Solution A is a .25M iron(III) nitrate solution solution D is a .001M KSCN solut

Question

Solution A is a .25M iron(III) nitrate solution solution D is a .001M KSCN solution

Solution A is a .25M iron(III) nitrate solution by adding 10.09 grams of Fe(NO3)3*9H2O and diluting to 100 mL

solution D is a .001M KSCN solution by adding 0.097181g of KSCN and diluting to 100 mL

one example for each

witl well etric ake areden and dry storage vesel. (Cover and lahel to the mark to a clean and dry s Using the centrations of solutions C and D Rask and dilute to mix. Transfer this solution label the solution " el the solutioner this solut ate used (above), calculate the exact con- as actitol amount of potassium thiocyanate used (ebow), actual amount of potassium thiocyanate used . Using Dart 3- Preparation of the Fe(SCN)2+ (aq) Standard Solutions Part 3 - Preparation of The standard solutions will be prepared specifically to ens is driven to completion. We will accomplish this task by forcing the equibium to the podtrea by the addition of large amounts of terric ion to low concentrations of thiocyanate ion. will be prepared specifically to ensure that the formation of Fe(SCN)?+ (g) equilibrium to the products . All four of the standards will be prepared in a 100 mL volumetric flask2. For each standard, begin by transferring a 20.00 mL aliquot of solution A. Then, for the first standard, add 5 mL of solution D. Dilute to the mark with dilute nitric acid. . Transfer the solution to a clean storage bottle, cover, and label the solution to be used later. Repeat the same procedure for aliquots of 10, 15, and 20 mL of solution D. Part 4- Preparation of the Equilibrium Assay Solutions Once again, all of these solutions will be prepared in the 100 ml wolumat cl cleaning and preparing storage bottles for these solutions. 00 mL volumetric fask. L volumetric flask. Take care in cleaning an ll of these solutions will e prepared in the 10 ml, olimetri ently, transfer each standard to a clean h the bottles thoroughl

Explanation / Answer

Part 3

Here we use total volume = 100 mL = 0.100 L

Calculation of concentration for solution number 1

Molarity of Fe3+ =0.25 M, volume of it used = 20.0 mL = 0.02 L

Step 1

calculation of moles of Fe3+ = Molarity of aliquote solution A x volume of A used in L

=0.25 M x 0.02 L = 0.005 mol Fe3+

Final concentration in solution = 0.005 mol / 0.100 L = 0.05 M

Step 2

1)Solution D [SCN-]=0.001; Volume of solution D used = 5.0 mL = 0.005 L

[SCN-]= 0.005 L x 0.001 mol/L = 0.000005 mol

Final concentration [SCN-] = 0.000005/0.100= 0.00005 M

2)Solution D [SCN-]=0.001; Volume of solution D used = 10.0 mL = 0.010 L

[SCN-]= 0.010 L x 0.001 mol/L = 0.00001 mol

Final concentration [SCN-] = 0.00001/0.100= 0.0001 M

3)Solution D [SCN-]=0.001; Volume of solution D used = 15.0 mL = 0.015 L

[SCN-]= 0.015 L x 0.001 mol/L = 0.000015 mol

Final concentration [SCN-] = 0.000015/0.100= 0.00015 M

4)Solution D [SCN-]=0.001; Volume of solution D used = 20.0 mL = 0.005 L

[SCN-]= 0.020 L x 0.001 mol/L = 0.000020 mol

Final concentration [SCN-] = 0.000020/0.100= 0.00020 M

Part 5

We will be calculating Fe(SCN)2+ calculation for each solution.

Obviously SCN- is the limiting reactant since there is so much less of it. According to the reaction, the

0.000020 moles of SCN- will produce 0.000020 moles of Fe(SCN)2+ with a lot of Fe3+ left over. That’s why we will be considering [SCN-] only

1)

Molarity Fe(SCN)2+ = moles / L = (0.00005 / 0.100) = 0.0005 M

2)

Molarity Fe(SCN)2+ = moles / L = (0.00010 / 0.100) = 0.001 M

3)

Molarity Fe(SCN)2+ = moles / L = (0.00015 / 0.100) = 0.0015 M

4)

Molarity Fe(SCN)2+ = moles / L = (0.00020 / 0.100) = 0.002 M

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