1
answer
0
watching
152
views
12 Dec 2019

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?

For unlimited access to Homework Help, a Homework+ subscription is required.

Irving Heathcote
Irving HeathcoteLv2
13 Dec 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Related questions

Weekly leaderboard

Start filling in the gaps now
Log in