3. The table below summarizes the volumes of Fe and SCN stock solutions that wil
ID: 1089045 • Letter: 3
Question
3. The table below summarizes the volumes of Fe and SCN stock solutions that will be mixed together to prepare the test solutions in Part A. Use the dilution equation to calculate the concentrations of Fet and SCN ions in each test solution before any reaction occurs. Enter the results of these calculations in scientific notation in the Test Solutions Data Table, on page 5. Hint: The final volume (V) of each test solution is 10.0 mL Volume of 0.0020 M Volume of 0.0020 M Volume of Distilled Fe(NO), Solution KSCN Solution Sample Test solution 6 Test solution 7 Test solution 8 Test solution 9 Test solution 10 5.0 mL 5.0 mL 5.0 mL 5.0 mL 5.0 mL 1.0 mL 2.0 mL 3.0 mL 4.0 mL 5.0 mL Water Added 4.0 mL 3.0 mL 2.0 mL 1.0 mL 0 mL 02015 Flinn Scientific, Inc. All Rights Reserved. No part of this material may be reproduced or transmitted in any form or by any means, electroaic or mechani- cal, without Inc. Student pages may be shared only by the instructor who purchased The Determination of K for FeSCN, Catalog No. AP7952, from Flinn Scientific, Inc and only by means of paper copies or a school website.Explanation / Answer
To calculate the concentration of Fe3+ and SCN- in the solution by using diultion equation which is
M1V1 = M2V2
where M1 is the concentration of the solution before mixing = 0.002M Fe(NO3)3
V1 = volume of solution before mixing
V2 = final volume of solution after mixing = 10 mL
M2 = concentration of Fe3+
for all test solution the volume of 0.002M Fe(NO3)3 added is 5mL so the concentration of Fe3+ will be
M1V1 = M2V2
0.002 x 5 = M2 x 10
M2 = 0.01 / 10
M2 = 0.001 M
The concentration of SCN - in each test solution is
Test solution 6
volume added is 1 mL of 0.0020M KSCN
M1V1 = M2V2
0.002 x 1 = M2 x 10
M2 = 0.002 / 10
M2 = 0.0002 M
Test solution 7
volume added is 2 mL of 0.0020M KSCN
M1V1 = M2V2
0.002 x 2 = M2 x 10
M2 = 0.004 / 10
M2 = 0.0004 M
Test solution 8
volume added is 3 mL of 0.0020M KSCN
M1V1 = M2V2
0.002 x 3 = M2 x 10
M2 = 0.006 / 10
M2 = 0.0006 M
Test solution 9
volume added is 4 mL of 0.0020M KSCN
M1V1 = M2V2
0.002 x 4 = M2 x 10
M2 = 0.008 / 10
M2 = 0.0008 M
Test solution 10
volume added is 5 mL of 0.0020M KSCN
M1V1 = M2V2
0.002 x 5 = M2 x 10
M2 = 0.010 / 10
M2 = 0.0010 M
Sample [Fe3+] [SCN-] Test Solution #6 0.001M 0.0002M Test Solution #7 0.001M 0.0004M Test Solution #8 0.001M 0.0006M Test Solution #9 0.001M 0.0008M Test Solution #10 0.001M 0.001MRelated Questions
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