to use in order that the titration will reqle 33.00 the Will you need 3. After p
ID: 590174 • Letter: T
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to use in order that the titration will reqle 33.00 the Will you need 3. After performing six titrations of tris with HCI, the followin calculated: 0.1014 M, 0.0989 M, 0.1008 M, 0.0970 M, 0.099 entratios o 0.1010 M. Use the Grubbs test to determine 95% CL. if any of the results should be discarded at the a. For the retained values, rep ort the average, standard deviation, and relative b. standard deviation. Three students (A, B, C) used bromothymol Student Indicator IH blue (BB) indicator for the titration of tris CI Molarity Avg Std D 0.099170.101395 0.001 0.1006Explanation / Answer
a) Among all the 6 molarities pick an outlier (value that is low when you arrange them in ascending order)
Data : 0.0970M,0.0989M, 0.0995M, 0.1008M, 0.1010M, 0.1014M
outlier = 0.0970M = X
Mean of the data = Xl = (0.0970+0.0989+0.0995+0.1008+0.1010+0.1014)/6 = 0.09976 M
Std deviation =2 = {[(0.0970-0.09976)2+(0.0989-0.09976)2+....+(0.1014-0.09976)2] / (6-1) }
2 = 2.74672*10-6 ; = 1.6573*10-3
Gcalc = |X - Xl | / = |0.0970 - 0.09976| / 1.6573*10-3 = 1.66533
For 95% CL at n = 6 Gtable = 1.822
Since Gcalc < Gtable the outlier 0.0970 should not be rejected.
If you repeat this step for any other outlier i.e) for any value of X from the data the result is same.
Hence there is no data that should be rejected among these values in data, no outliers in the data set.
b) Mean and std deviation are already found out in a)
Relative std deviation = ( / Xl)*100 = (1.6573*10-3 / 0.09976)*100 = 1.66128 %
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