Question

DETERMINATION OF AN EQUILIBRIUM CONSTANT FOR A CHEMICAL REACTION Name TA Name (posted on A2L under "Grades") Lab Bench (posted on A2L under "Grades") Lab Room (posted on A2L under "Grades") Date Complete the following pre-lab questions before coming to lab. Your TA will collect this page at the start of your lab period. In your laboratory notebook, prepare the axes of your calibration graph for Part A. The graph should take up an entire page with a y-axis range of 0.00-0.80, and an x-axis range of 0.00- 16.00. Be sure to follow the graphing guidelines given on page xxv for full marks. 1. 2. EXPLAIN why the spectrometer needs to be calibrated before measuring any absorbance of FESCN(aq) PRE-LAB QUESTIONS

Answer 1

1. This is fairly easy to do.

2. The spectrometer needs to be calibrated because it is not an absolute method to determine concentrations. We need to measure some known concentrations in order to know the response of the equipment for each of them and then we are able to do the inverse: obtain the response for unknown samples and, from that and the calibration curve, calculate their concentration.

3. I don't know exactly what part A is, but the explanation of why we need a calibration curve is given in the previous item.

4. This excercise makes reference to the equilibrium:

$Fe^{3+}+SCN^{-}\leftrightharpoons FeSCN^{2+}$

With the absorbance and the slope of the calibration curve we can calculate the concentration of FeSCN²⁺ in equilibrium. The relationship between absorbance and concentration is given by:

$A=\varepsilon \cdot l\cdot c$

Where epsilon is the molar absorptivity and l is the optical path length. the product of ε and l is the slope of the calibration curve, so we can calculate the concentration as:

$c=\frac{A}{\varepsilon \cdot l}=\frac{A}{slope}=\frac{0.246}{5140M^{-1}}=4.79x10^{-5}M$

We can now calculate the concentration of Fe(III) and SCN^- in equilibrium, using the dilution formula:

$C_{1}\cdot V_{1}=C_{2}\cdot V_{2}$

Where 1 and 2 indicate values before and after dilution. IN this case, our final volume is 10.0 mL (from adding all the volumes mixed), so we can calculate the final concentration of Fe(III):

$C_{2}=\frac{C_{1}\cdot V_{1}}{V_{2}}=\frac{0.00200M\cdot 5.00mL}{10mL}=0.00100M$

And for SCN^-:

$C_{2}=\frac{C_{1}\cdot V_{1}}{V_{2}}=\frac{0.00200M\cdot 2.00mL}{10mL}=0.000400M$

Of course, the concentration in equilibrium are given by the difference between these concentrations and the concentration of FeSCN(II), which has formed at expense of Fe(III) and SCN^-:

$Fe(III):0.00100M-4.79x10^{-5}M=9.52x10^{-4}M$

$SCN^{-}:0.000400M-4.79x10^{-5}M=3.52x10^{-4}M$

So now we have the concentration of the three species in equilibrium we can calculate the value of the constant:

$K=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]}=\frac{4.79x10^{-5}}{9.52x10^{-4}\cdot 3.52x10^{-4}}=143$