Question

Lecture 17 At-home Practice .Titration calculations: A chemist titrates 40.00 mL of 0.100M acetic acid with 0.200 M sodium hydroxide. Calculate the pH at the four titration points described below. 1.7x105 for acetic acid. If needed, use K 1. Before any sodium hydroxide is added. 2. After adding 15.00 mL of sodium hydroxide. 3. After adding 20.00 mL of sodium hydroxide. 4. After adding 35.00 mL of sodium hydroxide.

Answer 1

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Answer (a) befor any v=ome so added. Nach Ka = 1.75 x 105 for acetic aid. HA + H2O - Intial moh. Il 0.100 c mol/l = 2 x moell 0.100-x A + H₂ot. o o tx tx on x x - lyx10- THgot JTA -J THAT Ka = 4 x2 0.100-2 = 1.7x10 (& is neglul x² = 0.100 81.78155 x = a = x= T2 = 0.100 X 17 X 106 1.7x106 V1.78156 1. 30 x103 TH₃O+ J = x mol/h = 1. 30x103 moel L. pH = -log[r₃0+ ph = -log11.30 x 10 32 p. pn - flog (1.30) + leg 10 -

pH = - 0.114 + 3 [pH = 2.886 0 adding 1r me of Naoh No 15me excess moles of OH HA TOH- - A- +H2O Immol 4 1.5 c/mmol -1.5 -1.5 +1.5x2=3 E Immol! 1.0 15*2=3 finital moles of HA= 0.040h x 0.100 mol 0.004 mol - 4 mmol Moles of Nao H added = 0.035 LX 0.200 mos = 3.0 x103 mol Moles of HA remaning = (4-3.0mmol o 1.8 mmol pH = pka t leg (salty Lacia pH = 4,75 + log [1.50 mmo] o 90 mmold pH = 4.75 + rog a51 pH = 4.75 tor22 TpH = 4.926 01176

6 adding 20 me of Nath this is the half-heutralization point So, pH = pka = -log (1.78105 [ pn = pka - 4.75] addling 35 me Naot HA TOH A +H2O I'm mol {.25 7 e/mmolay - 7 E Immol 3 o Inital moles of HA = 0.040 4x 0.100ml = z 0.004 mol 4 mmol. Fusted mole of NaoH added = 0.035 KX 0.200 = 0.007 mol 7 mmol moles of HA remaning a 7-4 = 3 mnola ph = pka + leg [salt] lauid /

Date Ресе при раза 27 Цүка ме - Де ти та ) 2 - = 4.75 + (о. 995 - 0, 417) 4.75 + 0-369 5. } & | 10