BCHM461 Spring 2002 Solutions to ph and Buffers Problem Set

Transcription

1 BCHM461 Spring 00 Solutions to ph nd Buffers Problem Set 1. Here I will provide nswers for the pk = 6.5 exmple only () For this problem, s well s for problems (b) nd (c), it is convenient to use the Henderson-Hsselblch eqution: ph = pk + log([a - ]/[HA]) = pk =6.5 becuse when [A - ] = [HA], log([a - ]/[HA]) = log1 = 0. (b) now log([a - ]/[HA]) = log10 = 1, ph = pk + 1 = 7.5 (c) now log([a - ]/[HA]) = log0.1 = -1, ph = pk - 1 = 5.5; A ph pk (d) here we cn use the following eqution: = 10 [ HA] ph ph - pk [A - ]/[HA] 4.5-1/ / / / /1. It is convenient to use the sme eqution s in Problem 1(d). () when ph = pk, 10 ph-pk = 10 0 = 1, the molr concentrtions of cid nd conjugte bse re equl (b) when the ph is less thn the pk, ph-pk < 0, hence [A - ]/[HA] =10 ph-pk <1 HA is the predominnt species (c) when the ph is higher thn the pk, ph-pk > 0, hence [A - ]/[HA] = 10 ph-pk >1 A - is the predominnt species (d) When the ph is 1.0 ph unit bove the pk, ph-pk = 1, [A - ]/[HA] = 10, the rtio of [HA] to [A - ] is 1/10; when the ph is 1.0 ph unit below the pk, ph-pk = - 1, [A - ]/[HA] = 10-1 = 1/10, the rtio of [HA] to [A - ] is 10/1; (e) Likewise, when the ph is.0 ph units bove the pk, ph-pk =, [A - ]/[HA] = 10 =100, nd the rtio of [HA] to [A - ] is 1/100; when the ph is.0 ph units below the pk, ph-pk = -, [A - ]/[HA] = 10 - =1/100, nd the rtio of [HA] to [A - ] is 100/1; 1

2 3. () COOH COO - COO - + H 3 N C H + H 3 N C H H N C -- H CH 3 CH 3 CH 3 pk 1 =.34 pk = H 3 N-C H 4 -COOH + H 3 N-C H 4 -COO - H N-C H 4 -COO - fully protonted zwitterionic deprotonted (b) The reltive concentrtions of ll three forms of lnine re controlled by the two equilibrium equtions: [ + H 3 N-C H 4 -COO - ]/[ + H 3 N-C H 4 -COOH] = 10 ph-pk1 [] [H N-C H 4 -COO - ]/[ + H 3 N-C H 4 -COO - ] = 10 ph-pk [b] You cn use these equtions to clculte the rtios of the vrious forms for the prticulr vlues of ph nd pk s. However, the nswers to these problems cn be obtined from simple qulittive nlysis, s follows: (1) ph = 1.0 is smller thn both pk 1 nd pk, so + H 3 N-C H 4 -COOH is the predominnt form. Indeed, becuse ph < pk 1, the concentrtion of the fully protonted form, + H 3 N-C H 4 -COOH, is greter thn tht of its conjugte bse, + H 3 N-C H 4 -COO -. The concentrtion of H N-C H 4 -COO - ( conjugte bse for the zwitterionic form) is even smller (prcticlly negligible!) thn tht of. + H 3 N- C H 4 -COO - becuse the ph is lower thn the pk (pk ) for the equilibrium between these two forms. () Since 6. > pk 1, the concentrtion of the zwitterionic form is greter thn tht of the fully protonted form (its conjugte cid). Since 6. < pk, the concentrtion of the zwitterionic form is lso greter thn tht for its conjugte bse, H N-C H 4 -COO -. (3) Since pk 1 < 8.0 < pk, the nswer is sme s in Problem 3(): the zwitterionic form is predominnt. (4) ph 11.9 is higher thn both pk s. This mens tht the concentrtion of the deprotonted form (H N-C H 4 -COO - ) is significntly greter thn the concentrtion of the zwitterionic form (11.9 exceeds pk by more thn.0 ph units). The concentrtion of the fully protonted form is prcticlly negnigible

3 (compre pk 1 =.34 with ph = 11.9!). Therefore, the deprotonted form is predominnt t this ph. (c) Let the totl concentrtion of Al be c nd let x denote the frction of Al in the zwitterionic form t ph = pi. Recll tht pi = (pk 1 +pk )/. From eqution [] (see bove) the concentrtion of the fully protonted form is then [ + H 3 N-C H 4 -COOH] = x/10 ph-pk1 = x/ = x/473. The concentrtion of the deprotonted form cn be determined from eqution [b] s [H N-C H 4 -COO - ] = x. 10 ph-pk = x = x/473. From the mtter conservtion principle, the totl concentrtion of Al is the sum of the concentrtions of ll three forms (we neglect the unchrged form): [ + H 3 N-C H 4 -COOH] + [ + H 3 N-C H 4 -COO - ] + [H N-C H 4 -COO - ] = c or using the bove reltionships: x/473 + x + x/473 = c which leds to x (1 + /473) = c The frction of Al in the zwitterionic form is then x / c = , or 99.96%. 4. ph =.0 corresponds to [H + ] = 10 - M. When 1M HCl contins 1M H + nd 1M Cl -. This mens tht in order to mke 10 - M H +, the originl 1M HCl hs to be diluted 100 fold. Since we re dding HCl to 50 ml of wter, 100-fold dilution will be chieved if we tke 50mL/100 =.5 ml of 1M HCl. The OH - concentrtion in pure wter is [OH - ] = 10-7 M; t ph= (i.e. [H + ] = 10 - M) it cn be clculted from the ion product of wter: [OH - ] = M /[H + ] = 10-1 M. Now ssume we dded.5 ml 1M NOH insted of HCl. This mounts to 0.01 M concentrtion of NOH, which then provides 0.01 M OH -. The H + concentrtion cn be clculted from the ion product of wter: [H + ] = M /10 - M = 10-1 M, hence the ph = () [OAc - ] = [HOAc] = 0.1M; from the Henderson-Hsselblch eqution: ph = pk = 4.76 (becuse log([oac - ]/[HOAc]) = log1 = 0). [H + ] = 10 -ph M = M (b) 0.01 M HCl will dissocite completely into 0.01M H + nd 0.01M Cl - ; 0.01M H + will essentilly completely rect with the bse OAc - to form n cid, HOAc. As result, the concentrtion of OAc - will decrese by 0.01M to 0.09M; the concentrtion of the cid will increse to 0.11 M. The Henderson-Hsselblch eqution then gives: ph = pk + log(0.09/0.11) = 4.67, very smll chnge in ph! [H + ] = 10 -ph M= M. (c ) Initil conditions: [OAc - ]/[HOAc] = 0.0/0.18=1/9; the Henderson-Hsselblch eqution gives: ph = pk + log(1/9) = 3.81, nd [H + ] = M; ph = After 0.01 M HCl hs been dded: [OAc - ] will decrese to 0.01M, [HOAc] will increse to 0.19 M, so tht ph = pk + log(1/19) =3.48; nd [H + ] = M. Still smll chnge in the ph. 3

4 (d ) Initil conditions: [OAc - ]/[HOAc] = 0.18/0.0=9/1; ph = pk + log(9/1) = 5.71; [H + ] = M. After 0.01 M HCl hs been dded: [OAc - ] will decrese to 0.17M, [HOAc] will increse to 0.03 M, so tht ph = pk + log(0.17/0.03) =5.51; nd [H + ] = M. A smll chnge in the ph: there ws enough bse to resist the ddition of this mount of HCl.. (e) The initil concentrtions of protons nd the ph re sme. The dded 0.15M HCl is completely ionized, which results in 0.15M H +. (b) 0.1M H + will rect with the bse, the remining 0.05M H + will determine the ph: [H + ] = 0.05M; ph = The solution becme very cidic! (c) Similrly, only 0.0 M H + could be neutrlized by the vilble mount of OAc -, the remining 0.13 M H + will lower the ph to ph = 0.89, resulting in very cidic solution. (d) Now there is enough bse to completely neutrlize 0.15M H +, thus mking dditionl 0.15M of HOAc. The remining concentrtion of OAc - is 0.03 M: [OAc - ]/[HOAc] = 0.03/0.17 = 3/17; hence ph = log(3/17) = 4.01; [H + ] = M: this solution. 6. (1) Let us first clculte the ph for 1M cette buffer. HOAc OAc - + H + pk = 4.76 The concentrtion of H + ions before the cid ws dded ws [H + ] = 10-7 M. The totl cette concentrtion dded to the solution is 1.0 M. Let x be the totl number of moles HOAc per liter tht dissocited into H + nd OAc - ; the finl concentrtion of HOAc is then [HOAc] = 1M x. Since the dissocition of ech molecule of HOAc releses one cette ion nd one hydrogen ion, the finl concentrtion of OAc - will be [OAc - ] = x; nd the increse in the concentrtion of H + will lso be x, so tht the finl concentrtion of protons is now [H + ] = 10-7 M + x. Now we cn plug ll this into the equilibrium eqution (Eqution [1]): K = + [ H ][ OAc ] (10 = 7 [ HOAc] 1 x + x) x Here, for simplicity, I dropped the molr units ssuming tht x is lso expressed in M, s well s K. After some lgebr we get: x + x ( K ) K = 0; Solving this eqution for x gives: 7 7 ( 10 + K ) + ( 10 + K ) + 4K x = We cn now plug in prticulr vlues for K for cetic cid nd get: x = M. So, the equilibrium concentrtions re: [HOAc] = 1.0 M x = 1.0 M M = M [OAc - ] = x = M 4

5 [H + ] = 10-7 M + x = M hence ph =.38. Similr clcultions give For 1M dihydrogen phosphte (K = M): x= , nd ph = For 1M mmonium ion (K = M): x = ; nd ph = 4.63 () Now let s titrte our buffer. The strting concentrtions re: [HOAc] = M;[OAc - ] =0.004 M; [H + ] =0.004 M. () First we dd 0.05 M OH M OH - will be neutrlized by the vilble mount of H +. The remining 0.05 M M = M OH - will rect with HOAc, thus reducing [HOAc] nd incresing the mount of the conjugte bse, OAc -, so tht the equilibrium concentrtions will become: [HOAc] = M M = 0.95 M; [OAc - ] = M M = 0.05M; hence ph = log(0.05/0.95) = the ph incresed by more thn 1.0. (b) Now we dd 0.1M OH -. This mount of OH - will essentilly completely rect with HOAc. The equilibrium concentrtions will then be: [HOAc] = 0.95 M M = 0.85 M; [OAc - ] = 0.05 M M = 0.15M; hence ph = log(0.15/0.85) = Despite the two-fold increse in the mount of dded OH-, the increse in the ph is less thn one-hlf of the previous one. nd so on (see the tble). Totl OH- ph equivlents dded ph squres -- clcultions in Problem 6(1-3) line -- eqution in Problem 6(5) OH- equivlents 7. Method 1. (1) The molr mount of the cid (H PO 4 - ) in 50 ml initil solution is 0.1M L = moles. 5

6 From the eqution [H PO 4 - ] / [H PO 4 - ] = 10 ph-pk, we cn determine the molr rtio of bse/cid. Given the molr mount of the cid (H PO 4 - ) present in 50 ml of the initil solution ( moles), the molr mount of the conjugte bse to be dded is ph-pk moles. Dividing this by the molr concentrtion (0.1M = 0.1 mole/l) of the stock solution, we then find the mount of solution B tht we hve to dd: ph-pk L = ph-pk ml. Plugging in prticulr vlues for ph, we obtin: PH Added volume of solution B ml ml mL Note tht the chnge in the volume (e.g. when 69 ml of solution B re dded to 50 ml of solution A) s well s the subsequent ddition of wter, to bring the finl volume to 00 ml, will not chnge the molr rtio of the cid to bse, nd therefore will not ffect the ph. () The ph will be equl to pk = (3) When we dd 5 ml of B to 50 ml of A, the molr rtio of H PO 4 - to H PO 4 - is [H PO 4 - ] / [H PO 4 - ] = 5/50 = 1/10 (Since the numericl vlues of molr concentrtions in both stock solutions re equl, the number of moles of ech compound dded to the mixture will be proportionl to the dded volume of the corresponding stock solution). Then ph = pk + log(1/10 ) = Likewise, if we dd 100 ml of B to 50 ml of A, we get [H PO 4 - ] / [H PO 4 - ] = 100/50 =, nd ph = pk + log( ) = Method. 100 ml of 0.1M TRISbse contin 0.1M. 0.1L = 0.01 moles of TRISbse. Assume we dded x moles of HCl, tht completely ionized in wter to give x moles of H +. These protons rected with x moles of TRISbse to mke x moles of TRIS. H +. The remining mount of moles of TRISbse is then x. The rtio of molr mounts of TRISbse nd TRIS. H + must stisfy the following eqution: ph - pk (0.01- x)/x = 10 Solving it for x, we obtin: x = 0.01/( ph pk ). For ph = 9.0, x = 0.01/( ) = moles. This corresponds to moles / 0.1M = L = 10.7 ml of 0.1M HCl. Similrly, for ph = 8.0, x = 0.01/( ) = moles, 54.6 ml of 0.1M HCl. For ph = 7., x = 0.01/( ) = moles, 88.4 ml of 0.1M HCl. The results re summrized in the tble below. 6

7 PH Added volume of 0.1M HCl ml ml mL 8. According to Problem 7 Method 1 exmple (), we mixed 50 ml 0.1 M H PO 4 - nd 50 ml 0.1 M HPO 4 - nd then djusted the totl volume to 00 ml by dding wter. The molr concentrtions of both cid nd bse re now 0.05M (four-fold dilution). 1 ml 1M HCl dded to this solution will be diluted ~00 fold. It will completely ionize into M H + nd M Cl - ions. These dditionl protons will essentilly completely rect with HPO 4 - to produce H PO 4 -. The finl concentrtions then will be [H PO 4 - ] = 0.05M M = 0.03 M nd [HPO 4 - ] = 0.05M M = 0.0 M. Using the Henderson-Hsselblch eqution, we obtin: ph = pk + log(0.0/0.03) = If we dd 10 ml HCl, the dded concentrtion of H + will be 0.05 M. There is not enough bse to neutrlize this mount of hydrogen ions: 0.05M HPO 4 - will essentilly completely rect with H +, the remining 0.05 M H + will then determine the ph: ph = -log(0.05) =