Question

oncentration of NaOH solution, MOOCOM oncentration of HICHO solution, M O L UM AAAAA Data Sheet 2 LT CI Solution Son first d i Initial buret reading, ml. 0.00ML HO'rt PHOTO tre NOH (1) before iting action 0 .9 Zwerg 232. el before equivalence point 210 288 8 99 000990 200 400 4.02 LLM 2 a 4.42 4.56 4.10 4.83 4.98_ 01 - 8.00 8.00 1000 10.00 12.00 12.00 100_14.00 1600 16.00 1800 18.00 2000 20.00 2100 21.00 23.co 22.00 23 00 23.00 2400 24.00 2450 24 SO .00 35.00 bad Naruto heading 10.00 ml 0 0.20 ml before equivalence po 5:21 5.39 5.5ah 5.68 5.91 6.09 636 40 quilme point Golpevalence point 080 ate last din

how willI find the first determination of the lll. acetic acid I am really confused

Answer 1

3 For eq. pt mmoles of chroom = mnoles of Nach 0.0984xloo - 0.096 XV = v= 102.5 mL = Volume of NaOH = 102.5- 0.2 = 102.3 m2 mmoles of NaOH = 0.096 X 102.3 Cu cool & Noot z cu Loona tho = 9.8208 9.84 9-84-9.8208 9.8208 - 9.8208 = 7.44 pH = 4.74 t log 9.8208 0.0192 [not] = 3.56 x 108 [on] =2-81X. 107 choo Na + H₂O cy, coor & NaOH 9. 84 - 9.84 - 9.84 = [choo] = 9.84 = 0.0486 100 + 102.5 pH = 7+ 1 (pka + logc) = 1 + 1 / 4.74 + log 0.0486) - (salt hydsolyers) » M = 8.71 [rzot] = 1.94 X10 M [0] = 5.17x106m

Acetic acid k = 1.8x105 ro ol and pk = 4.74 let say we have 100 mL acetic acid Before adding NaOH Only Choot present a pH = 1 (bka - logc) pH = 1 (4.74 - log (00.00 984)) py = 9.87 pH = -log [ot ] - [hot] - 10% = 4.33 10 M [hot] [or] = 10-14 » [on] = 10th 5746xlM 133x103 2) After adding 10 m2 Naoh е см, СаъNo +цо mmoles Слл, Соъи + NODң 0.0984x100 0.00 6x 10 - 9.84 0.96 9.84-0.96 4 = 0.96 Butler en pH = pka + log [cacool (Acidic encor] pH = 4.74 + log 0.96 = 3.77 => [hot] = 1031 = 1.68x104m Lor ] = 10-14 5.94 x 10" 1.686104