Question

c) Calculate the pH if 95 ml of 10.0M HC (ag) is added to 200.0 ml of this buffer. For full creds, show your work in detail. (4) aj Calculate the pH if 9.5 mL of 10.0 M NaOH (ag) is added to 200.0 mL ofa freshly prepared solution of this buffer. For full credit, show your work in detail (4) e) You should see from the calculations that this buffer is better able to buffer against added acid than against added base. Looking at the relative amounts of H,CO, and NaHCO, used to prepare 200.0 mL of this buffer system and the resulting (base form]/[acid form] ratio in the Henderson-Hasselbalch Equation in part a), explain why the buffer is better able to buffer against added acid than against added base. (2) Does this data show that a buffer resists changes in pH caused by (compared to an "unbuffered" solution)? 5.A carbonic acid (H;CO)/bicarbonate (HCO, ) buffer system is a major factor keeping human blood pH in the range of 7.35 to 7.45. This is an interesting buffer system, and we will represent it here as the following equilibrium: Module 6-Lecture 2 H&O' (aq) HCO, (aq) H2O () H2CO, (aq) + Suppose 200.0 mL of a carbonic acid/bicarbonate buffer solution is prepared using 0.100 moles H2CO3 and 1.12 moles of sodium bicarbonate (NaHCO). a) Use the Henderson-Hasselbalch Equation to calculate the pH of this buffer. (It should be somewhere between 7.35 and 7.45.) (Ka of H2CO3 4.3 x 10) (3) b) Is the pH of this buffer system outside of the normally recommended range (pK, + 1) for a buffer? (1)

Answer 1

Here all pH values are calculated using Henderson's equation for buffer solution. If there is any problem plz let me know.

HA2 Co)t30 Ca) H, cog (a) + H20() -thie butfen to paupaned b ing 1.12 moles of Naticos and 2 00 ml. volume f buffer- Dipcos) mo. of molos (meentration o H2 co Tolad Volume O-1 youe 2 gv 0-028(M) mol/200 and 2 00 buffer sot Honderson For 7Ka oy Liau Taid PH - O.a28 4.3 X10 =7.42. bo10m7-3rt 7.95. see Aoe pHrange So can r0c0mmen ded pt 0v pka t omag Le.9 p arye should oe bekaen but fare 1o7-27 5.37 So cam Say that pH .42 mo iommended pH naye in #li boordentine 14 in

in thn with aud TUat buffer hen it Pnoduee Consponliny asa congugatabane 40 it's dhe and back eangnga te noith will nac For eanpu Rere Heaz o ase e (a1) 442 3 (ag) 1-12m1 0-095m01 no. molss o.1+0.095) mo (1-12-0-09) aftan addirs 43A o.19S mol 1.025 mol 41 ane Hu ia Sol HC Contoina sed 1 0! 10 mol of L1 - '. mo I u5.b oilh Hu Peads Now of 412003 Produce ana tonc TH co mol (200+9.5) mL "1Ka tos Heo) THe co +10s G 00+945) 1.025 O.195 709.

we add 9.5 ml contains and 9.SX10 0.095mol 1g00 ith o .015mo) auid 5ill oana +0 prddur Car) 4-12 eo3 (ar) initias 1.12 mol 0.095ma1 0.1 mol finat (1.12 to.0T ma 1-9.095)m mol - .21 Smol. =0.005 m1 New conCA . 44 TAzcos 0,005 mol SO JHC05) 1-21Smal 2o095)m pka PH 1 M15/2078) 6.37+ 8.76 thol p alue changes ApH -6.6-4-42) 7.09-7-42) baohen MaoH is add ed) -194. -O.33 wohen He ts adaea) nesioto the charge ot pH S0 the bu beter in auidic The eaconit buffr -hin buffen is basic oro a ( 2c03is He lative trons gneaten eamcentratim , s the pl change when aid is value 7.4 woak auiN O and base and hora com estst added to 11.