(Turn in to your Instructor with the post laboratory report for this experiment)
ID: 944115 • Letter: #
Question
(Turn in to your Instructor with the post laboratory report for this experiment) 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)? 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)? 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). 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 lg/mLExplanation / Answer
a- NaOH concentrations would be higher since there would already be some base in the flask.If we were to put the KHP in the buret, then the concentration would be lower, since it would neutralize the NaOH present.
b-Since there is still undissolved KHP in flask it means a smaller mass reacting with the NaOH than recorded. So the amount of NaOH would be lower than expected, and the concentration would be higher since our denominator decreases.
c- concentration of KHP will too low because we take lesser amount of NaOH for titration.Equivalence point will reach very soon.
4-CH3COOH(aq) + NaOH(aq) = CH3COONa(aq) + H2O(l)
V1=25ml, V2=35.75ml M2=0.1950
M1V1=M2V2=>M2 = M2V2 / V1 = 35.75 x 0.1950 / 25 = 0.27885=molarity of acetic acid
b-The molar mass of acetic acid = 60
M=0.27885
mass of acetic acid in 1000ml= 60g x 0.27885g/L=16.731g
1 16.731/1000
25 ml 16.731x25/1000=0.41827g
total volume of solution= 25 +35.75= 60.75ml
mass of solution =60.75ml x1 = 60.75g
% of acetic acid = 0.41827 / 60.75 x100 = 0.688%
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