Question

5. Neville Shortop, a budding bioorganic chemist, extracted ascorbic acid (Vitamin C, H2C6H606) from several lemons and oranges. If his purified solution of vitamin C was found to be 0.325 M, Calculate the pH, POH, (H2CH 061. (HC.H606) and [CH6061 at equilibrium. Make sure to CLEARLY CIRCLE your final answer for each item asked. The base ionization constants are as follows: Kb1 = 3.73 x 10"; Kb2 = 1.50 x 10-10

Answer 1

Date classmate concentration of Ascorbic (e page auid (th CHO y = 0.3 25 m till 4 The base lonisation Constants are 4 = 3.73x105 I K2 = 15 to 10 i The base boom of Astostic acid when two portons are removed ng cho 2 - It is a difasic acidi do Kb, indicate the base ionisation constant of SH602- all CH 02- +HO HC Hol- tot M y aid = 3.73710 Kaz Ky xka, to 4 T - Ka - 2014 2014 101 kb 3.073 105 3.73 Aence kao = 2.680 10- kb, indicate the base ionisation Constant of HC Hol HEGH trzo & H G A6 of K tu acid = 15x10 to Hence Kap. - - lol4 10 4 Since ka x Kb2 =10147 kb2 1sxlolo Tas- Hence Kq = 6.67X1-5. -6.67X105 base Hence Kbi = ty Kaz kb, kas sare He Hence Kai and Ka of ascorbic and the Go Hoo are Kai - 6.674105 kg = 2 68 1016 This inde cates that firstionisation Constant of ascostic acid (kas) y verry

high than Constant second sonisation classmate of alid (kant. e page Date Hence almost cur, all the 4t Tons will be obtained from binst conisation constant of acidita, because Scond lonisation. Constant Kas y Very small - Now the jonisation of concentration equibrium in the asconec alic or 325 m are of ho о o и кач sol 6 8 t h at of t760 ² HECHO +ho. initial conc. 0.325in final conce 10.325-60 m "at equilibrium HEHO +40 ² Cho thot xy yu gem of ascorbic alid ionise at equilibrum to form um HICHO and aim of thot, in borst ionisation constant kay Then in the and step ie in second step ionisation constant kan ym of HG Hol- sonise to boom. Ym of Hoat equilibrium and fu-ym of HGHOST Here at 0.325m and ay x K x , hence we Can neglect X w.got 0.325m in Kay and y w rot x in Kas

classmate Date Page | Now Kay = Herol- boty th 8767 .2 x2 -=6.67x105 = 2 0.325 0.325-20 Meglecting worit 0.325m Хосе иско • 325 м 2, РРХ (m tion, calculation . 42 42 in > Kai = - = 2 =00225 xkan - 02 -0.325X6.67410 2:1687105 - x = 1/2.165 410 25 - 4.654 103m. Hence that y = 4.65*10-3m. H6400 g = {0.325-n . - 0.325 - 4.657 1035 0.3 204m. \so Ka - (<b,22 Се ух ч со 1 - 10 Now meglecting by wort i bor approximation calculation. = Ka = you/ - 2.68x10 to => y = 2.68 810 10m concentration of EH 2 ] = 2.68410 and concentration of HGH of ty=x-Y = 4.65X163 2.68x1 to = 0.0046514 Hence PH=2.333 and potl=14-PH_14-2.332 3-0.66722352 =11-667 Now PH = -log(120ty -log(4.658103 -3-log4.65
Therefore from the above calculation the

pH is=2.333, pOH=11.667

[H2C6H6O6]=0.3204M

[HC6H6O6^-] =0.00465M , [C6H6O6^2-]=2.68×10^-10M