The chemicals that will be used in this experiment are: 2.0 x 10-4 M KSCN Potass
ID: 492660 • Letter: T
Question
The chemicals that will be used in this experiment are: 2.0 x 10-4 M KSCN Potassium thiocyanate is the source of SCN- .
0.20 M Fe(NO3)3 Iron(III) nitrate is the source of Fe3+. Clean and dry 6 large test tubes, then label them #1 - #6. Use a 5 mL Pipetman to transfer 5.00 mL of 2.0 x 10-4 M KSCN solution into each test tube.
Each of the six test tubes should have 5.00 mL of potassium thiocyanate in it at this point. Test Tube #1 To the potassium thiocyanate solution in test tube #1, add 5.00 mL of the 0.20 M Fe(NO3)3 solution.
This solution will be used as the standard; SCN- is the limiting reactant, and the large excess of Fe3+ converts all of the SCN- into FeSCN2+. This sample will have the maximum concentration of FeSCN2+ complex and the most intense color.
Test Tube #2 Pipet 10.0 mL of the 0.20 M Fe(NO3)3 solution into a 25 mL graduated cylinder. Add distilled water to get 25.0 mL of solution. Mix thoroughly.
What is [Fe3+] in this 25.0 mL solution? Add 5.0 mL of the 25.0 mL Fe3+ solution to test tube #2. Remember that the tube already contains KSCN. What is [Fe3+]o for test tube #2? What is [SCN- ]o for test tube #2? Test Tube #3 Discard all but 10.0 mL of the Fe3+ solution in the graduated cylinder. To this 10.0 mL, add water to get 25.0 mL of solution.
Mix thoroughly. What is [Fe3+] in this 25.0 mL solution? Add 5.0 mL of the 25.0 mL Fe3+ solution to test tube #3. Remember that the tube already contains KSCN. What is [Fe3+]o for test tube #3? What is [SCN- ]o for test tube #3? Test Tube #4 Again, discard all but 10.0 mL of the Fe3+ solution in the graduated cylinder.
To this 10.0 mL, add water to get 25.0 mL of solution. Mix thoroughly. What is [Fe3+] in this 25.0 mL solution? Add 5.0 mL of the 25.0 mL Fe3+ solution to test tube #4. Remember that the tube already contains KSCN. What is [Fe3+]o for test tube #4? What is [SCN- ]o for test tube #4?
Explanation / Answer
Use the relation M1*V1 = M2*V2 to calculate the concentrations in the dilute solutions. M1 = concentration of the concentrated solution; M2 = concentration of the dilute solution; V1= volume of the concentrated solution taken; V2 = volume of the dilute solution obtained.
Final volume of all test tubes is (5.0 + 5.0) mL = 10.0 mL; initial concentration of KSCN = 2.0*10-4 M and initial volume = 5.00 mL; therefore, the final concentration will be the same for all the test tubes.
(5.00 mL)*(2.0*10-4 M) = [SCN-]0*(10.00 mL)
===> [SCN-]0 = (5.00*2.0*10-4 M)/(10.00) = 1.0*10-4 M (ans).
Test tube#1
Fe3+: initial volume = 5.00 mL; final volume = 10.00 mL. Initial concentration = 0.20 M. Therefore,
(5.00 mL)*(0.20 M) = (10.00 mL)*[Fe3+]0
===> [Fe3+]0 = (5.00*0.20 M)/(10.00) = 0.10 M (ans).
[SCN-]0 = 1.0*10-4 M (ans).
Test tube#2
Fe3+: initial volume = 10.00 mL; intermediate volume = 25.00 mL; initial concentration = 0.20 M. Find out the intermediate concentration as
(10.00 mL)*(0.20 M) = (25.00 mL)*[Fe3+]
===> [Fe3+] = (10.00*0.20 M)/(25.00) = 0.08 M (ans)
Volume of concentrated solution taken = 5.00 mL; volume of dilute solution = 10.00 mL. Concentration of concentrated solution = 0.08 M. Therefore,
(5.00 mL)*(0.08 M) = (10.00 mL)*[Fe3+]0
===> [Fe3+]0 = (5.00*0.08 M)/(10.00) = 0.04 M (ans).
[SCN-]0 = 1.0*10-4 M (ans).
Test tube#3
Fe3+: initial volume = 10.00 mL; intermediate volume = 25.00 mL; initial concentration = 0.08 M. Find out the intermediate concentration as
(10.00 mL)*(0.08 M) = (25.00 mL)*[Fe3+]
===> [Fe3+] = (10.00*0.08 M)/(25.00) = 0.032 M (ans)
Volume of concentrated solution taken = 5.00 mL; volume of dilute solution = 10.00 mL. Concentration of concentrated solution = 0.032 M. Therefore,
(5.00 mL)*(0.032 M) = (10.00 mL)*[Fe3+]0
===> [Fe3+]0 = (5.00*0.032 M)/(10.00) = 0.016 M (ans).
[SCN-]0 = 1.0*10-4 M (ans).
Test tube#4
Fe3+: initial volume = 10.00 mL; intermediate volume = 25.00 mL; initial concentration = 0.032 M. Find out the intermediate concentration as
(10.00 mL)*(0.032 M) = (25.00 mL)*[Fe3+]
===> [Fe3+] = (10.00*0.032 M)/(25.00) = 0.0128 M (ans)
Volume of concentrated solution taken = 5.00 mL; volume of dilute solution = 10.00 mL. Concentration of concentrated solution = 0.0128 M. Therefore,
(5.00 mL)*(0.0128 M) = (10.00 mL)*[Fe3+]0
===> [Fe3+]0 = (5.00*0.0128 M)/(10.00) = 0.0064 M (ans).
[SCN-]0 = 1.0*10-4 M (ans).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.