Question

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Period Name 20.00 ml of 0.2000 M acetic acid is reacted with 0.1000 M NaOH. (a) (b) Calculate the number of millimoles of acetic acid reacted. Calculate the number of milliliters of NaOH needed to reach the equivalence point. For each of the following conditions, calculate the following: [H''), [OH'), pH, POH, VnNaOH (c) (d) (e) After adding zero milliliters of NaOH. After adding 10.00 mL of NaOH. After adding 20.00 mL of NaOH. After adding 39.00 mL of NaOH. After adding 39.90 mL of NaOH. After adding 40.00 mL of NaOH. After adding 40.10 mL of NaOH. After adding 41.00 mL of NaOH. After adding 50.00 mL of NaOH. After adding 60.00 mL of NaOH. (h) (k) (0) After you have calculated each of the above, graph the pH versus the amount of base added. Oh yes, by the way, the Ka for acetic acid is 1.8 x 10 and you will also graded on how well this problem is set up and how easy it is to follow your logic/roblem solving skills -- if I can't read it -- it's wrong.

Answer 1

- R omy of o- 2 mol Acoll Ummol Aloh Ka=1.8xlos pka= logka pro=4.75 Encon} = 0.1M (a) MacOH = lummol b) ACOH + NagH na DAC tt 0.1 Patel raoh = MACOH = 4mmol a n c) Oml NaOH is ACOH ² Aro-tht egini c-cd un ca Cd untouchs Ka=cd² d = 1 1. Bxlos = 9.49x103 0.2 • [4+1 = (d = 0.2(9.49x103) = 1.9 X10 M pH = -log [ht] = (pH = 2.72] 2 pol=14-pH = ( pon = 11:281 1 [H+] = 1.9X163 milloh7 Kwie Tako klovy ht] TOH-] = 5.2781612/ Innaoh=011 ķ=20me d) loml/of oolmar Nach INNGOH = immol/ ALOH + NaOH > Nao AC + H₂O 3. pH=pkatlog. [NCOAC] =4.75+ [Acom /pH = 4.273 | ipon = 9:73] pH = -log[ht] [on] =1, 1878/6 tom! [H+] = 5 SMT VT = 20+10mL - V=30ml