Question

Calculate the pH for each case in the titration of 50.0 mL of 0.130 M HClO(aq) with 0.130 M KOH(aq). Use the ionization constant for HCIO. What is the pH before addition of any KOH? pH = What is the pH after addition of 25.0 mL KOH? pH = What is the pH after addition of 35.0 mL KOH? pH = What is the pH after addition of 50.0 mL KOH? pH = What is the pH after addition of 60.0 mL KOH? pH =

Answer 1

The ka value of Helo is 3.0 x10-8 So Pkaz – log, 6 (3.0 x 108) 7.523 So before addition of RoH [elo Pka 2 PH - logio [Helo - ③ Now [elo] 2 [Ht] for the reaction Helo Kali Httelo Then equation ② changes to Pka 2 2pH + lsg [helo] 2 pH 7.523 +0.886 4,2045 = [ht] = 6:2445x10 m * Addition of 25.0 ml KOH Moles of oth on asml 0.130 m kohool", 25 x 0-130 1000 myles 23:25 x 103 modes Moles Htion in 0.130 M Helo soin, 50x6:2445x ios 1000 - = 30122x106 moles Moles of exeers olf (3.25 x10 3 - 3,122x1066) 23,2468x10 3 moles [OH-] = 3.2468x10-3 0.05+0025 0.0433 Molto Pon 2 1.363 2) | PH = 12.637)

Addition of 35 ml kot 1000 Moles of oh in 35 ml kott 35x 0.13 smoles= 4.55x10 mole - Moles of it in some Helo z 3.128 x 10 moles, moles of ot in exeen, (4.551103_3-122x106) moes 24.5468x10 3 moles Protes 190 log tu. sacerro [OH-] 4.5468x10 3 - M 20.05349 m 0.05+0.035) PH 2 14+ logo (0.05349) = 12.7 28 50XD: 130 : 1000 moks 6.5x10 moles 1 mole Addition of so me koh Moles of ot in some KOH= 5 moles of Ht in some Helo = 3,122x10 moles moles of ot in exeensa (6.5x10 3 3-122x106) moles 26.4968x10-3 moles 6.4968X10 3 ooostoios a 0.064968 m PH₂ 14+ log (0.064968) = 12,812

- 3 1006 - Moles, 6.7.8 X10 moles Addition of Gooo ml koh Moles of ot in 60 ml KOH 60X0-130 Moles of At in so ml Helo = 3,122 x 10-6 moles Moles of ot in exeers= (7.8x10 3_ 3.122 x 106 )moles = 7.79680103 [OH-] = 7.7962x153 M 2 0.07088 M - 0.05+0.06 PH2 14 + logo (0.07088) = 12.850