Question

Phosphorous acid, H2PO2 (aq), is a diprotic oxyacid that is an important compound in industry and agriculture. pKalp Ka2 1.30 6.70 но —р —он Calculate the pH for each of the points in the titration of 50.0 mL of 1.5 M H2PO2 (aq) with 1.5 M KOH(aq). before addition of any KOH: after addition of 25.0 mL KOH: after addition of 50.0 mL KOH: after addition of 75.0 mL KOH: after addition of 100.0 mL KOH:

Answer 1

given, pka So Kai = 1.30 = 5x10-2 plaz z 6.70 u. Ka = 2x10-7 Mapoz = some Ckon a 1. 59. си, бо 4- = 1.5 L Since Hapozis diperbic acid and Ka, Kaz so [ut] will be considered due do dissociations & of Po3 only. So let a motllus dissociation of epos takes place ou, napoz Kay Hapoge & HO c(t =o) 1.5 CeC+=0) 1-5** * * Kai - Ch2 por ⓇJ CH®] [M₂ Po ] К = x x (1-5-x) 5x16-2 (1-5-x) = x2 0 x2 + 5x10-2 x -1.5X5 X10 = 0 + x 2 + 5x10-2x - 7-5810-2 zo x² +0.05x - 0.075 0 -0.os & Ico.us)? 4x1x (-0.075] 2x1 = 0.25 pha tag [ne]= -log(0.25) = 0.602

Case-I Ckong is added = when no base 0.602 126602 Case-II - After adding 25ml of Koh wish kon. H₂PO2 will reacts as, H₂PO & KOH KIPO3 +40 n (t=0) 75 5 37. o melt=te) 37.5 37.5 37.5 How in the solution only H₂PO2 and kn poz is present so it will bct like a buffer solutions [H₂ pog] Tacid = 37.5 mmol (n= not milliml) and, [Kh2po, a [Sall] 375 manele How from henderson hasselbalch equation. pn=pka, & leg -sall I where the pos] [AC s) 1 pH = 1.3 I case in After adding so me kon Base will neutralise the acid to Ist equivalent and sclution bare only behare like! sothof amphipontic Species (kn, peg. sense, ph of pka, +pler amphipoodic species = -2 1.3 + 6.7 2 11 = 4

Case- II Adding Ts me of KOH Con Kon = 112.5 mmol) у кн, род Иую Р на РМ + он my 5 112.5 n (to) a (4=to o 375 75 75 Kon further senects with KH, PO3. KH₂PO3 + Kon 75 375 K₂ HPO 3 + H₂O 0 0 nct=t n(+=te) 37.5 37. 5o so this is a buffer solf, so ph will be | [salt] = [Kanpog] pH = 100 (sall] pka z - 37.5 mm 125 me. 2012 1 and Acid] = [ 4000] pH = 6-2 Case Is Adding loome of kon. Now on addling loome od kon it will neutralises 2 mol all the hydropen to form salt with no replacable hydrogen [H3 Poz] 20 [colleghPog] = 75 1 so on hydrolysis of d. Kahpo, ph of sole will be pH = 7+ Cakes plage) z 7 + £ £ 6.7 + leg (2)] - I pn = 1002 So