Question

4. (20 pts) 25 mL of 0.12M solution of HClO2 (pKa=1.92) is titrated with a 0.087M solution of KOH. Find the pH - Initially before adding any KOH After 8.5 mL of KOH is added At the equivalence point of the titration After 39 ml (total) of KOH is added

Answer 1

Initiall 4). 25 ml of 0.12M HClO2 a) Before adding any Koh HClO (106 + H+ t=0 C 0 0 at eaim c-ca ca ca 2 degree of dissociation Ci concentration = 0.12M pka = 1.92 -log Ka = 1.92 [Ka=0.012 L 8. 25m C- Cd HO Ka - kaa = ca ² c2² tkad – ka = 0 Putting values; 0.12d²+ 0.0128 - Ra 0.012=0 Solving quadratic equation ; d = – 0.012 56.012) 2 - 4 (012) I. 2(0.12)

solving for a, we got d = -0.012 20.07683 260.12) Neglecting the value of a & cannot be negative Id= 0.27016? M [ht] = C&= 0. 12X 0.27016 CH+] = 0.0324M ) pH = -log [ht] pH = 1.489] Initially before adding any kol PH is 1.489 b) 8.5 mL of 0.087 M KOH 8.5mt x 0.087 mol = 0.7395 monol 8.5ml xo7 KOH 25mL of 0.12M HCla 25 myx 0.12 mol 3 mmol of HClO The reacties taking place; HC40₂ +KOH CIOK + H₂O. 1. 1 mol Kol react with Imo Hllo to produce 1 mol Clok (salt) will 0.7395 mmd Kol react with 0.7395 mmol Hlly to produce 0.7395 mmol El

Moles of Helo Remaining = Original - Reacted = 3 mmol - 0.7395 mmol = 2.2605 monol of Heloz final volume = 8.5ml +25mL = 33.5mL Conc. of Hclo₂ = moles = 2.2605 mmol Volume 33.5 mL At equivalence point, 3 mmol of HClo, will react with 3 mmol of koh to froduce 3 mmol of cloak (salt) * Only 3 mmol of clą will remain at equivalence point. Totod Volume of Kol = moles added molarity 3 mmol 0.087 poll Volume Koll = 34-4827mL = 0.067478 M Moles of Koh remaining = 0 (All reacted) 0.7395 mmol Total Volume = KOH + HCla = 34.4827+ 25 m2 M moles of ClO₂K = Pooded Conc. of Clo, K = = 59.4827mL 0.7395 phonol = 0.022M 33.5 phL Using Handerson equation, cląk PH=pka + log [salt Treid) x HClO2 = 1.92 +log/0.022 10.067478 . Conc of clok (salt) = 3 phmol 59.4827AL [ Conc. of Clok = 0.0504 MI - At equivalence froint only 59.4827 m2 of ClO₂K is present. "ClO₂k is a salt of weak acid and strong base pH = 7+ 1 [lka tlogo] conc. of sat + + ++ [1-42 +63(0.150P] (pH = 7.3114 TpH = 1.434] So, pH after addition of KOH was found to be 1.434 & At equivalence point; KOH + HClO kloak 3 mnd 3 mmo 3mmel + H₂O pH at equivalence point

d) After 39mL Koh added After equivalence, we have addad 0.393 mmol KDH 1 mol OH react with Imm) wild (Hot). to form neutral H₂O To reach equivalence pt. Kot used = 34.4827mL . We added = 39mL Excess Koh added = 139-34.4027m = 4.5173mL excees moles= 4.5173 mKX 0.087ml 2.904 Not H + react with 2.904x106 to form neutral water [OH] remaining = 0.393-2.904x160 = 0.39299 mmolot = 0.393 mmol Conc. ef excerskoH= moles of KOH to tal volume lonco = 0.39299 59.4827 +4.5173 [or] = 6.140579 80 3M = 0.393 mmol 59.4827 +4.5173 ML Vol exer Volume at equivalena pt. Koll added pOH = -log[ole] = 2.2118 pH = 14-poh TPH = 11.788 .. PH aff after excers addition of KOH is 11.788 Torben 6.1406 xlo-3 matt At equivalent pt; PH= [ht] = 7.3114 antilog (-7.3114) 4.882816-OM Ht tot ho (neutral) At equivalent point, moles of H+ = molarity x volume (equivalent pt.) = 4.682x108 mol x 59.4927 m/ = 2.904 xloo mmol.