Question

pH PP he following Table to obtain a summary of various points in the titration of a Complete the following Table to o Triprotic Acid and the resulting PH. pH is calculated by Point in the titration of НЗА No Base added Major ions present in the aqueous solution HA Use K, and an ICE table Before the first equivalence point At the first equivalence point pH = (pK 1 + PK 2)*0.5 Between the 1st and 2nd equivalence points At the 2nd equivalence point Between the 2nd and 3rd equivalence points At the 3rd equivalence point As, OH Past the third equivalence point

Answer 1

(ao-hol (6) Before 1st equivalence paint Hz, Boy - H, A + НО ao hoo o Car-ho o ho ho → Buffer solution р, рк 1 (re (И) - "" () 6. [Hz A) ious posent = H₂ Á A H ₂ A (1) 1st yuivalnu point H₂A + BOH - H A + tho . на H2A > Att 13 или Колоо оч . На вип, как на поигри. (рға 14 pk12 ) , ious present are > Ht, OH, HAR, H₂ A, H₂A) (1) Between 1st and 2nd equivalence point on 14 вон — и, и и, о 'bo 544) o bo-ao ao ao и , gor -- НА 1 НО - < «) пер-46ъо Zarbo o pH p?» («ч Гия?) - Fa", I has ) ° pas - bo ) Monox lous о ao ho qo b bo- a TH₂A) ims are H2A , HA-21

(1) At and equivalence point 5. — 240 H₂ A & BOH M₂ Ä 200 H₂O H₂H + Boh HA 2 460 до 2 ио оололоо o ao 01 ao o o ao do TO HA? + H20 Khz He Atohi || 4.2 KM3 A3 + Ht ions prestat are; ut, oh HA-2, H₂A, A-3 (o) Betwound and 1.30d equivalence point: b. 2 и • РВОН — Н, А И,0 . Наи + Ra -НА АН,0 ao hoo ao ho-9000 4, bao o Goo во b. — 20. 0 . не? ? ? 0 —> A + И . . - 200 г . ) до , - 24 соо Saotho o h or 2003 hur2ao - Buffis solution r, p?s 3 Is A 1 = p = la (bar 2 ) o Н-2 J 13 aorhol 10м х (и от НА” , A3

« И А4 RUH — И, А + H, о л о 2 со vil at god equivalnu point 1) - Зао Н ОН — и, дио. Н, А + қой » И Н ,0 До 3dooo 246 долоо хооло • НА 2, 30И АЗАН, О. оосоо о oo a do " At will undergo hydeanlysis kha -2 our Go ao-x | Kha= kw Kha= [HA-2) Con] ГА) 10 ч - х 2 , үй <a aa - ао - Хя ак1с14 - Гали) - ро (ак"):(-м 1 ас - у нас ) . ho, 24 • 4. - lenker) | "14. 24 ( 1 м = 1 кг ) і см, ром и (rt : ои, ин, А-

(vii) Part 3 od equivalence point The > 390) : Moles of buse left= [bo- 3901 [OH-] = (bor 200) : pon - log (be - 390) PM 14- polt - 14 Alog (bo-3a0) |