1 Chpter 17: Additionl Aspects of Aqueous Equilibri Khoot! 1. Adding Br to sturted queous solution of decreses its solubility in wter. BSO 4, Li CO 3, PbS, AgBr. Which of the following mitures could be used to prepre n effective buffer? HCl nd KCl, HNO 3 nd KNO 3, HCl nd NH 4 Cl, NH 3 nd NH 4 Cl 3. Which 1 L solution hs the gretest buffer cpcity? 0.1 M NH 3 nd 0.1 M NH 4 Cl, 0.05 M NH 3 nd 0.05 M NH 4 Cl, 0.1 M NH 3 nd 0.01 M NH 4 Cl, 0.5 M NH 3 nd 0.5 M NH 4 Cl 4. Select the best cid or bse to pir with its conjugte slt to prepre 8.5 ph buffer. cetic cid; K = 1.810 5, mmoni; K b = 1.810 5, hydroylmine; K b = 1.110 8, citric cid; K = 7.510 4 5. Select the correct representtion of the Henderson Hsselblch eqution. K = [H + ][A ], K w = [H + ][OH ] = 10 14, ph = pk + log [bse]/[cid], pk = ph + log[bse]/[cid] 6. Which indictor is preferble when titrting wek bse with strong cid? Methyl red (color chnge ph = 5), bromothymol blue (ph = 7), phenolphthlein (ph = 9), None of the bove 7. Which indictor is preferble when titrting wek cid with strong bse? Methyl red (color chnge ph = 5), bromothymol blue (ph = 7), phenolphthlein (ph = 9), None of the bove 8. For BCO 3, K sp = 5.0 10 9. Wht is [B + ]? 7.1 10 5 M, 1.0 10 8 M,.5 10 9 M, 5.0 10 9 M 9. For BF, K sp = 1.7 10 6. Wht is [B + ]? 1.7 10 6 M, 3.4 10 6 M, 7.6 10 3 M, 1.5 10 M 10. Which of the following regents will reduce the solubility of BF? NCl, B(C H 3 O ), KOH, NH 4 Br 11. Which of the following regents will increse the solubility of BF? HCl, HF, KOH, NH 4 Br Whitebord Emples Emple: Clculte the ph of solution prepred by miing equl volumes of 0.0 M CH 3 NH nd 0.60 M CH 3 NH 3 Cl (K b = 3.7 10 4 ). Wht is the ph of 0.0 M CH 3 NH without ddition of CH 3 NH 3 Cl? ssuming we hve 1 L of ech 0.0 moles CH3NH 1L L CH3NH 0.10 M CH 0 3NH.0 L 0.60 moles CH3NH3 1L CH L 3NH 3 0.30 M CH3NH 3 0.0 L I 0.1 0.3 0 C + + E 0.1 0.3 + + [ CH3NH3 ][ OH ] (0.3 ) 4 Kb 3.710 [ CH3NH] 0.1 5 0.30043.710 0 5 8.9710 M OH 5 ph 14 log(8.9710 ) 9.95 ph of 0.0 M CH 3 NH :
[ 3 3 ][ ] b CH3NH 4 5 CH NH OH K 3.710 [ ] 0.1 3.710 3.710 0 3 8.410 M OH 3 ph 14 log(8.410 ) 11.9 Buffer Emple I: Clculte the ph nd poh of 500.0mL solution contining 0.5M HPO 3 4 nd 0.5M PO 4 t 5C where the K ( HPO 4 ) = 4.10-13. [ A ] 13 0.5M ph pk log log(4.10 ) log [ HA] 0.5M ph 1.38 Buffer Emple II: How would we prepre ph = 4.44 buffer using CH 3 CO H nd CH 3 CO N? K = 1.8 10 5 [ A ] [ CH3CO ] 5 ph pk log log 4.44 log(1.810 ) 0.305 [ HA] [ CH3COH ] [ CH3CO ] 0.305 10 0.496 [ CH3COH ] Therefore, in order to mke 4.44 buffer solution we need 0.496 moles of CH 3 CO N for every mole of CH 3 CO H Comprehensive Emple:Clculte the ph of ech of the following solutions: 5. 0.100 M HC3H 5O, K 1.3 10 4 H C H O 3 5 3 5 3 5 HC3H 5O 5 3 HC H O H C H O K 1.310 1.110 M H ph.96 0.100 b. 0.100 M NC3H5O NC3H5O N C3H5O K OH HC H O CHO HO OH HCHO K CHO w 3 5 3 5 b K 10 6 3 5 3 5 7.710 8.810 0.100 0.100 M OH ph 14 poh 8.94 c. miture contining. & b. H C3H5O HC3H 5O H C3H 5O K HC H O 3 5 5 0.100 5 1.310 1.310 M H ph 4.89 0.100 A using Henderson-Hsselbch, ph pk log HA 5 0.100 ph log1.310 log 4.89 0.100 d. miture contining c. nd 0.00 mol of NOH when strong bse is dded to n cid contining solution it will
3 neutrlize the cid nd so ll of the OH will rect completely with our propnoic cid HC3HO 5 OH HO CHO 3 5 HC3HO 5 OH CHO 3 5 B 0.100 0.00 0.100 C 0.00 0.00 +0.0 A 0.080 0 0.10 5 0.10 ph log1.310 log 5.06 0.080 e. miture contining c. nd 0.00 mol of HCl H CHO 3 5 HCHO 3 5 CHO 3 5 H HC3HO 5 B 0.100 0.00 0.100 C 0.00 0.00 +0.0 A 0.080 0 0.10 5 0.080 ph log1.310 log 4.71 0.10 Titrtion Emple I: Wht is [NH 3 ] if.35ml of 0.1145 M HCl were needed to titrte 100.0mL smple? NH3 HCl NH 4 Cl 0.1145 moles HCl 1 mole NH3 1 [ NH3] 0.035L 0.059M L 1mole HCl 0.1000L Titrtion Emple II Strong with Strong: A 15.0 ml smple of 0.00 M NOH is titrted with 0. 50 M of HCl. Clculte the ph of the miture fter 10.0, nd 0.0 ml of cid hve been dded. If you re not given dissocition constnt this should remind you tht the cid/bse is strong. For 10.0 ml of HCl & 15.0 ml of NOH: 1 L 0.50 mol Hq 10.0 ml 0.0050 mol Hq 1 L 0.00 mol OHq 15.0 ml 0.00300 mol OH q H OH H O q q l B 0.0050 0.00300 - C -0.0050-0.0050 - A 0 0.00050-0.0005 moles OH 0.000M 0.05L ph 14 log(0.0) 1.30 For 0.0 ml of HCl 1 L 0.50 mol Hq 0.0 ml 0.00500 mol Hq H OH q q H Ol B 0.00500 0.00300 - C -0.00300-0.00300 - A 0.0000 0 -
4 0.00 moles H 0.05714M 0.035L ph log(0.05714) 1.4 Titrtion Emple III Strong with Wek: A 5.0 ml smple of 0.100 M cetic cid (HC H 3 O ) is titrted with 0.15 M of NOH. Clculte the ph of the miture fter 0.0,10.0, 0.0, nd 30.0 ml of bse hve been dded. (K = 1.8 10 5 ) For 0.0 ml it is the sme s wek cid: H CHO 3 3 HCHO 3 3 H CHO 3 3 K HC H O For 10.0 ml: 3 3 5 1.810 0.100 ssume << 0.100 5 3 1.810 1.3410 M H 0.100 3 1.3410 verify ssumption: 100% 1.34% 5% 0.100 3 ph log 1.110.87 Initilly we hve: HC H O NOH NC H O H O 3 q q 3 q l N is specttor ion so we re relly looking t: HC H O OH C H O H O 3 q q 3 q l This is in fct neutrliztion rection. For 10.0 ml of NOH: 1 L 0.100 mol HCH3Oq 5.0 ml 0.0050 mol HC H O 1 L 0.15 mol OH q 10.0 ml 0.0015 mol OHq Now we set up BCA tble or before, chnge, fter tble HC H O OH C H O H O 3 q 3 q q 3 q l B 0.0050 0.0015 0 C 0.0015 0.0015 A 0.0015 0 +0.0015 So, we hve neutrlized the cid with the given mount of bse nd re now redy to pply the Henderson Hsselblch eqution: 0.015 A 5 ph pk log log 0.0350 1.8 10 log L 4.74 HA 0.015 0.0350L Time for 0.0 ml of NOH: 1 L 0.15 mol OHq 0.0 ml 0.0050 mol OHq
5 HC HO OH CHO HO 3 q q 3 q l B 0.0050 0.0050 0 C 0.0050 0.0050 A 0 0 +0.0050 All of the cid hs rected with ll the bse nd so we re now t the equivlence point. However, since we hd wek cid nd strong bse we should epect our ph to be higher thn 7. To find the ph we need n ICE tble. CHO 3 0.0050 moles/ 0.045L 0.0556M q CHO 3 HO HCHO q l 3 q OHq I 0.0556 0 0 C + + E 0.0556 + + HC 14 H3O OH q q Kw 10 10 Kb 5.610 5 CHO K 1.8 10 0.0556 3 q ssume 0.0556 10 6 5.610 OH 5.6 10 M q 0.0556 ck ssumption : 6 5.610 100% 0.01% 5% 0.0556 6 ph 14 log(5.610 ) 8.75 Finlly, 30.0 ml 1 L 0.15 mol OHq 30.0 ml 0.00375 mol OH Now, we hve more strong bse thn cid so we cn set up the BCA tble nd then determine hydroide concentrtion nd ph directly. HCHO 3 OH CHO q q 3 q HO l B 0.0050 0.00375 0 C 0.0050 0.0050 A 0 0.0015 +0.00375 0.0015 moles OH 0.07M 0.055L ph 14 log(0.07) 1.36 Solubility Emple: Determine the equilibrium concentrtions (nd solubilities) of BF (s), K sp = 1.710 6. BF B F 6 1.710 ( s) ( q) ( q) s s [ S][ S] S S M [ B ] 0.0075 M nd [ F ] 0.0075 0.015M 0.0075moles 137.37 g g B : 1.03 L mol L 0.015moles 18.998 g g F : 0.85 L mol L 6 1.710 4 3 0.0075 Common Ion Solubility Emple: Clculte the solubility of clcite (CCO 3 ) in 0.00100 M of N CO 3 nd in just plin wter (K sp = 4.510 9 t 5C). q
6 CCO C CO 3( s) ( q) ( q) C ( q) CO ( q) Initil 0 0.0010 Chnge +s +s Eq s 0.0010+s [][0.001 s s] ssume 0.001 >> s 9 6 4.510 0.001s s 4.510 M ck : 6 4.510 100% 0.45% 5% 0.001 Therefore the solubility of CCO 3 is 4.510 6 M with [C + ] = 4.510 6 M & [CO 3 ] = 0.0010 M How does this compre with the solubility of just CCO 3? CO C ( q) ( q) Initil 0 0.0010 Chnge +s +s Eq s +s [][] s s 9 5 4.510 s s 6.7110 M 5 C CO 3 6.7110 M 5 6 6.7110 M 4.510 M thereby demonstrting how common-ion represses solubility