Solution 1 contains 10.00 mL of 0.002 M iron (III) nitrate solution, 4.00 mL of 0.002 M potassium thiocyanate solution, and 6.00 mL of 0.05 M nitric acid solution.

Calculate the concentration of iron (III) nitrate in the final solution in Solution 1. SHOW ALL WORK!

Calculate the concentration of potassium thiocyanate in the final solution in Solution 1. SHOW ALL WORK!

Summarize the concentration of iron (III) nitrate and concentration of potassium thiocyanate in Solution 1 from your calculations in questions 1a and 1b above. You can fill in the same for Solutions 2-3 later after doing the calculations on your own time.

Solution #	mL of iron (III) nitrate	mL of potassium thiocyanate	mL of nitric acid	M of iron (III) nitrate in the final solution	M of potassium thiocyanate in the final solution	Absorbance of the solution
1	10 ml	4.00ml	6.00ml	0.0001M	0.000004M
2
3

You cannot do the limiting reagent thing here because you do not know that one reagent is in so much excess like before that the reaction will go to completion. So you must calculate the concentration of the [FeSCN²⁺] from the graph you created before.

Is the absorbance value of Solution 1 within the range of values on the graph you created using the data from Phase 2? So can you use this absorbance value to determine the concentration of [FeSCN²⁺] from the graph? (think about the definition of the slope and what the x and y axis are in this case and how it relates to the Beer-Lambert law).

Doing the calculation to find the equilibrium constant

Consider Solution 1 of Part 3,

What is the concentration of iron (III) nitrate aka iron (Fe³⁺) ions in this tube?

What is the concentration of potassium thiocyanate aka thiocyanate ions (SCN^-) ions in this tube?

Refer back to the graph produced for the Pre-lab Assignment to solve for [FeSCN²⁺]

According to Beer and Lamberts law

A =e l C

A = absorbance

e = molar absorptivity

l = optical path length in cm (usually 1 cm is the length of the cuvettes used)

C = concentration

The equation of a line is y = mx + b

Where

y = value for the Y axis

x = value of the X axis

m = slope

b = y intercept

When the graph passes through the origin, the y intercept (b) is zero.

So your equation becomes y = mx

What is the y axis in this case?

What is the x axis in this case?