Question

The accompanying data were obtained in a slope-ratio investigation of the complex formed between Ni²⁺and 1-cyclopentene-1-dithiocarboxylic acid (CDA). The measurements were made at 530 nm in 1.00-cm cells.

(a) Determine the formula of the complex. Use linear least-squares to analyze the data.

(b) Find the molar absorptivity of the complex and its uncertainty.

Short answer

a) Formula of the complex is Ni(CDA)₃.

b) The molar absorptivity of the complex and its uncertainty is$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{average}}=10276.912{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \\ \mathrm{SD}=21.27{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \end{gathered}$$

Step-by-step solution

Part (a): Step 1: Given Information

Measurements are made in 1.00 cm cell at 530 nm.

Part (a) Step 2 : Explanation

The plot of absorbance at 530 nm versus C_Ni,

Least squares analysis data

Slope = 10277.12

Intercept = -0.00038

Sm= 5.674341

Sb= 0.000258

The plot of absorbance at 530 nm versus concentration of CDA,

Least square analysis data

Slope = 3426.568

Intercept = -0.00021

S_m= 6.8333575

S_b= 0.000294

If the stoichiometric ratio of Ni:CDA is x:y

$$\begin{gathered} \frac{m_{Ni}}{m_{CDA}}=\frac{y}{x} \\ \frac{10277.12}{3426.568}=\frac{y}{x} \\ \frac{y}{x}=2.99\approx 3 \\ x:y=1:3 \end{gathered}$$

Therefore, the formula of the complex is Ni(CDA)₃.

Part (b) Step 1: Given Information

The molar absorptivity of the complex and its uncertainty should be determined.

Part (b) Step 2: Explanation

Slope of absorbance versus concentration of Ni,

$$\begin{gathered} 10277.12=\frac{\mathrm{\epsilon l}}{\mathrm{x}} \\ 10277.12{\mathrm{M}}^{-1}=\frac{\mathrm{\epsilon }\times 1.00\mathrm{cm}}{1} \\ \mathrm{\epsilon }=\frac{10277.12{\mathrm{M}}^{-1}}{1.00\mathrm{cm}} \\ \mathrm{\epsilon }=10277.12{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \end{gathered}$$

Standard deviation for ε =$\frac{\pm 5.674341}{1.00}=5.67$

Slope of absorbance versus concentration of CDA.

$$\begin{gathered} 3426.568=\frac{\mathrm{\epsilon l}}{\mathrm{y}} \\ 3426.568{\mathrm{M}}^{-1}=\frac{\mathrm{\epsilon }\times 1.00\mathrm{cm}}{3} \\ \mathrm{\epsilon }=\frac{3\times 3426.568{\mathrm{M}}^{-1}}{1.00\mathrm{cm}} \\ \mathrm{\epsilon }=10276.704{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \end{gathered}$$

Standard deviation for ε =$\frac{\pm 6.8333575\times 3}{1.00}=20.5$

$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{average}}=\frac{10277.12+10276.704}{2}=10276.912{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \\ \mathrm{SD}=\sqrt{{\left({\mathrm{SD}}_1\right)}^2+{\left({\mathrm{SD}}_2\right)}^2} \\ \mathrm{SD}=\sqrt{{\left(5.67\right)}^2+{\left(20.5\right)}^2} \\ \mathrm{SD}=21.27{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \end{gathered}$$