Question

Consider the titration of 30.00 mL of 0.200 M AsO.. with 0.150 M HCI. K for H AsO4 = 3 x 102 K., for H:AsO. = 6 x 10 K, for H:AsO4 = 1.1 x 10-7 a) How many milliliters of HCl are needed to reach the 1, 2nd, and 3rd equivalence points in this titration? b) What is the pH of the 0.180 M AsO solution before any HCl is added. c) What conjugate pair is present at the first midpoint? d) Determine the approximate pH at the first midpoint. e) What conjugate pair is present at the second midpoint? f) Determine the approximate pH at the second midpoint. g) Estimate the pH at equivalence point #1. h) What conjugate pair is present at the third midpoint? i) Determine the approximate pH at the third midpoint. i) Estimate the pH at equivalence point #2. k) Sketch the titration curve using the calculated volumes and your estimated pH values. Label the axis. Label midpoints 1-3 and equivalence points 1-3 with the titrant volumes and their corresponding pH values.

Answer 1

Solution As per given in the question, Volume of Asoy?- solution = 30ml Concentration of Asou? == 0.2m Volume of 0.150 MHCl used = ? (ka. of thasoy are-ka, z6x103, kq2 = 11x167, kaz=341012 ] 0 Volume of hcl used =) As Oy²- + HCl → Hasou ter ( Titration for 2st Equivalent moles of. As Oq² = moles of her point) - 30x 0.20 = ', x0.150 4,2 60ml for que equivalent point HAS042-+ HC) → Haasoj te 30x0,20 = V2X0.150 (for Hel) ķ= 40ml

Volume for 3rd Titration - Azou thoi - bpoytin 30x0.2= 0.15 x V3 (for Hel) - Y32 40ml ри« o-12м :s o4° - - Ax04 - koлoute A8043 +@ HASO? oHº. В 4 بیا k we know .- = we Esopha 1 , P4, +4+3° 7 :7+ ' llog34102; +lyxboy (0-18) 57+ 5176 - 0.37 pH = 12.38 © at ast midpoint Asons thai - Haso ter As 043-/ HAS OG Conjugated pairs wiu present.

Vounne for 3rd titration - Hy Asoy the → brouter 30x0.2 = 0:15*12 - V3=40ml O At ist mid point: (A50437401 -HASO? +ci] [Asoy?"] = (1 as oy?] (Bufter solution) + log on to? - pliga -boy @elisme), 1152 for and mid point HASO42- that Hats ou ter HASOG 2- | LAS OG =) Conjugate base pair. ② pH at and mid point (Haso%] = (hasou] piz = plaz + log 1430 = pkk2= -log 11x157 = 6.95 O pH at Equivalent pointi :- at 1- Equivalent point Haso,&- will present thus plz Pkez tpk93, -log 1-4X107+(-log3x16'2) 2 9.24 conjugated Acid-base pair at 3rd Equivalent point HASO 2 + HCl → 113 poutal Стуя<te Pair 2 1, so - /h roy I (1) pH at 3rd mid point - H2Aso i tha bisoytcr [thaso6 1=(BASO4] . Ha pka, + logle Asoc] = pica, = -log 6x153= 2122 oy Asoy s pH at zrel Equivalent point pH = pka, tek&26 26 log ex 103 _ loglilx107%= 4.59 Log 2.22