Question

I need help calculating values for [Fe3+] and [FeSCN2+]. I don't know where to begin.

Sample Volume(mL) of 0.002 M Fe3+ Volume(mL) of 0.10 M SCN Volume(mL) of 0.1 M HNO3 Volume(mL) of H20 0.00 10.00 5.0 5.0 030 10.00 50 47 0.60 10.00 5.0 4.4 1.20 10.00 5.0 3.8 2.00 10.00 5.0 3.0 2. Calculate the values for [Fe3+] in the samples and record in the data table below. Remember to account for the dilution of each of these ions. 3. In these standard solutions, we have forced the equilibrium far to the right by having the [SCN) much higher than the [Fe3+] so that at equilibrium, [FeSCN2+Jeg = [Fe3+] Sample [Fe3+] [FeSCN2+Jeg Absorbance N- w a Spectrophotometry

Absorbance:

1. 0.0619

2. 0.3953

3. 2.0000

4. 1.599

5. 1.249

Answer 1

To calculate the concentration of Fe³⁺ in the mixtures, you have to use the dilution formula:

$C_{diluted}=\frac{C_{stock}\cdot V_{stock}}{V_{final}}$

Where the C_stock = 0.002 M, V_stock is the volume of this solution added and V_final is the volume after the addition of all the components, which is 20 mL in all cases. We can then calculate:

$Sample\, 1:\: C_{diluted}=\frac{0.002M\cdot 0}{20mL}=0$

$Sample\, 2:\: C_{diluted}=\frac{0.002M\cdot 0.30mL}{20mL}=0.00003M$

$Sample\, 3:\: C_{diluted}=\frac{0.002M\cdot 0.60mL}{20mL}=0.00006M$

$Sample\, 4:\: C_{diluted}=\frac{0.002M\cdot 1.20mL}{20mL}=0.00012M$

$Sample\, 5:\: C_{diluted}=\frac{0.002M\cdot 2.00mL}{20mL}=0.0002M$

Since the excercise is telling us that the conditions are such that we can consider that the concentration of FeSCN²⁺ in equilibrium is the same as the initial concentration of Fe³⁺, the values for [FeSCN²⁺] are the same we have already calculated.