Worksheet 0 Polyprotic Acids nd Slt Solutions Acid Bse strong cid HNO, HI, HCl, etc NO, I, Cl, etc negligile sicity 1. HSO SO 7.7 1 7.1 HNO NO 1. 11 6.8 HF F 1.5 11 1.8 5 CH COOH CH COO 5.6.5 7 H CO HCO. 8 9 8 H S HS 1.1 7 5.6 NH NH 1.8 5 6. HCN CN 1.6 5.7 11 HCO CO 1.8 1 17 HS S 1 negligile cidity Li O, NOH O, OH strong se Slts re ionic compounds which dissocite in wter to produce ions. They re formed in the neutrliztion rection etween cids nd ses. Depending on the nture of the cids nd ses (strong or wek), the solutions of the slts will e cidic, sic or neutrl. 1. Decide which of the following slts will form cidic, sic or neutrl solutions when dissolved in wter. (Hint: look t the cids nd ses tht formed them) For emple: CH COO ws formed in the rection etween OH nd CH COOH OH is strong se. Its conjugte cid,, hs negligile cidity nd will leve the ph t 7.00, neutrl solution. CH COOH is wek cid, mking its conjugte se, CH COO reltively strong se. It will produce sic solution. ) F ) CN sic sic c) NNO d) RI neutrl neutrl e) NH NO f) N CO cidic sic
. Rnk the following 0.1 M queous slt solutions in order of incresing ph. (Hint: write out the rections of the slts wter) ) NO SO S neutrl sic ( 7.7 1 ) sic ( 1 ) strongest se ph rnk: NO < SO < S ) NH NO NHSO NHCO N CO cidic cidic sic sic 5.6 1.. 8 1.8 strongest cid strongest se ph rnk: NHSO < NH NO < NHCO < N CO. In n eperiment, it is found tht the phs of three slts, X, Y nd Z re 7.0, 9.0 nd 11.0. Arrnge the cids, HX, HY nd HZ in order of incresing cid strength. The strongest cid would hve the wekest conjugte se. Weker ses (t equl concentrtions) will hve lower ph vlues (they re less sic). The se strengths must e X < Y < Z The conjugte cids will hve the reverse trend: HZ < HY < HX. Clculte the ph of 0.1 M NCN solution. N does not ffect the ph of the solution CN is wek se with 1.6 5 CN H O HCN OH [CN ] [HCN] [OH ] Initil 0. 0 0 Chnge Equil. 0. 1.6 5 1. poh log(1. ph 1.9 11.1 [ HCN][ OH ] [ CN ] ).9 0.
5. Suppose tht 50.00 ml of 0.M CH COOH is comined with 50.00 ml of 0.M NOH. Wht is the ph of the resulting system? The eqution for this rection is: CH COOH OH CH COO H O (neutrliztion rection) ) How mny moles of CH COO re formed? (ssume the rection goes to completion) 0. mol CH COOH mol CH COOH 0.050L 0.0050 mol 1L solution 0. mol OH mol OH 0.050L 0.0050 mol 1L solution moles CH COOH OH CH COO Initil 0.0050 0.0050 0 Chnge 0.0050 0.0050 0.0050 Equil. 0 0 0.0050 This rection would form 0.0050 moles of CH COO ) Wht is the finl volume of the solution? Finl volume is the sum of the components: 50 ml 50 ml 0 ml c) Wht is the concentrtion of CH COO? [CH COO ] 0.0050 mol / 0.0 L 0.050 M d) Chrcterize CH COO (i.e. wek/strong cid/se). wek se e) Write the rection of the product (CH COO ) nd wter. CH COO H O CH COOH OH f) Set up n ICE tle for the rection nd clculte the ph of the solution. [CH COO ] [CH COOH] [OH ] Initil 0.050 0 0 Chnge Equil. 0.050 5.6 5. poh log(5. ph 1 5. 8.7 [ CH COOH ][ OH ] [ CH COO ] ) 5. 0.050
6. A 0. M solution of OC 6 H 5 hs ph of 11.0. Clculte the vlue for HOC 6 H 5. [OC 6 H 5 ] [HOC 6 H 5 ] [OH ] Initil 0. 0 0 Chnge Equil. 0. poh 1 ph 1 11.0.60 poh.60 [ OH ].5 [ HOC H ][ OH ] 6 5 [ OC H ] 6 5 (.5 ) 0. 0..5 1 w 1.0 1.6 5 6. 7. Wht is the ph t the end of the following neutrliztion rections? 6. ) 50.00 ml of 0. M CH COOH comined with 50.00 ml of 0. M NOH see #5; ph 8.7 ) 50.00 ml of 0. M NOH comined with 50.00 ml of 0.1 M HCl H OH H O Since we egin with equl moles of oth the strong cid nd the strong se, they will neutrlize ech other. The finl ph 7 c) 50.0 ml of 0. M NH comined with 50.00 ml of 0. M HCl This is numericlly ectly the sme s prt. The strong cid will e consumed entirely y recting with NH to form the conjugte cid (NH ). The conjugte cid will then rect with wter to yield n cidic solution. So 0.0050 moles of NH will form, with concentrtion of 0.050M nd then rect with wter ccording to the eqution NH H O NH H O [NH ] [NH ] [H O ] Initil 0.050 0 0 Chnge Equil. 0.050 5.6 5. ph log(5. [ NH ][ H O ] [ NH ] ) 5. 0.050 5
8. Four different ses, ll t 0. M concentrtions nd 1.0 L volumes, re rected with 0 ml of 1.00 M HNO. Wht re the ph vlues of the solutions fter the rection? mol se mol H 0. mol CH COOH 1.0L 0. mol 1L solution 1.00 mol H 0.0L 0. mol 1L solution ) CH NH (. ) CH NH HNO CH NH NO This initil rection produces 0. mol CH NH (wek cid) with concentrtion of 0. mol / 1.1 L 0.0909 M By now, we should see the generl trend (without drwing out the full ICE tle), nd solve immeditely for the [H O ] 1 w 1.0 11.. [ H 11. 0.0909 O ] 1. ph log 1. ) NH ( 1.8 5 ) ( ) 5. 8 [ H O w ph log 1.0 1.8 1 5 ] 7.1 5.6 ( 7.1 ) 5. 1 c) C 5 H 5 N ( 1.7 9 ) w 1.0 1.7 [ H O ph log 1 9 ] 7. 5.9 ( 7. ). 1 0.0909 0.0909 d) NOH Strong cid Strong se H O (ssuming equl moles of ech) ph 7
Polyprotic cids re those with more thn one cidic proton. One emple is rsenic cid, H AsO, triprotic cid. It hs three equilirium epressions ssocited with its rection with wter: H AsO H O H AsO H O 1 5.0 H AsO H O HAsO H O 8.0 8 HAsO H O AsO H O 6.0 It is wek cid, whose ioniztion constnts decrese mrkedly for the second nd third dissocition rections. The concentrtions of ech of the four species cn e clculted s follows: The vlue of 1 will e used to clculte the concentrtion of H AsO, H AsO nd H O. These concentrtions will not e ffected y the other equiliri. The vlue of will e used to clculte the concentrtion of HAsO, using the previously clculted vlue of H AsO nd H O. The vlue of will e used to clculte the concentrtion of AsO using the previously clculted vlue of HAsO nd the concentrtion of H O, from 1. 9. If we hve 5.0 M solution of H AsO, wht re the concentrtions of ll of the species present? ) First, clculte the following concentrtions using the first ioniztion: [H AsO ] 5.0 [H AsO ] _0.158_ [H O ] _0.158_ [H AsO ] [H AsO ] [H O ] Initil 5.0 0 0 Chnge Equil. 5.0 1 5.0 0.158 [ H AsO ][ H O ] [ H AsO ] ) Wht is the ph of this solution? ph log( 0.158) 0.80 5.0
c) Clculte the [HAsO ] y solving for. 8.0 8 [H O ][HAsO ] [H AsO ] Use [H AsO ] nd [H O ] from prt ) [H AsO ] [HAsO ] [H O ] Initil 0.158 0 0.158 Chnge y y y Equil. 0.158 y y 0.158 y 8.0 8 8 0.158y 8.0 0.158 8 y 8.0 [ HAsO ][ H O ] [ H AsO ] ( y ) ( 0.158 y) y 0.158 y [HAsO ] _8.0 8 d) Clculte the [AsO ] y solving for. 6.0 [H O ][AsO ] [HAsO ] Use [HAsO ] from prt c) nd [H O ] from prt ) [HAsO ] [AsO ] [H O ] Initil 8.0 8 0 0.158 Chnge z z z Equil. 8.0 8 z z 0.158 z 6.0 0.158z 6.0 8.0 16 z.0 [ AsO ][ H O ] [ HAsO ] 8 ( 0.158 z) z 8.0 8 z [AsO ] _.0 16
. Ascoric cid, H C 6 H 6 O 6 is diprotic cid, with 1 1.0 5 nd 5.0 1. It is often revited s H Asc. Using this revition: ) Write out the equiliri of this cid with wter. H Asc H O HAsc H O 1 1.0 5 HAsc H O Asc H O 5.0 1 ) Write ll of the species tht will eist in 0.500 M solution of this wek cid, nd lel them s cids, ses or oth species cid/se [species] e i) H Asc cid ii) HAsc oth iii) Asc se iv) H O oth v) H O cid (oth?) vi) OH se (oth?) ) Clculte the equilirium concentrtions of ll of these species (ecept H O) nd enter them in the tle ove. Using the previous results [H Asc] eq [H Asc] 0 0.500 [HAsc ] [H O ] 1 1.0. [Asc ] y 5.0 5 1 [ HAsc ][ H O ] [ H Asc] 0.500 This leves only [OH ], which cn e found using the product of hydronium nd hydroide ions: [H O ][OH ] 1.0 1 [ OH ] 1 1 1.0 1.0 1.5 [ H O ].