Question

Calculate the pH for each case in the titration of 50.0 mL of 0.100 M HCIO(aq) with 0.100 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 30.0 mL KOH? pH =

Answer 1

PH for weak acide For weak wear acids [+*]= /kac where ka= ionisation constant @ - Concentration. PH for buffer Sowtions : For wear acid buffer pH = pka + Log [salt) [weak alid] By using these Simulaes, we can calculate PH M 4. Given values are concentration of Holweak acid) = O volume of Helo - 50 ml Concentration of KOH = 0.1M Қ, go wlo - 3х10 1. PH before addition of koh. C - 01 M ka = 3x108

[s*). Skaic - 13x16 € x 1 No3 x10% = 0.547x10" PH = -log[**] = -log (0.547x104) --[ Log (10%)+log (0.547)] --[(-4701 -0.262) - - £4.26] = 4.26 2. after addition of 35 ml of OIM KOH Holo ton clot H₂O From balance equation i male of Hilo reacts with I mole on to form Imole of clo PH = pkat Log [clo-] [helo]

No.of mores - Molotily a Volonel nd) 1000 No of initid moles of HUO = 01x50 1000 0.005 OH = 0.12 No.of molek af NOOH = 0.1 x 25 1000 0.0025 Initial Chonge fina! HClO + oH - 0.005 0.0075 -0.0025 -0.0025 0.0025 0 CO + Ho 0 0 +0.0025 - 0.0025 Final molex of Helo = 0.0025 Final moles of clo = 0.0025 Fina voimme = von 0 H + vol. of WOOH - 50+25 = 75 ml pha= -log ka s-69 (37109 8-L03/2) =8-0-98 = 7.52 [wilo) = moles x 1000 Volome But here total volume is same for Helo & clo so we com directly use mules pH = pka +log (molesl volume) (moler og Hela / volume) = pkatlog (melek of llo) (moles of Hela) = 7.52 +log 10.0025) 70.0025) = 7.52 thog(1) : 7.52 +10) TPH = 7.527

Volume 3. afto addition of 20 m job molek 08 KOH = 0.1x30 = 0.003 1000 - tot Helo 0.005 -0003 F 0.002 toh 0.003 -0.003 Clo thao 0 - +0.003 0.003 0.0032 pH = 7.52 tlog ay To.002 = 7.52 + log (1.5) = 7.52 +0.18 = 7.7 4. after addition of 50 m) KOH molek og KOH = 0.1*50 = 0.005 1000 Hilo + uha - clo + to 0.005 0.005 F 0 0 0.005

clo tiho e Helo toh 1 C o-005 - + + Co-o05-) К = Снід Сон:] [clo kb can be calculated som Ka&rw Kх Кь: ku -4 -1 го:33х10 = 23х10 kb = to ka c 1x 15 ч 321080 : 3:3xоб”, (1967) 0.005-L k, j& Very Smill, S 4 -005-х о 05 3:3x161 - 0065 - э т”- о• 005г 3.3 xь" 1= | o-o5х3-3 хіба - 0-41 x1.5 = 4.1x105 [OH-] = 2 = 4.18105 From table

pOH = -log Com'] = log(41*10) = 5- Log(4.1) = 5-0.61 = 4.39 , pH = 14-poH = 1474.39 = 9.61 5. after addition of 60 ml oH moles of 101 = 0.1*60 = 0.006 1000 Helo + oH clotho 1 0.005 0.006 0 - C -0.005 -0.005 +0.005 - E o 0.001 after titration moles of or remaining = 0.001 Total volume = vol. Hclot Vol NaOH - 50 + 60 = 110 nel row-]= = 0.0091 0.00 x 1000 moles x 1000 volume (hal) 110 pOH = -log[om-] = -log(0-0091) = -12.04) = 2:04 pH = 14- POH = 14-2.04 - 11.96