how would your results have been affected if? a) each of the flasks you used for
ID: 1016387 • Letter: H
Question
how would your results have been affected if? a) each of the flasks you used for the standardization of NaOH had initially turned pink with the addition of phenolphthalein, but the color disappeared when you added KHP to each? (and you did not rinse the flasks and start over) b) there was still visible, undissolved KHP in the bottom of each flask when you finished with your titration? c) you stopped the titration immediately after the solution first became pink (the color disappeared almost immediately)?Name: athal SectionvtaroupCsw-3ou Instructor: ructor Section/Group: Chem1-30 Post-Lab Questions (Turn in to your Instructor experiment) with the post laboratory report for this . How would your results (calculated concentration of NaOH) have been affected if (provide an explanation for each answer): a. each of the flasks you used for the standardization of NaOH had initially turned pink with the addition of phenolphthalein, but the color disappeared when you added KHP to each (and you did not rinse the flasks and start over)? b. there was still visible, undissolved KHP in the bottom of each flask when you finished with your titration? you stopped the titration immediately after the solution first became pink (the color disappeared almost immediately)? c. 4. You are given 25.00 mL. of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). a. What is the molarity of the acetic acid solution? (Show your work.) What is the percentage of acetic acid in the solution? Assume the density of the solution is 1 g/mL. b. Experiment9 Acid-Base Titration 127
Explanation / Answer
a) each of the flasks you used for the standardization of NaOH had initially turned pink with the addition of phenolphthalein, but the color disappeared when you added KHP to each
Ans: IT means we have already NaOH in the flask so it will consume some of the KHP added which will not be available to be neutralized by NaOH. So it will need less NaOH solution then acutal for neutralization.
the cocnentration of NaOH calculated will be More than acutal
b) As the KHP is not dissolved so it will not react with NaOH so the volume of NaOH will be less consumed as expected so it will give higher concentration of NaOH than actual
c) Again less volume will be used then expected and it will give higher concentration of NaOH than actual
4) Moles of NaOH = Moles of acetic acid at point of neutralization
Moles = Molarity X volume
M1V1 = M2V2
M1 = molarity of NaOH
V1 = Volume of NaOH
M2 = molarity of acetic acid
V2 = Volume of acetic acid
35.75 X 0.195 = 25 X M2
M2 = 0.279 M
b) Moles of acetic acid = Molarity X volume = 0.279 X 25 / 1000 = 0.00698 moles
Mass = Moles X molecular weight = 0.00698 X 60 = 0.419 grams
So this is the amount of acetic acid in 25 mL or 25 grams of water
% of acetic acid = 0.419 X 100 / 25 = 1.68%
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