1. For both of the following experimental procedures, determine the absolute unc
ID: 1022930 • Letter: 1
Question
1. For both of the following experimental procedures, determine the absolute uncertainty, relative uncertainty and % relative uncertainty associated with the final answer.
A. You are given a piece of metal and told to determine the density of the metal to help in its identification. You measure the object’s mass using an analytical balance and determine the mass to be 25.0135g. You take a 100 mL graduated cylinder and fill the cylinder with 50.00 mL of water. After placing the object into the same graduated cylinder, you read the new water level to be 82.50 mL. Calculate the density of the object and the uncertainty associated with that density.
B. You carry out a simple titration to determine an unknown concentration of HCl in a solution. You use a 25.00 mL Class A pipette to take an aliquot of the HCl solution. You then use a Class A 50.00 mL burette to deliver 25.10 mL of a 0.10000 (+/- 0.00001 M) NaOH solution to reach the endpoint. Determine the concentration of HCl and the uncertainty associated with that concentration in the unknown solution.
Explanation / Answer
A) Since the analytical balance can measure 4 significant figures, the uncertainty of the measurement of mass using this analytical balance is 0.0001 gm. Therefore, the mass of the solid weighed out is 25.0135 gm ± 0.0001 gm.
We will assume a non-standard measuring cylinder is used, since nothing is mentioned. On a 100 mL measuring cylinder the uncertainty in measurement is 0.5 mL. Therefore, the change in volume due to the solid, i.e, the volume of water displaced = (82.50 – 50.00) mL = 32.50 mL.
The density of the solid = 25.0135 gm/32.50 mL = 0.7696 gm/mL.
The absolute uncertainty in the measurement = 0.07696 gm/mL[(0.5 mL/32.50 mL) + (0.0001 gm/25.0135 gm)] = 0.0118 gm/mL.
Therefore, the density of the solid is 0.7696 gm/mL ± 0.0118 gm/mL (ans).
The relative uncertainty in the measurement = (0.0118 gm/mL)/(0.07696 gm/mL) = 0.0153 (ans).
The percent relative uncertainty = 0.0153*100 = 1.53 (ans).
B) For a 25 mL class A pipette, the uncertainty in measurement = 0.03 mL while for a class A 50 mL burette, the uncertainty is 0.05 mL.
The reaction is
HCl + NaOH -----à NaCl + H2O.
There is a 1:1 molar ratio, hence we can use the law of molarity as
Volume (acid)*concentration (acid) = volume (base)*concentration (base)
Therefore, (25.00 mL)*concentration (acid) = (25.10 mL)*(0.10000 M)
Therefore, concentration (acid) = 0.1004 M
There is an uncertainty of 0.00001 M in the concentration of NaOH. Therefore, uncertainty in the concentration of HCl is given as below:
([acid]/[acid])2 = ([base]/[base])2 + (V,base/Vbase)2 + (V,acid/Vacid)2
=[(0.00001 M)/(0.10000 M)]2 + [(0.03 mL/25.00 mL)]2 + [(0.05 mL/50.00 mL)2]
= 2.45*10-6
Therefore, [acid]/[acid] = 1.565*10-3
===è acid = [acid]*1.565*10-3 = (0.1004 M)*1.565*10-3 = 1.571*10-4
The absolute error in the concentration of the acid = 0.1004 M ± 1.571*10-4 M (ans)
The relative error = 1.571*10-4 M/0.1004 M = 1.565*10-3 (ans)
The percent relative error = (1.565*10-3)*100 = 0.1565 (ans)
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