Question

Phosphorous acid, H, PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. =0 pKalp Ka2 1.30 6.70 HOM -B- OH Calculate the pH for each of the points in the titration of 50.0 mL of 2.1 M H2PO2 (aq) with 2.1 M KOH(aq). before addition of any KOH: .523 after addition of 25.0 mL KOH: 1.3 after addition of 50.0 mL KOH: 1.48 after addition of 75.0 mL KOH: 6.70 after addition of 100.0 mL KOH: 1.94

Answer 1

page for - Answers According to to determine the given the ph for problem, each points. Given data, pkan pkal - 130 6. to. where, some of 2.4MHz PO, (99) with 2.1 MKOM (294). @ - Before addition of any KOH :- Before base is added the solution contains just weak acid. ton, Following the pH is governed by the following Equilibrium Het = HA+ gal - - nitial (M) x xx xx change (M) 9.1+52 X equilibrium (M).

page:02 then, Kai ? [HA] [M+ [H2A} li kar = 10 prail 10:30 XX0 (2.1-x) 101-30 (2.1-x) DEV = 107301911-X) 2.0 XV + 101.30 - 9:1515.30) = 0 by solving this equation, X0 300 then, den (H+) - 0.300M pH = -log[44] pH = -log(0.300) 201523 i before addition of any KOH, PH : 0.593

page:03 aften addition of asicom Koh :- at is first equilence point pH = pkai ptt a 4:30 1. after addition of 25. Om KOH.pH = 7:30) @ after addition og sooma kott :- It is first Equilence point. then, pH = 1/2 (pka + pkar) illa 11:30+ 6.70) = 1/2 (8) = 4.0 pH his after addition of so-om. Kot , pH 7.4.0 after addition of is. Omd Koh :- It is second Equilince point. : then,

page sou pt a Pkaz . 6.40 1. often addition of 75.0ml Kott, pH = 6-70 @ after additton os 100.0ml koht:- Given, mmol (M) = (10.0ml) X 9.1M a 210 mmol Initial mmol H2A = ($0.omh) * .1M - 105 mmol. then, HAT + H₂O Initial H2A + oH 105 mmol alommo! ~0 105 mmol cosmmol final. HA + OH A2 + H₂O lommo 20 105mmol o Initial final. 1o5mmol

page :05 affen veachon : mmol Abla - 105 mmol [Adm) = 108 mmol 60.0ml + 100 ML 105mmol 150 ml 0.700M .:(Ad-] 2. 0. 700M thena Art Hoo HA + oH Initial (M) 0.400 +2 +8 change (M) Equilibrium (M). 2 > 0700-X then, [HA] [OH-] - Kibr [A2

where, page06 Kb. : Ku) KA2 Κω και ο , Ко x x x Kal (0.400-x) 10-670 (0.700-1) [:: Kaz a 10 Pkar] . 10-6070 10 C . X -15 10 -14 10 - x 10670 + 10x - 7x105 = o. by solving this equation, ax + bu + C 20 Hen, X 1087 X 104 80, [OH-] = 1 878104 then,

page :07 ptta 140-poh = 14- (-logfont]] c!u- 3.728 - 10. 972 Ti after addition of 100.md of KOH, pt - 10.21a;)