Question

The accompanying absorption data were recorded at 390 nm in 1.00-cm cells for a continuous-variation study of the colored product formed between Cd²⁺ and the complexing reagent R.

(a) Find the ligand-to-metal ratio in the product.

(b) Calculate an average value for the molar absorptivity of the complex and its uncertainty. Assume that in the linear portions of the plot the metal is completely complexed.

(c) Calculate K_f for the complex using the stoichiometric ratio

Short answer

a) The ligand-to metal ratio in the product is 1:1.

b) The average value of molar absorptivity of the complex is$13612\pm 510{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1}$and its uncertainty is 510.

c) The value for Kf for the complex is,$K_f=6.0\times {10}^5$

Step-by-step solution

Part (a) Step 1: Given Information

Part (a) Step 2: Explanation

VCd	VR	VCd/(VCd+VR)	A390
10.00	0.00	0.00	0.000
9.00	1.00	0.10	0.174
8.00	2.00	0.20	0.353
7.00	3.00	0.30	0.530
6.00	4.00	0.40	0.672
5.00	5.00	0.50	0.723
4.00	6.00	0.60	0.673
3.00	7.00	0.70	0.537
2.00	8.00	0.80	0.358
1.00	9.00	0.90	0.180
0.00	10.00	1.00	0.000

The plot between absorbance and volume fraction is:

At the point of intersection y and x values are equal,

$$\begin{gathered} 1.7\mathrm{x}+0.0058=-1.703\mathrm{x}+1.712 \\ 3.403\mathrm{x}=1.7062 \\ \mathrm{x}=0.50 \end{gathered}$$

Therefore, volume fraction of R at the maximum point of the graph = 0.50

Volume fraction of Cd at the maximum point of the graph = $\mathrm{I}-0.50=0.50$

$\frac{{\displaystyle \frac{V_R}{V_{cd}+V_R}}}{{\displaystyle \frac{V_{cd}}{V_{cd}+V_R}}}=\frac{V_R}{V_{cd}}=\frac{0.50}{0.50}=1$

Therefore, ligand-to metal ratio in the product is 1:1

Part (b) Step 1: Given Information

Average molar absorptivity of the complex and its uncertainty should be calculated.

Part (b) Step 2: Explanation

For the linear portion from solution 0 to 4

$$\begin{gathered} \mathrm{\epsilon }=\frac{\mathrm{Slope}}{{\mathrm{C}}_{\mathrm{cd}}}=\frac{1.7}{1.25\times {10}^{-4}}=13600{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \\ \mathrm{SD}=\frac{0.040}{1.25\times {10}^{-4}}=320 \end{gathered}$$

For the linear portion from solution 7 to 10

$$\begin{gathered} \mathrm{\epsilon }=\frac{\mathrm{Slope}}{{\mathrm{C}}_{\mathrm{cd}}}=\frac{1.703}{1.25\times {10}^{-4}}=13624{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \\ \mathrm{SD}=\frac{0.04960}{1.25\times {10}^{-4}}=397 \\ {\mathrm{\epsilon }}_{\mathrm{average}}=\frac{13600+13624}{2} \\ {\mathrm{\epsilon }}_{\mathrm{average}}=13612{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \\ \mathrm{SD}=\sqrt{{\left(320\right)}^2+{\left(397\right)}^2} \\ \mathrm{SD}=510 \end{gathered}$$

Therefore, the average molar absorptivity =$13612\pm 510{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1}$

Part (c) Step 1: Given Information

K_f for the complex should be determined

Part (c) Step 2: Explanation

Absorbance at the point of intersection = 0.723

$$\begin{gathered} \mathrm{A}=\mathrm{\epsilon }\times \left[\mathrm{CdR}\right]\times \mathrm{l} \\ 0.723=13612{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1}\times \left[\mathrm{CdR}\right]\times 1.00\mathrm{cm} \\ \left[\mathrm{CdR}\right]=\frac{0.723}{13612{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1}\times 1.00\mathrm{cm}} \\ \left[\mathrm{CdR}\right]=5.31\times {10}^{-5}\mathrm{M} \\ \left[{\mathrm{Cd}}^{2+}\right]=\frac{1.25\times {10}^{-4}\mathrm{mol}\times 5.00\mathrm{mL}-5.31\times {10}^{-5}\mathrm{mol}\times 10.00\mathrm{mL}}{10.00\mathrm{mL}} \\ \left[{\mathrm{Cd}}^{2+}\right]=9.4\times {10}^{-6}\mathrm{M} \\ \left[\mathrm{R}\right]=9.4\times {10}^{-6}\mathrm{M} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{f}}=\frac{\left[\mathrm{CdR}\right]}{\left[{\mathrm{Cd}}^{2+}\right]\left[\mathrm{R}\right]}=\frac{5.31\times {10}^{-5}}{9.4\times {10}^{-6}\times 9.4\times {10}^{-6}} \\ {\mathrm{K}}_{\mathrm{f}}=6.0\times {10}^5 \end{gathered}$$