Question

Buffers Lab (I ONLY NEED HELP WITH #5)

The 2 buffers are (acetic acid + sodium acetate) and (acetic acid + sodium hydroxide)

3:The two buffers that you will make in class are identical. If 5.00mL of a 1.0 M NaOH solution is added to one of these buffers, what would be the resulting pH after addition?

---Had to make a IACE table and what I got was pH=5.20

4: What is the total volume of NaOH (1.0 M) you should add to the other buffer solution (not the same one as in problem #3) so that it is no longer a buffer?

18.0 mL

5: The procedures for the lab instruct you to take the volume calculated in problem #4 and added it in 4 aliquots (divide the total volume by 4 to get the volume added per each addition). What will the pH be after each addition?

Please answer this question and explain steps if possible.

Answer 1

ANSWERS I Faking buffer, of solution of question number & anscurer, because both buffer are identical, the volume you calcelated is also taking from problems # 4. Butter of 200 mL of '9.12 M Cu, CoolH & 0.23M of cinco8 No & pkai 3-74 Problem #4 , volume = 50mL ( of os Om Na 1) Now this volume divide in 4 aliquats lil = 50 ml = 12.5 mL Now add ist aliquot = 12.5 mL, Now see who th calculate pH changing mol of Naou adding - 1215 ML X ( to md = 0.0125 mol Mole of Chycoon be forse adeling beseff ss. 1.0M Nagh 2 - 200 mL X0-12M = 0.024 mol De After adding NaOH ineet mo of CCOOH = 0-024 -0.0125 = 0.0 15 mol Mole of Chz co je na before adding Naoy - 200 ml x 0.23 M = 0.046 mol - After adding Nach, het mol of Cucosna z 0.046 +0.0125 mill 0.0575 mol New. volume of Butter = 200+ 125 ml: 2:12.5 m2 Now asing Hendersen Haselbalch equation : phia 3-74 +lega 2013 SME = 3-74 + log on to 0.0585 met [pH = 4.446) Die Somer -

ence pH = 4.448) after frost aliquets addihon. Now seca R 0.024mal 2 io fecond aliquet adding to same soketin cchero lume add first aliquot Male of Naon = 12 me x 0.18 M = 0.0 24 mol 7 Before adeling Na ok , the more of Cuscuela = 0.0115 mol Afles adding Na OH, the mole of cccom = 0.0115 mila -20. 012 5 mol Means only if precesent = -0.0125 mol Hence your solution is now, nd putter, LOH-] = 0.0125 mil = 0.0125 mol 12:54:25ML 2.25 ML Molarity of (ou.] : 5.555*10PM poke -log [S-55xii°2] = 4255 pH = 14- port = 14–4255 =18745 Now third aliquot of 12.5 ml Naut Total Md of Na OH = 0.0125 + 0.024 me) =0.0315 me! Total volume of mistere = 225ml + 12.5 m2 = 237.5mL 0.0365 m B 237.5mL poha - log Cou) - - 0.15369) 5 0.813370 pH= 14- port = 14-0-813370 13.186623 Now fourth aliquot of 12.5 mL Waoll Total mol of Naok= 0.0365 mol +0.0125 ml 50.049 mol Total volume = 237.5 mL +12.5 me = 250 ml [ok] = 0;081 mil = 0.196 M. 250 ml & FO3 = = 0.15368

pobt :-1 og cong = -log [0.1963 = 0.70774 pH = 14-poll = 14-0-70774 13.292 [PH= 13.292)