Problem : Buffer solutions. The equilibrium, hich governs the concentrtion of H ithin the solution is HCOOH! HCOO H [HCOO ] 4 Hence. [HCOOH] nd since [HCOOH] 0.00 M nd [HCOO ] 0.50 M -4 0.00 4..8 M 0.50 nd ph.55.. Since sodium hydroxide rects ith formic cid: HCOOH OH HCOO H O the concentrtion of formic cid in the solution is reduced to [HCOOH] 0.00 M 0.00 M 0.90 M nd the concentrtion of formte is incresed to [HCOO ] 0.50 M 0.00 M 0.60 M -4 0.90 4 Therefore:..5 M 0.60 nd ph.60 Note tht the ddition of sodium hydroxide, hich is strong bse, cuses very smll increse of the ph of the solution.. Let the volume of the solution of sodium hydroxide. Therefore, the finl volume of the solution ill be (0.0 ml nd the number of mmol of CH COOH nd OH hich re mixed re 0.0 ml 0.50 mmol/ml 5.00 mmol nd ml 0.00 mmol/ml 0.00 mmol, respectively. From the rection: CH COOH OH CH COO H O it is obvious tht the mount of cette produced is 0.00 mmol nd the mount of cetic cid hich remins unrected is (5.00 0.00 mmol. Hence, the concentrtion of ech constituent in the buffer solution is: (5.00-0.00 [CH COOH] M 0.0 0.00 nd [CH COO ] M 0.0 From the cid dissocition constnt expression of cetic cid [CH COO ].8 [CH COOH] it cn be derived CH COO ] [CH COOH] [H [ 0.00 M.8 nd 0.0 (5.00-0.00 M.0 0.0 from hich 48. cm -. 4. 5. (i c, (ii b 6. (i b, (ii d 7. (i c, (ii c ] 54
8. (i b, (ii c Problem : Titrtions of ek cids The titrtion rection is CH COOH OH - CH COO - H O Initil ph The ph of the solution before the titrtion begins, is clculted by the cid dissocition constnt nd the initil concentrtion of CH COOH: CH COOH CH COO - H From the cid dissocition constnt expression: - [CHCOO ] -5. 8 [CHCOOH] the concentrtion of H cn be clculted: -5.8 0. 00. 4 M nd ph.87 b ph fter the ddition of.00 cm of titrnt The solution contins cetic cid nd sodium cette. Therefore it is buffer solution. The concentrtion of ech constituent is clculted: (50.00 cm 0.00 M - (.00 cm 0.00 M [CHCOOH] 0. 0667 M 60. 00 cm -. 00 cm 0.00 M [CH COO ] 0. 0667 M 60.00 cm These concentrtions re then substituted into the dissocition constnt expression of cetic cid for clculting the concentrtion of [H ]: 0.0667.8 7.0 M nd ph 4.4 0.0667 c ph t the equivlence point At the equivlence point, ll cetic cid hs been converted to sodium cette nd the ph is clculted from the hydrolysis of cette ions: CH COO - H O CH COOH OH - The volume of titrnt required for the equivlence point ( ep is clculted: 50.00 cm 0.00 M ep 50.00 cm 0.00 M nd the totl volume of solution is 0.0 cm. Therefore, t this point of the titrtion [CH COOH] [OH - ] nd - 50. 00 cm 0.00 M - [CH COO ] [OH ] 0. 0500 M 0.0 cm - 4 [ OH ].00 5.56 0.0500 M.8-6 [OH ] 0.0500 5.56 5.7 M poh 5.8 nd thus ph 4-5.8 8.7 d ph fter the ddition of 50. cm of titrnt At this stge, ll cetic cid hs been converted to sodium cette nd the ph of the solution is clculted by the excess of sodium hydroxide, hich hs been dded: - (50. cm 0.00 M - (50.00 cm 0.00 M 4 [OH ]. 0 M 0. cm Therefore poh 4.00 nd ph.00 55
ph 4 8 6 4 0 0 0 40 60 80 olume of NOH, cm Titrtion curve of 0.00 M cetic cid ith 0.00 M NOH e Selection of indictor Since the ph t the equivlence point is 8.7, the pproprite cid bse indictor is phenolphthlein.. (i b, (ii c, (iii, (iv b, (v c, (vi d Problem 4: Seprtion by extrction Strting ith n mount 0 of S in phse, fter the extrction this mount is distributed beteen the to phses s follos: 0 (C S (C S Since D (C S / (C S, e hve 0 (C S D(C S (D (C S Therefore, fter removing phse, the remining mount of S in phse is: (C S 0 D By repeting extrction ith fresh portion of volume of phse, the mount of S is similrly distributed. After removing phse, the remining mount of S in phse is: (C S 0 D D nd so on. Therefore fter n extrctions ith fresh portion of volume of phse, the remining mount of S in phse ill be: n n 0 D. ( The remining frction of S fter extrction ith 0 ml of chloroform is clculted using Eqution.4: 50 f 0. 0 50 0.5, therefore the percentge of S extrcted is 0.5 86.5% (b The remining frction of S fter 4 extrctions ith 5 ml of chloroform ech time is similrly clculted: 4 4 50 f 4 0. 5 50 0.0, 56
therefore the percentge of S extrcted is 0. 97.8%. The result is indictive of the fct tht successive extrctions ith smller individul volumes of extrctnt re more efficient thn single extrction ith ll the volume of the extrctnt.. Using Eqution -4 e hve: n 0.0 0.0 9.5 5.0 0.0 or 0.0 0.96 n hence n log(0.0/log(0.96.78, therefore t lest 4 extrctions re required. 4. The equilibri involved re represented schemticlly s follos: e hve (subscripts nd o denote concentrtions in queous nd orgnic phse, respectively (CHA o [HA] o D (CHA [HA] [A ] D [HA] o / [HA] nd [H ] [A ] / [HA] Combining ll three equtions e finlly obtin: D D (.5 Lst eqution predicts tht if [H ] >> (strongly cidic queous phse, then D D (i.e. D cquires the mximum possible vlue nd the cid is extrcted (or prefers to sty in the orgnic phse. On the other hnd, if [H ] << (strongly lkline queous phse, e hve D D [H ] /, nd becuse of the smll vlue of D the cid is then extrcted (or prefers to sty in the queous phse. In this y, by regulting the ph of the queous phse, the course of extrction is shifted tords the desired direction. 5. ( By using the previously derived Eqution.5, e obtin the folloing plots of the D/ D rtio vs. ph. (b From these plots it is cler tht t the ph region 7-8 the distribution rtio for benzoic cid ill be prcticlly 0, heres tht of phenol ill cquire the mximum possible vlue. Therefore, phenol cn be efficiently seprted from n queous solution of both compounds by extrction ith diethylether, provided tht the ph of this solution hs been djusted in the rnge 7 to 8 (e.g. by the presence of excess of NHCO. 57
6. ( The equilibri involved re represented schemticlly s follos: e hve the expressions (COxH o o D (COxH [OxH ] D o / 70 [OxH ] [Ox [Ox ] Combining ll four equtions, e hve the sought-for expression o D [OxH ] o D (b Using lst eqution e obtin the folloing D-pH plot: ] (c e clculte the st nd nd derivtive of the denomintor, i.e., F([H ] hereupon e hve the st derivtive F'([H ] [H nd for the nd derivtive F''([H ] ] 58
Since lys F''([H ] > 0, then hen F'([H ] 0, F([H ] is minimum under these conditions. Consequently, the distribution rtio is mximum hen 0 or [H] ( ( 4.5 8 M or ph 7.5 Problem 5: Mss spectroscopy The ionic frgment SiCl ill be represented by the folloing peks: M 98 8 Si 5 Cl M 99 9 Si 5 Cl M 0 8 Si 5 Cl 7 Cl 0 Si 5 Cl M 9 Si 5 Cl 7 Cl M4 0 Si 5 Cl 8 Si 5 Cl 7 Cl M5 9 Si 7 Cl M6 4 0 Si 7 Cl Therefore, the correct nser is 7. The expected peks nd the corresponding probbilities re: m/z 45 B 5 Cl : 0.99 0.7577 0.5 m/z 46 B 5 Cl : 0.80 0.7577 0.607 m/z 47 B 7 Cl : 0.99 0.4 0.048 m/z 48 B 7 Cl : 0.80 0.4 0.94 Hence, the bse pek hs nominl mss M 46 nd the reltive intensities re: M 45 (0.5/0.607 0 4.9% M 46 0% M 47 (0.048/0.607 0 7.9% M 48 (0.94/0.607 0.0% Therefore, the correct nser is C. For the ion N e hve: 4 N 4 N (0.9964 0.997 4 N 5 N 5 N 4 N (0.9964 0.0066 0.0079 hence, (M/M 0.0079/0.997 0.0075 or 0.75% For the ion CO e hve: C 6 O 0.989 0.9976 0.9866 C 7 O C 6 O (0.989 0.0008 (0.0 0.9976 0.05 hence, (M/M 0.05/0.9866 0.05 or.5% For the ion CH N e hve: C H 4 N 0.989 (0.99985 0.9964 0.985 C H 4 N C H H 4 N C H H 4 N C H 5 N 0.0 (0.99985 0.9964 0.989 0.99985 0.0005 0.9964 0.989 (0.99985 0.0066 0.0487 hence, (M/M 0.0487/0.985 0.05 or.5% For the ion C H 4 e hve: C H 4 (0.989 (0.99985 4 0.9775 C C H 4 C C H 4 C H H C H H H C H H H C H H 0.0 0.989 (0.99985 4 4 0.989 0.0005 (0.99985 0.04 hence, (M/M 0.04/0.9775 0.09 or,9% Therefore the correct nser is (b CO 59