 5 NH 2 / C 6 H 5 NH conjugte pir comes close with K 1.0 x / 4.0 x x b) For ph 7.0, [H O ] x 10 7 M. Two choices re the H 2 PO 4 / HPO 4 2 conjugte pir with K of 6. x 10 8 nd the H 2 AsO 4 / HAsO 4 2 conjugte pir with K of 1.1 x Select conjugte pirs tht hve K or K b vlues close to the desired [H O ] or [OH ]. ) For [H O ] 1 x 10 9 M, the HOBr / OBr conjugte pir comes close with K equl to 2. x From the bse list, K b 1.0 x / 1 x x 10 5, the NH / NH 4 conjugte pir comes close with K b 1.76 x b) For [OH ] x 10 5 M, the NH / NH 4 conjugte pir comes close; lso, it is possible to choose K 1.0 x / x x 10 10, the C 6 H 5 OH / C 6 H 5 O comes close with K 1.0 x Select conjugte pirs with pk vlues close to the desired ph. Convert ph to [H O ] for esy comprison to K vlues. Determine n pproprite bse by [OH ] K w / [H O ]. ) For ph.5 ([H O ] 10 ph x 10 4 ), the best selection is the HOCH 2 CH(OH)COOH/ HOCH 2 CH(OH)COOH conjugte pir with K 2.9 x The CH COOC 6 H 4 COOH / CH COOC 6 H 4 COO pir, with K.6 x 10 4, is lso good choice. The [OH ] 1.0 x /.2 x x 10 11, results in no resonble K b vlues from the ppendix

13 b) For ph 5.5 ([H O ] 10 ph x 10 6 ), no K 1 gives n cceptble pir; the K 2 vlues for dipic cid, mlonic cid, nd succinic cid re resonble. The [OH ] 1.0 x / x 10 6 x 10 9, the K b selection is C 5 H 5 N / C 5 H 5 NH Select conjugte pirs tht hve K or K b vlues close to the desired [H O ] or [OH ]. ) For [OH ] 1 x 10 6 M, no K b vlues work. The K vlues re [H O ] K w / [OH ] 1.0 x / 1 x x 10 8, giving the following cceptble pirs H 2 PO 4 / HPO 4 2 or HC 6 H 5 O 7 2 / C 6 H 5 O 7 or HOCl / OCl. b) For [H O ] 4 x 10 4 M, the HF / F conjugte pir comes close with K equl to 6.8 x From the bse list, K b 1.0 x / 4 x x 10 11, there re no resonble choices The vlue of the K from the ppendix: K 2.9 x 10 8 pk log 2.9 x (unrounded) Use the Henderson-Hsselblch eqution to determine the ph. [ClO ] ph pk log [HClO] [0.100] ) ph log [0.100] [0.150] b) ph log [0.100] [0.100] c) ph log [0.150] d) The rection is NOH HClO N ClO H 2 O. The originl moles of HClO nd OCl re both (0.100 mol / L) (L) mol NOH HClO N ClO H 2 O Initil: mol mol mol Chnge: mol mol mol Finl: 0 mol mol mol ph [0.105] log [0.095] 19.8 The vlue of the K from the ppendix: K 6. x 10 8 (We re using K 2 since we re deling with the equilibrium in which the second hydrogen ion is being lost). Determine the pk using pk log 6. x (unrounded). Use the Henderson-Hsselblch eqution: 2 ph pk log [HPO 4 ] [H PO ] log [HPO 4 ] [H PO ] log [HPO ] [H PO ] [HPO 4 ] [H PO ] You need to know the pk vlue for the indictor. (Its trnsition rnge is pproximtely pk ± 1.) If the indictor is diprotic cid, it will hve two trnsition rnges, one for ech of the two H O ions lost. 19-1

14 19.40 To see distinct color in mixture of two colors, you need one color to be bout 10 times the intensity of the other. For this to tke plce, the concentrtion rtio [HIn] / [In ] needs to be greter thn 10:1 or less thn 1:10. This will occur when ph pk 1 or ph pk 1, respectively, giving trnsition rnge of bout two units This is becuse the concentrtion of indictor is very smll The equivlence point in titrtion is the point t which the number of moles of OH equls the number of moles of H O (be sure to ccount for stoichiometric rtios, e.g.,1 mol of C(OH) 2 produces 2 moles of OH ). The endpoint is the point t which the dded indictor chnges color. If n pproprite indictor is selected, the endpoint is close to the equivlence point, but not normlly the sme. Using n indictor tht chnges color t ph fter the equivlence point mens the equivlence point is reched first. However, if n indictor is selected tht chnges color t ph before the equivlence point, then the endpoint is reched first ) The rections re: OH (q) H PO 4 (q) H 2 PO 4 (q) H 2 O(l) K x 10 OH (q) H 2 PO 4 (q) HPO 2 4 (q) H 2 O(l) K 2 6. x 10 8 The correct order is C, B, D, A. Scene C shows the solution before the ddition of ny NOH. Scene B is hlf-wy to the first equivlence point; Scene D is hlfwy to the second equivlence point nd Scene A is t end of the titrtion. b) Scene B is the second scene in the correct order. This is hlfwy towrds the first equivlence point when there re equl mounts of the cid nd conjugte bse, which constitutes buffer. [H2PO 4 ] ph pk log. [H PO 4 ] Determine the pk using pk log 7.2 x [] ph log [] c) ml of NOH is required to rech the first hlf-equivlence point. Therefore, n dditionl ml of NOH is required to rech the first equivlence point, for totl of 20 ml for the first equivlence point. An 2 dditionl ml of NOH will be required to rech the second equivlence point where only HPO 4 remins. A totl of ml of NOH is required to rech Scene A ) The initil ph is lowest for flsk solution of the strong cid, followed by the wek cid nd then the wek bse. In other words, strong cid-strong bse < wek cid-strong bse < strong cid-wek bse. b) At the equivlence point, the moles of H O equl the moles of OH, regrdless of the type of titrtion. However, the strong cid-strong bse equivlence point occurs t ph 7.00 becuse the resulting ction-nion combintion does not rect with wter. An exmple is the rection NOH HCl H 2 O NCl. Neither N nor Cl ions dissocite in wter. The wek cid-strong bse equivlence point occurs t ph > 7, becuse the nion of the wek cid is wekly bsic, wheres the ction of the strong bse does not rect with wter. An exmple is the rection HCOOH NOH HCOO H 2 O N. The conjugte bse, HCOO, rects with wter ccording to this rection: HCOO H 2 O HCOOH OH. The strong cid-wek bse equivlence point occurs t ph < 7, becuse the nion of the strong cid does not rect with wter, wheres the ction of the wek bse is wekly cidic. An exmple is the rection HCl NH NH 4 Cl. The conjugte cid, NH 4, dissocites slightly in wter: NH 4 H 2 O NH H O. In rnk order of ph of equivlence point, strong cid -wek bse < strong cid-strong bse < wek cid -strong bse In the buffer region, comprble mounts of wek cid nd its conjugte bse re present. At the equivlence point, the predominnt species is the conjugte bse. In strong cid-wek bse titrtion, the wek bse nd its conjugte cid re the predominnt species present At the very center of the buffer region of wek cid-strong bse titrtion, the concentrtion of the wek cid nd its conjugte bse re equl, which mens tht t this point the ph of the solution equls the pk of the wek cid

15 19.47 The titrtion curve for diprotic cid hs two breks i.e., two regions where the ph increses shrply. For monoprotic cid, only one brek occurs Indictors hve ph rnge tht is pproximted by pk ± 1. The pk of cresol red is log (.5 x 10 9 ) 8.5, so the indictor chnges color over n pproximte rnge of 7.5 to Indictors hve ph rnge tht is pproximted by pk ± 1. The pk of ethyl red is log (.8 x 10 6 ) 5.42, so the indictor chnges color over n pproximte rnge of 4.4 to Choose n indictor tht chnges color t ph close to the ph of the equivlence point. ) The equivlence point for strong cid-strong bse titrtion occurs t ph 7.0. Bromthymol blue is n indictor tht chnges color round ph 7. b)the equivlence point for wek cid strong bse is bove ph 7. Estimte the ph t equivlence point from equilibrium clcultions. At the equivlence point, ll of the HCOOH nd NOH hve been consumed; the solution is M HCOO. (The volume doubles becuse equl volumes of bse nd cid re required to rech the equivlence point. When the volume doubles, the concentrtion is hlved.) The wek bse HCOO undergoes bse rection: Concentrtion, M COOH (q) H 2 O(l) COOH(q) OH (q) Initil: M 0 0 Chnge: x x x Equilibrium: x x x The K for HCOOH is 1.8 x 10 4, so K b 1.0 x / 1.8 x x [ ][ ] (unrounded) HCOOH OH [ x][ x] K b [ HCOO ] [ x ][ x ] x 10 [ x] [ 0.050] 11 [OH ] x x 10 6 M poh log ( x 10 6 ) (unrounded) ph poh Choose thymol blue or phenolphthlein ) Determine the K (of the conjugte cid) from the K b for CH NH 2. K K w / K b (1.0 x ) / (4.4 x 10 4 ) x (unrounded) An cid bse titrtion of two components of equl concentrtion nd t 1:1 rtio gives solution of the conjugtes with hlf the concentrtion. In this cse, the concentrtion of CH NH M. HO CHNH K x x x CHNH x x [H O ] x 10 6 M (unrounded) ph log [H O ] log ( x 10 6 ) Either methyl red or lizrin is cceptble. b) This is strong cid-strong bse titrtion; thus, the equivlence point is t ph The best choice would be bromthymol blue; lizrin might be cceptble ) The equivlence point for wek bse strong cid is below ph 7. Estimte the ph t equivlence point from equilibrium clcultions. At the equivlence point, the solution is 0.25 M (CH ) 2 NH 2. (The volume doubles becuse equl volumes of bse nd cid re required to rech the equivlence point. When the volume doubles, the concentrtion is hlved.) K K w / K b (1.0 x ) / (5.9 x 10 4 ) x (unrounded) Concentrtion, M (CH ) 2 NH 2 (q) H 2 O(l) (CH ) 2 NH (q) H O (q) Initil: 0.25 M 0 0 Chnge: x x x Equilibrium: 0.25 x x x 19-15

16 HO (CH) 2NH 2 2 K x x x (CH ) 2NH 0.25 x x [H O ] x 10 6 M (unrounded) ph log [H O ] log ( x 10 6 ) Methyl red is n indictor tht chnges color round ph 5.7. b) This is strong cid strong bse titrtion; thus, the equivlence point is t ph Bromthymol blue is n indictor tht chnges color round ph ) Determine the K b (of the conjugte bse) from the K for C 6 H 5 COOH. K b K w / K (1.0 x ) / (6. x 10 5 ) x (unrounded) An cid-bse titrtion of two components of equl concentrtion nd t 1:1 rtio gives solution of the conjugtes with hlf the concentrtion. In this cse, the concentrtion of C 6 H 5 COO M (unrounded). CHCOOH 6 5 OH K b [ x][ x] [ x ][ x ] x CHCOO [ x] [ ] 6 5 [OH - ] x x 10 6 M poh log ( x 10 6 ) (unrounded) ph poh The choices re phenolphthlein or thymol blue. b) The titrtion will produce 0.25 M NH solution t the equivlence point. Use the K b for NH from the Appendix. NH 4 OH K b [ x][ x] [ x ][ x ] 1.76 x 10 5 NH [ 0.25 x] [ 0.25 ] [OH ] x x 10 M poh log ( x 10 ) (unrounded) ph poh The best choice would be lizrin yellow R; lizrin might be cceptble The rection occurring in the titrtion is the neutrliztion of H O (from HCl) by OH (from NOH): HCl(q) NOH(q) H 2 O(l) NCl(q) or, omitting specttor ions: H O (q) OH (q) 2 H 2 O(l) For the titrtion of strong cid with strong bse, the ph before the equivlence point depends on the excess concentrtion of cid nd the ph fter the equivlence point depends on the excess concentrtion of bse. At the equivlence point, there is not n excess of either cid or bse so the ph is 7.0. The equivlence point occurs when ml of bse hs been dded. Use (M)(V) to determine the number of moles. The initil number of moles of HCl ( mol HCl / L) (10 L / 1 ml) (40.00 ml) x 10 mol HCl ) At 0 ml of bse dded, the concentrtion of hydronium ion equls the originl concentrtion of HCl. ph log ( M) b) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (25.00 ml) x 10 mol NOH HCl(q) NOH(q) H 2 O(l) NCl(q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HCl is (1.500 x 10 mol HCl) / ( L) M (unrounded) ph log (0.0207) (Note tht the NCl product is neutrl slt tht does not ffect the ph)

17 c) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (9.00 ml).900 x 10 mol NOH HCl(q) NOH(q) H 2 O(l) NCl(q) Initil: x 10 mol x 10 mol 0 Chnge:.900 x 10 mol.900 x 10 mol.900 x 10 mol Finl: x 10 4 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HCl is (1.00 x 10 4 mol HCl) / ( L) M (unrounded) ph log ( ) d) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (9.90 ml).990 x 10 mol NOH HCl(q) NOH(q) H 2 O(l) NCl(q) Initil: x 10 mol.990 x 10 mol 0 Chnge:.990 x 10 mol.990 x 10 mol.990 x 10 mol Finl: x 10 5 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HCl is (1.0 x 10 5 mol HCl) / ( L) M (unrounded) ph log ( ).90 e) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (40.00 ml) x 10 mol NOH. HCl(q) NOH(q) H 2 O(l) NCl(q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 mol The NOH will rect with n equl mount of the cid nd 0.0 mol HCl will remin. This is the equivlence point of strong cid-strong bse titrtion, thus, the ph is Only the neutrl slt NCl is in solution t the equivlence point. f) The NOH is now in excess. It will be necessry to clculte the excess bse fter recting with the HCl. The excess strong bse will give the poh, which cn be converted to the ph. Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (40.10 ml) x 10 mol NOH The HCl will rect with n equl mount of the bse, nd 1.0 x 10 5 mol NOH will remin. HCl(q) NOH(q) H 2 O(l) NCl(q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 5 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess NOH is (1.0 x 10 5 mol NOH) / ( L) M (unrounded) poh log ( ).906 (unrounded) ph poh g) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (50.00 ml) x 10 mol NOH The HCl will rect with n equl mount of the bse, nd x 10 mol NOH will remin. HCl(q) NOH(q) H 2 O(l) NCl(q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess NOH is (1.000 x 10 mol NOH) / ( L) M (unrounded) poh log ( ) (unrounded) ph poh

18 19.55 The rection occurring in the titrtion is the neutrliztion of OH (from KOH) by H O (from HBr): HBr(q) KOH(q) H 2 O(l) KBr(q) H O (q) OH (q) 2 H 2 O(l) For the titrtion of strong bse with strong cid, the ph before the equivlence point depends on the excess concentrtion of bse nd the ph fter the equivlence point depends on the excess concentrtion of cid. At the equivlence point, there is not n excess of either cid or bse so ph is 7.0. The equivlence point occurs when 0.00 ml of cid hs been dded. The initil number of moles of KOH ( mol KOH / L) (10 L / 1 ml) (0.00 ml).000 x 10 mol KOH ) At 0 ml of cid dded, the concentrtion of hydroxide ion equls the originl concentrtion of KOH. poh log ( M) ph poh b) Determine the moles of HBr dded: Moles of HBr ( mol HBr / L) (10 L / 1 ml) (15.00 ml) x 10 mol HBr The HBr will rect with n equl mount of the bse, nd x 10 mol KOH will remin. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess KOH is (1.500 x 10 mol KOH) / ( L) 0.0 M (unrounded) poh log (0.0) ph poh c) Determine the moles of HBr dded: Moles of HBr ( mol HBr / L) (10 L / 1 ml) (29.00 ml) x 10 mol HBr The HBr will rect with n equl mount of the bse, nd 1.00 x 10 4 mol KOH will remin. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess KOH is (1.00 x 10 4 mol KOH) / ( L) M (unrounded) poh log ( ) ph poh d) Determine the moles of HBr dded: Moles of HBr ( mol HBr / L) (10 L / 1 ml) (29.90 ml) x 10 mol HBr The HBr will rect with n equl mount of the bse, nd 1.0 x 10 5 mol KOH will remin. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess KOH is (1.0 x 10 5 mol KOH) / ( L) M (unrounded) poh log ( ) ph poh e) Determine the moles of HBr dded: Moles of HBr ( mol HBr / L) (10 L / 1 ml) (0.00 ml).000 x 10 mol HBr The HBr will rect with n equl mount of the bse nd 0.0 mol KOH will remin. This is the equivlence point of strong cid strong bse titrtions; thus, the ph is f) The HBr is now in excess. It will be necessry to clculte the excess bse fter recting with the HCl. The excess strong cid will give the ph. Determine the moles of HBr dded: Moles of HBr ( mol HBr / L) (10 L / 1 ml) (0.10 ml).010 x 10 mol HBr The HBr will rect with n equl mount of the bse, nd 1.0 x 10 5 mol HBr will remin. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HBr is (1.0 x 10 5 mol HBr) / ( L) M (unrounded) ph log ( ) g) Determine the moles of HBr dded: Moles of HBr ( mol HBr / L) (10 L / 1 ml) (40.00 ml) x 10 mol HBr The HBr will rect with n equl mount of the bse, nd x 10 mol HBr will remin. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HBr is (1.000 x 10 mol HBr) / ( L) M (unrounded) ph log ( )

19 19.56 This is titrtion between wek cid nd strong bse. The ph before ddition of the bse is dependent on the K of the cid (lbeled HBut). Prior to reching the equivlence point, the dded bse rects with the cid to form butnote ion (lbeled But ). The equivlence point occurs when ml of bse is dded to the cid becuse t this point, moles cid moles bse. Addition of bse beyond the equivlence point is simply the ddition of excess OH. The initil number of moles of HBut (M)(V) ( mol HBut / L) (10 L / 1 ml) (20.00 ml) x 10 mol HBut ) At 0 ml of bse dded, the concentrtion of [H O ] is dependent on the dissocition of butnoic cid: HBut H 2 O H O But Initil: M 0 Chnge: x x Equilibrium: x x K HO But 2 x [ HBut] x 2 x x 10 5 x [H O ] x 10 M (unrounded) ph log [H O ] log ( x 10 ) b) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (10.00 ml) x 10 mol NOH The NOH will rect with n equl mount of the cid, nd x 10 mol HBut will remin. An equl number of moles of But will form. HBut(q) NOH(q) H 2 O(l) But (q) N (q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HBut is (1.000 x 10 mol HBut) / ( L) 0.0 M (unrounded) The molrity of the But formed is (1.000 x 10 mol But ) / ( L) 0.0 M (unrounded) Using rection tble for the equilibrium rection of HBut: HBut H 2 O H O But Initil: 0.0 M 0.0 M Chnge: x x Equilibrium: 0.0 x 0.0 x K HO But [ HBut] x( 0.0 x) 0.0 x x ( 0.0) x 10 5 x [H O ] 1.54 x 10 5 M (unrounded) ph log [H O ] log (1.54 x 10 5 ) c) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (15.00 ml) x 10 mol NOH The NOH will rect with n equl mount of the cid, nd 5.00 x 10 4 mol HBut will remin, nd x 10 moles of But will form. HBut(q) NOH(q) H 2 O(l) But (q) N (q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 4 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HBut is (5.00 x 10 4 mol HBut) / ( L) M (unrounded) The molrity of the But formed is (1.500 x 10 mol But ) / ( L) M (unrounded) Using rection tble for the equilibrium rection of HBut: HBut H 2 O H O But Initil: M M Chnge: x x Equilibrium: x x 19-19

20 HO But K HBut [ ] x ( x) x x ( ) x 10 5 x [H O ] 5.1 x 10 6 M (unrounded) ph log [H O ] log (5.1 x 10 6 ) d) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (19.00 ml) x 10 mol NOH The NOH will rect with n equl mount of the cid, nd 1.00 x 10 4 mol HBut will remin, nd x 10 moles of But will form. HBut(q) NOH(q) H 2 O(l) But (q) N (q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 4 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HBut is (1.00 x 10 4 mol HBut) / ( L) M (unrounded) The molrity of the But formed is (1.900 x 10 mol But ) / ( L) M (unrounded) Using rection tble for the equilibrium rection of HBut: HBut H 2 O H O But Initil: M M Chnge: x x Equilibrium: x x K HO But [ HBut] x ( x) x x ( ) x 10 5 x [H O ] x 10 7 M (unrounded) ph log [H O ] log ( x 10 7 ) e) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (19.95 ml) x 10 mol NOH The NOH will rect with n equl mount of the cid, nd 5 x 10 6 mol HBut will remin, nd x 10 moles of But will form. HBut(q) NOH(q) H 2 O(l) But (q) N (q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 6 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess HBut is (5 x 10 6 mol HBut) / ( L) M (unrounded) The molrity of the But formed is (1.995 x 10 mol But ) / ( L) M (unrounded) Using rection tble for the equilibrium rection of HBut: HBut H 2 O H O But Initil: M M Chnge: x x Equilibrium: x x K HO But [ HBut] x( x) x x ( ) x 10 5 x [H O ] x 10 8 M (unrounded) ph log [H O ] log ( x 10 8 ) f) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (20.00 ml) x 10 mol NOH The NOH will rect with n equl mount of the cid, nd 0 mol HBut will remin, nd x 10 moles of But will form. This is the equivlence point

21 HBut(q) NOH(q) H 2 O(l) But (q) N (q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 mol The K b of But is now importnt. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the But formed is (2.000 x 10 mol But ) / ( L) M (unrounded) K b K w / K (1.0 x ) / (1.54 x 10 5 ) x (unrounded) Using rection tble for the equilibrium rection of But : But H 2 O HBut OH Initil: M 0 0 Chnge: x x x Equilibrium: x x x [ ] K b HBut [ x][ x] [ x] [ x][ x] [ ] OH x But [OH ] x x 10 6 M poh log ( x 10 6 ) (unrounded) ph poh g) After the equivlence point, the excess strong bse is the primry fctor influencing the ph. Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (20.05 ml) x 10 mol NOH The NOH will rect with n equl mount of the cid, 0 mol HBut will remin, nd 5 x 10 6 moles of NOH will be in excess. There will be x 10 mol of But produced, but this wek bse will not ffect the ph compred to the excess strong bse, NOH. HBut(q) NOH(q) H 2 O(l) But (q) N (q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 6 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess OH is (5 x 10 6 mol OH ) / ( L) x 10 4 M (unrounded) poh log ( x 10 4 ).906 (unrounded) ph poh h) Determine the moles of NOH dded: Moles of NOH ( mol NOH / L) (10 L / 1 ml) (25.00 ml) x 10 mol NOH The NOH will rect with n equl mount of the cid, 0 mol HBut will remin, nd 5.00 x 10 4 moles of NOH will be in excess. HBut(q) NOH(q) H 2 O(l) But (q) N (q) Initil: x 10 mol x 10 mol 0 Chnge: x 10 mol x 10 mol x 10 mol Finl: x 10 4 mol x 10 mol The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess OH is (5.00 x 10 4 mol OH ) / ( L) x 10 2 M (unrounded) poh log ( x 10 2 ) (unrounded) ph poh This is titrtion between wek bse nd strong cid. The ph before ddition of the cid is dependent on the K b of the bse ((CH CH 2 ) N)). Prior to reching the equivlence point, the dded cid rects with bse to form (CH CH 2 ) NH ion. The equivlence point occurs when ml of cid is dded to the bse becuse t this point, moles cid moles bse. Addition of cid beyond the equivlence point is simply the ddition of excess H O. The initil number of moles of (CH CH 2 ) N ( mol (CH CH 2 ) N) / L) (10 L / 1 ml) (20.00 ml) x 10 mol (CH CH 2 ) N 19-21

22 ) Since no cid [( hs been ) dded, ][ only ] the wek bse (K b ) is importnt. CH CH NH OH [ x][ x] [ x][ x] 2 K b 5.2 x 10 [( CH CH ) N] 2 [ x] [ ] 4 [OH ] x x 10 M poh log ( x 10 ) (unrounded) ph poh b) Determine the moles of HCl dded: Moles of HCl ( mol HCl / L) (10 L / 1 ml) (10.00 ml) x 10 mol HCl The HCl will rect with n equl mount of the bse, nd x 10 mol (CH CH 2 ) N will remin; n equl number of moles of (CH CH 2 ) NH will form. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess (CH CH 2 ) N is (1.000 x 10 mol (CH CH 2 ) N) / ( L) 0.0 M (unrounded) The molrity of the (CH CH 2 ) NH formed is (1.000 x 10 mol (CH CH 2 ) NH ) / ( L) 0.0 M (unrounded) ( CHCH2) NH OH K b [ x ][ 0.0 x ] [ x ][ 0.0 ] 5.2 x 10 ( CHCH2) N [ 0.0 x] [ 0.0] 4 [OH ] x 5.2 x 10 4 M poh log (5.2 x 10 4 ) (unrounded) ph poh c) Determine the moles of HCl dded: Moles of HCl ( mol HCl / L) (10 L / 1 ml) (15.00 ml) x 10 mol HCl The HCl will rect with n equl mount of the bse, nd 5.00 x 10 4 mol (CH CH 2 ) N will remin; nd x 10 moles of (CH CH 2 ) NH will form. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess (CH CH 2 ) N is (5.00 x 10 4 mol (CH CH 2 ) N) / ( L) M (unrounded) The molrity of the (CH CH 2 ) NH formed is (1.500 x 10 mol (CH CH 2 ) NH ) / ( L) M (unrounded) ( CHCH2) NH OH K b [ x ][ x ] [ x ][ ] 5.2 x 10 ( CHCH2) N [ x] [ ] 4 [OH ] x 1.7 x 10 4 M poh log (1.7 x 10 4 ) (unrounded) ph poh d) Determine the moles of HCl dded: Moles of HCl ( mol HCl / L) (10 L / 1 ml) (19.00 ml) x 10 mol HCl The HCl will rect with n equl mount of the bse, nd 1.00 x 10 4 mol (CH CH 2 ) N will remin; nd x 10 moles of (CH CH 2 ) NH will form. The volume of the solution t this point is [( ) ml] (10 L / 1 ml) L The molrity of the excess (CH CH 2 ) N is (1.00 x 10 4 mol (CH CH 2 ) N) / ( L) M (unrounded) The molrity of the (CH CH 2 ) NH formed is (1.900 x 10 mol (CH CH 2 ) NH ) / ( L) M (unrounded) ( CHCH2) NH OH K b [ x][ x] [ x ][ ] 5.2 x 10 ( CHCH2) N [ x] [ ] 4 [OH ] x x 10 5 M poh log ( x 10 5 ) (unrounded) ph poh