APEF Acids and Bases - Answers

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APEF Acids nd Bses - Answers 1. It requires 3.0 ml of 0.0500 mol/l NOH(q) to neutrlize 100.0 ml of gstric juice. We cn ssume tht HCl(q) is the only cid present in gstric juice. ) Clculte the concentrtion of HCl(q) present in gstric juice. HCl(q) NOH(q) NCl(q) H O(l) C CbVb R V R ( 0.0500 mol L)( 0.030 L)( 1) ( 0.1000 L)( 1) b 0.0185 mol L b) Clculte the ph of gstric juice. [H ] 0.0185 mol/l ph -log(0.0185) 1.3 c) Clculte the number of grms of HCl(q) per 100 ml of gstric juice. n CV (0.0185 mol/l)(0.100l) 0.00185 mol mss (mol)(mol mss) (0.00185 mol)(36.460 94 g/mol) 0.065 g APEF Acids nd Bses - Answers Pge 1 of 15

. Ascorbic cid (vitmin C) is wek monoprotic cid tht hs the formul HC 6 H O 6 nd K 8.0 x 10-5. Wht is the ph of 0.035 mol/l solution of scorbic cid? HC 6 H O 6 H - C 6 H O 6 [initil] 0.035 M 0 0 [ ] -x x x [equilibrium] 0.035 x x x K 8.0 10 check : [ H ][ C H O ] [ HC H O ] ( x)( x) ( 0.035 x) [ HA] 0.035 ( 8.0 10 )( 0.035 x).8 10 6 6 K 6 6 6 8.0 10 8.0 10 x x x 43.5 < 500 must keep x in the deno min tor 0 x x 8.0 10 8.0 10 ± x.8 10 6 6 ( 8.0 10 ) ( 4)()( 1.8 10 ) ( )( 1) x x x 0.001 [ H ] 8.0 10 8.0 10 ± or 0.0016 mol L 1.1 10 ± 0.00335 0.0016 ph log 1.6 ( 10 ). 8 APEF Acids nd Bses - Answers Pge of 15

3. The ph of 0.0 mol/l solution of benzoic cid (HC H 5 O ) is.68. Clculte the numericl vlue of the K for this cid. HC H 5 O H - C H 5 O [initil] 0.0 M 0 0 [ ] -x x x [equilibrium] 0.0 x x x x [H] [H] 10 -.68.09 10-3 x K K [ H ][ C H O ] [ HC H O ] 5 5 (.09 10 )(.09 10 ) ( 0.0.09 10 ) 6. 10 4. Wht is the ph of solution formed by mixing 100.0 ml of 0.150 mol/l HCl(q) with 150.0 ml of 0.0900 mol/l NOH(q)? HCl(q) NOH(q) NCl(q) H O(l) mol H (0.150 mol/l)(0.1000 L) 0.0150 mol H mol OH - (0.0900 mol/l)(0.1500 L) 0.0135 mol OH - excess mol H 0.0150 mol 0.0135 mol 0.0015 mol n 0.0015 mol L C 0.0060 mol L [H ] V 0.50 L ph -log(0.0060). APEF Acids nd Bses - Answers Pge 3 of 15

5. Acetylslicylic cid (ASA) is the most widely used drug in the world. ASA hs the chemicl formul, C 8 H O COOH, nd its K equls 3. x 10-4. Clculte the ph of 0.50 mol/l ASA solution. C 8 H O COOH H C 8 H O COO - [initil] 0.50 M 0 0 [ ] -x x x [equilibrium] 0.50 x x x K 3. 10 3. 10 8.18 10 [ H ][ C H O COO ] [ C H O COOH] 8 4 4 8 x x 9.04 10 ph log 9.04 ( x)( x) ( 0.50 x) ( x)( x) ( 0.50) ( 10 ). 04 6. Clculte the H 3 O (q) ion concentrtion of the resulting solution if 0.0 ml of 0.00 mol/l HCl(q) is mixed with 30.0 ml of 0.350 mol/l NOH(q). HCl(q) NOH(q) NCl(q) H O(l) mol H (0.00 mol/l)(0.000 L) 0.00400 mol H mol OH - (0.350 mol/l)(0.0300 L) 0.0105 mol OH - excess mol OH - 0.0105 mol 0.00400 mol 0.0065 mol n 0.0065 mol L C 0.130 mol L [OH - ] V 0.0500 L poh -log(0.130) 0.88 ph 14.0 0.88 13.1 [H] 10-13.1. 10-14 mol/l APEF Acids nd Bses - Answers Pge 4 of 15

. Clculte the [H ] in solution of 0.10 mol/l hypochlorous cid, HOCl. The K vlue of HOCl is.9 x 10-8. HOCl H OCl - [initil] 0.10 M 0 0 [ ] -x x x [equilibrium] 0.10 x x x K [ H ][ OCl ] [ HOCl] 8.9 10 8.9 10 9.9 10 x x 5.4 10 ( x)( x) ( 0.10 x) ( x)( x) ( 0.10) [ H ] 8. A student titrted 45.0 ml of nitric cid solution of unknown concentrtion with 0.10 mol/l strontium hydroxide solution. A ph meter ws used to obtin the following titrtion curve: ) Bsed on the results shown on the grph, clculte the concentrtion of the nitric cid solution. HNO 3 (q) Sr(OH) (q) Sr(NO 3 ) (q) H O(l) C CbVb R V R ( 0.10 mol L)( 0.050 L)( ) ( 0.0450 L)() 1 b 0.11 mol L b) Wht would be n pproprite indictor for this titrtion nd why? litmus (4.5 8.3), bromothymol blue (6.0.6), phenol red (6.6 8.0), phenolphthlein (8. 10.0) A suitble indictor must chnge colour during the verticl section of the grph. APEF Acids nd Bses - Answers Pge 5 of 15

9. ) A chemistry lbortory hs solutions of four indictors: methyl ornge, phenol red, phenolphthlein, nd lizrin yellow R. Use the informtion given in the Chemistry Dt Booklet to choose n pproprite indictor for ech of the following titrtions, nd give your resons in ech cse. (The concentrtions of ll cid nd bse solutions re in the 0.10 to 0.30 mol/l rnge.) i) CH 3 COOH(q) with NOH(q) wek cid & strong bse endpoint ph > phenolphthlein (8. 10.0), or lizrin yellow R (10.1 1) ii) HCl(q) with NH 3 (q) strong cid & wek bse endpoint ph < methyl ornge (3. 4.4) iii) HCl(q) with NOH(q) strong cid & strong bse endpoint ph phenol red (6.6 8) or phenolphthlein (8. 10.0) b) For which titrtion would you hve more thn one choice mong the bove indictors? Which indictors could be used in this titrtion? i) CH 3 COOH(q) with NOH(q) phenolphthlein or lizrin yellow R iii) HCl(q) with NOH(q) phenol red or phenolphthlein 10. Originlly, Arrhenius believed tht there must be the sme number of positive nd negtive ions in solution. However, the modern theory of electrolytes sttes tht the totl positive ionic chrge must equl the totl negtive ionic chrge. Is there difference between these two sttements? If so, why did Arrhenius originl theory hve to be modified? Yes, there is difference between the two sttements. The Arrhenius theory would hve been correct if ll positive ions nd negtive ions hd equl but opposite chrges. However, different positive ions cn hve chrges of 1,, 3, or 4, nd different negtive ions cn hve chrges of 1-, -, or 3-. Thus, in the cse of solution of slt like clcium chloride (CCl ), there would be twice s mny chloride ions s clcium ions: however, the totl positive ionic chrge does equl the totl negtive ionic chrge. APEF Acids nd Bses - Answers Pge 6 of 15

11. Butnoic cid, C 3 H COOH, is found in rncid butter nd hs well-deserved, mlodorous reputtion (it stinks). In well-ventilted room, 0.0 moles of the cid re dissolved in wter to give 500 ml of solution whose ph is found to be.60. Assuming tht butyric cid is monoprotic (it hs only one ionizble proton), clculte K for the cid. C 3 H COOH H C 3 H COO - [initil] 0.040 M 0 0 [ ] -x x x [equilibrium] 0.40 x x x 0.40 -.51 10-3.51 10-3.51 10-3 n V 0.0 mol 0.500 L [ H COOH] 0.40 mol L C 3 x [H] [H] 10 -.60.51 10-3 x K K [ H ][ C H COO ] [ C H COOH] 3 3 (.51 10 )(.51 10 ) ( 0.40.51 10 ) 1.6 10 1. The K w of wter t 5 C is 1.0 x 10-14, nd the thermochemicl eqution for the rection is: H O(l) H (q) OH - (q) H 56 kj Predict, giving your reson, how the Kw will chnge if the temperture is increses. The rection is endothermic. If the temperture is incresed, the equilibrium shifts to the right, producing more H (q) nd OH - (q). Since Kw [H ][ OH - ], Kw will increse s the concentrtions of H (q) nd OH - (q) increse. 13. The HCO 3 - (q) ion is cpble of cting s either Brønsted-Lowry cid or bse. Illustrte this behviour by writing blnced equtions for its rections with CN - (q) ions nd with HNO 3 (q). HCO 3 - s n cid: HCO 3 - s bse: HCO 3 - (q) CN - (q) CO 3 - (q) HCN(q) HCO 3 - (q) HNO 3 (q) H CO 3 (q) NO 3 - (q) APEF Acids nd Bses - Answers Pge of 15

14. Three solutions re known to hve ph vlues of 4.5, 6.0, nd 8.0. Using two cid-bse indictors (Chemistry Dt Booklet, pge 8), describe how you would distinguish mong them. The experiment design involves testing ech solution with two indictors. One indictor must distinguish the solution tht hs ph 4.5 from the other two use bromocresol green, litmus or methyl red. The other indictor must distinguish solution tht hs ph 8 from the other two use bromothymol blue or phenol red. The third solution cn be identified by elimintion. 15. The following dt show how the ph chnges during titrtion when 0.100 mol/l NOH(q) is dded to 10.0 ml of queous cetic cid (CH 3 COOH). 14 ml NOH 0.0 1.0.0 3.0 4.0 6.0.0 8.0 10.0 14.0 ph.9 4.0 4.3 4. 5. 5.5 6.4 11. 1.0 1.4 ) Using grph pper, plot the ph titrtion curve. Determine the vlue of the end-point ph. b) Clculte the concentrtion of CH 3 COOH(q) ph 1 10 8 6 4 0 0 4 6 8 10 1 14 16 volume of NOH (ml) CH 3 COOH(q) NOH(q) N CH 3 COO(q) H O(l) C CbVb R V R ( 0.100 mol L)( 0.005 L)( 1) ( 0.010 L)( 1) b 0.05 mol L c) Use the informtion provided in this question nd your nswer in prt (b) to clculte the numericl vlue of K for CH 3 COOH(q) CH 3 COOH H CH 3 COO - [initil] 0.05 M 0 0 [ ] -x x x [equilibrium] 0.05 x x x 0.05 1.3 10-3 1.3 10-3 1.3 10-3 x [H], K K [ H ][ CH COO ] 3 [ CH3COOH] ( 1.3 10 )( 1.3 10 ) ( 0.05 1.3 10 ) [H] 10 -.9 1.3 10-3 x.1 10 APEF Acids nd Bses - Answers Pge 8 of 15

16. A set of six solutions ws prepred for demonstrtion. The solutions were: 0.05 mol/l B(OH) (q) 0.10 mol/l NOH(q) 0.05 mol/l H SO 4 (q) 0.10 mol/l HCl(q) 0.10 mol/l CH 3 COOH(q) 0.10 mol/l NH 3 (q) Before the solutions were lbeled, they becme scrmbled, nd the technicin ws forced to lbel the bottles A to F. The technicin tested ll six solutions nd four mixtures (prepred by mixing equl volumes of two solutions) first in conductivity pprtus nd then with bromothymol blue. The technicin knew tht of the possible mixtures of cids nd bses only the following would produce n insoluble slt: B(OH) (q) H SO 4 (q) BSO 4 (s) H O(l) Solution(s) Ability to Colour of Conduct Electricity Bromothymol Blue A only good yellow B only poor blue C only poor yellow D only good blue E only good blue F only good yellow A & D mixed good green A & E mixed good green D & F mixed good green E & F mixed poor green ) Clssify the solutions A to F s: strong cids, wek cids, strong bses, or wek bses. Strong Acids: A & F Strong Bses: D & E Wek Acids: C Wek Bses: B b) Explin the differences in the tests results of mixture E & F compred with the other mixtures, nd identify solutions E nd F. The mixture of E nd F ws the only mixture of n cid nd bse tht ws poor conductor of electricity. This must hve been mixture of queous solutions of brium hydroxide (strong bse, E) nd sulfuric cid (strong cid, F). The product of this mixture would be brium sulfte, nd it is n insoluble slt tht is poor conductor of electricity. c) Mtch the letters A to D with the pproprite solutions. A hydrochloric cid (HCl) B mmoni (NH 3 ) C cetic cid (CH 3 COOH) D sodium hydroxide (NOH) APEF Acids nd Bses - Answers Pge 9 of 15

1. ) Give one similrity nd one difference between the Arrhenius nd the Brønsted-Lowry definitions of cids nd bses. Both the Arrhenius nd the Brønsted-Lowry theories predict tht substnces tht hve ionisble hydrogen toms re cidic nd tht ionic hydroxide compounds re bsic. However, the Brønsted-Lowry definition is broder one in which cids nd bses re defined in terms of how they rect nd not merely whether they produce hydrogen ions or hydroxide ions in queous solution. b) Explin how well CH 3 COOH fits into ech definition. CH 3 COOH fits well in both definitions. The hydrogen tom tht is bonded to the oxygen tom does ionize in queous solutions, nd CH 3 COOH is proton donor. CH 3 COOH H O CH 3 COO - H 3 O c) Explin how well NH 3 fits into ech definition. NH 3 fits well into the Brønsted-Lowry definition since it cn ccept proton: NH 3 H O NH 4 OH - However, it does not fit well into the Arrhenius definition since it is not n ionic hydroxide nd does not dissocite in queous solution to produce hydroxide ions. ph versus Volume of NOH dded 1 10 8 ph 6 4 0 0 5 10 15 0 5 30 35 40 Volume APEF Acids nd Bses - Answers Pge 10 of 15

18. Construct grph of ph versus volume of bse dded for the titrtion of 50.0 ml of 0.0500 M HI with 0.050 M NOH, bsed on the clcultions below. ) How mny ml of bse re required to neutrlize the HI? NOH HI NI H O V CV R C R ( 0.0500 mol L)( 0.0500 L)( 1) ( 0.050 mol L)() 1 b b b 0.0333 L or 33.3 ml b) Wht is the ph before ny bse is dded? ph -log(0.0500) 1.30 c) Wht is the ph fter the ddition of 5.0 ml of NOH? mol H (0.0500 mol/l)(0.0500 L) 0.0050 mol H mol OH - (0.050 mol/l)(0.050 L) 0.00185 mol OH - excess mol H 0.0050 mol 0.00185 mol 0.00065 mol n 0.00065 mol L C 0.00833 mol L [H ] V 0.050 L ph -log(0.00833).08 d) Wht is the ph fter 30.0 ml of NOH hs been dded? mol H (0.0500 mol/l)(0.0500 L) 0.0050 mol H mol OH - (0.050 mol/l)(0.0300 L) 0.005 mol OH - excess mol H 0.0050 mol 0.005 mol 0.0005 mol n 0.0005 mol L C 0.0031 mol L [H ] V 0.0800 L ph -log(0.0031).5 e) Wht is the ph t the equivlence point? Titrtion of strong cid with strong bse, therefore the ph is seven t the equivlence point. APEF Acids nd Bses - Answers Pge 11 of 15

f) Wht is the ph fter 35.0 ml of NOH hve been dded? mol H (0.0500 mol/l)(0.0500 L) 0.0050 mol H mol OH - (0.050 mol/l)(0.0350 L) 0.0065 mol OH - excess mol OH - 0.0050 mol 0.0065 mol 0.00015 mol n 0.00015 mol L C 0.0014 mol L [OH _ ] V 0.0850 L poh -log(0.0014).8 ph 14.0.8 11. 19. Using the K tble in your Dt Booklet, determine the [OH - ] of 0.068 M of HCN. HCN H CN - [initil] 0.068 M 0 0 [ ] -x x x [equilibrium] 0.068 x x x K [ H ][ CN ] [ HCN] 10 6. 10 10 6. 10 11 4. 10 x 6 x 6.5 10 ( x)( x) ( 0.068 x) ( x)( x) ( 0.068) [ H ] ph -log(6.5 10-6 ) 5. poh 14.0 5. 8.8 [OH-] 10-8.8 1.5 10-9 mol/l APEF Acids nd Bses - Answers Pge 1 of 15

0. ) Write equtions for the rections of the mphoteric substnce, HS -, with HCl nd with KOH. HS - with HCl: or HS - with KOH: or HS - (q) HCl(q) H S(q) Cl - (q) HS - (q) H(q) H S(q) HS - (q) KOH(q) S - (q) K(q) H O(l) HS - (q) OH - (q) S - (q) H O(l) b) Write the formul for the conjugte bse of HSO 4 -. SO 4 - c) Write the formul for the conjugte cid of HSO 4 -. H SO 4 1. ) Clculte the ph of.5 x 10-3 mol/l NOH solution. [OH - ].5 x 10-3 [H 3 O ] K w 1.0 x 10-14 4.0 x 10-1 [OH - ].5 x 10-3 ph -log[h 3 O ] -log(4.0 x 10-1 ) 11.4 or poh -log[oh-] -log(.5 x 10-3 ).6 ph 14.00 poh 14.00.6 11.4 b) Clculte the ph of solution of HCl(q) mde by diluting 3.50 ml of 1.0 mol/l HCl to finl volume of 500.0 ml. C 1 V 1 C V [H 3 O ] (1.0 mol/l)(0.003 50 L) 0.0840 mol/l 0.5000 L ph -log(0.0840) 1.08 APEF Acids nd Bses - Answers Pge 13 of 15

. ) Write equtions for the rections of the mphoteric substnce, HPO 4 -, with i) with HCl HCl HPO 4 - Cl- H PO 4 - or: H 3 O HPO 4 - Cl- H PO 4 - ii) with KOH KOH HPO 4 - K H O PO 4 3- or: OH- HPO 4 - K H O PO 4 3- b) Write the formul for the conjugte bse of ech of these species: i) H SO 4 HSO 4 - ii) PH 4 PH 3 c) Write the formul for the conjugte cid of ech of these species: i) HPO 4 - H PO 4 - ii) NO 3 - HNO 3 APEF Acids nd Bses - Answers Pge 14 of 15

3. ) Wht is the ph of 0.155 mol/l HCl solution? HCl is strong cid, therefore, [H] 0.155 mol/l ph -log(0.155) 0.810 b) Clculte the volume of 0.10 mol/l NOH needed to neutrlize 5.0 ml of 0.155 mol/l HCl HCl NOH NCl H O V b C V R b C b (0.155 mol/l)(0.050 L)(1) 0.033 L or 3.3 ml C b R (0.10 mol/l)(1) c) Explin why the endpoint of n cid-bse titrtion does not lwys coincide with the equivlence point. The equivlence point represents the stoichiometric quntity of titrnt required by the blnced chemicl eqution. The endpoint of titrtion is indicted by sudden chnge in some property such s the colour chnge of n cid-bse indictor. If suitble indictor is chosen nd the titrtion is performed crefully, the endpoint should be the equivlence point. If n unsuitble indictor is chosen (one which does not complete its colour chnge within the verticl region of the ph curve), or sloppy titrtion techniques re used, the endpoint will differ from the equivlence point. 4. The dt given in the tble hs been compiled for 0.10 mol/l queous solutions of these compounds: CH 3 COOH, HBr, HCN, H SO 4, N HBO 3, NHCO 3, nd NH PO 4. Complete the COMPOUND column of the tble. In the tble, n/ mens not pplicble. (Pges 8 nd 9 of the Chemistry Dt Booklet my be useful.) Compound ph Ornge IV Colour Rection with zinc ml needed to neutrlize 0 ml 0.10 mol/l NOH ml needed to neutrlize 0 ml 0.10 mol/l HCl H SO 4 red fst 10 n/ NHCO 3 yellow 0 0 NH PO 4 yellow 10 0 HCN 5.1 slow n/ N HBO 3 yellow 0 10 HBr red fst n/ CH 3 COOH.9 slow n/ APEF Acids nd Bses - Answers Pge 15 of 15