Question

The accompanying data ($1.00-\mathrm{cm}$cells) were obtained for the spectrophotometric titration of $10.00\mathrm{mL}$of Pd(II) with data-custom-editor="chemistry" $2.44\times {10}^{-4}\mathrm{M}$Nitroso R (O. W. Rollins and M. M. Oldham, Anal. Chem., 1971, 43, 262, DOI: 10.1021/ac60297a026).

Calculate the concentration of the Pd(II) solution, given that the ligand-to-cation ratio in the colored product is 2:1.

Short answer

The concentration of the Pd(II) solution is $3.79\times {10}^{-7}\mathrm{mol}/\mathrm{L}$.

Step-by-step solution

Step 1. Given information

The accompanying data ($10-\mathrm{cm}$cells) were obtained for the spectrophotometric titration of $10.00\mathrm{mL}$of Pd(II) with $2.44\times {10}^{-4}\mathrm{M}$Nitroso R (O. W. Rollins and M. M. Oldham, Anal. Chem., 1971, 43, 262, DOI: 10.1021/ac60297a026).

Calculate the concentration of the Pd(II) solution, given that the ligand-to-cation ratio in the colored product is 2:1.

Step 2. Corrected absorbance data 

The absorbance data must be corrected for the diluted volume.

$A_{corrected}=A_{500}\times \left(\frac{10\mathrm{mL}+v}{10\mathrm{mL}}\right)$

For 1.00 mL of nitroso R added,

$$\begin{gathered} A_{corrected}=0.147\times \left(\frac{10\mathrm{mL}+1\mathrm{mL}}{10\mathrm{mL}}\right) \\ =0.162 \end{gathered}$$

Volume of Nitroso R, mL	A500	Acorrect
0	0	0
1.00	0.147	0.162
2.00	0.271	0.325
3.00	0.375	0.488

Step 3. Corrected absorbance versus volume graph

The corrected absorbance versus volume of titrant added is plotted and the point of intersection is determined. This is the end point of the titration.

At the point of intersection, y and x values are equal. So, we can solve the two equations for the two linear portions of the graph to obtain the volume of titrant spend at the endpoint.

$$\begin{gathered} y=0.0001x+0.5194\stackrel{}{\to }\left(1\right) \\ y=0.167x+0.0003\stackrel{}{\to }\left(2\right) \end{gathered}$$

Equation (1) is equal to (2).

$$\begin{gathered} y=0.0001x+0.5194=0.167x+0.0003 \\ 0.1669x=0.5197 \\ x=3.11\mathrm{mL} \end{gathered}$$

The volume of Nitroso R spent at the endpoint = data-custom-editor="chemistry" $3.11\mathrm{mL}$

Number of moles of Nitroso R:

data-custom-editor="chemistry" $\frac{2.44\times {10}^{-4}\mathrm{mol}}{1000\mathrm{mL}}\times 3.11\mathrm{mL}=7.59\times {10}^{-7}\mathrm{mol}$

Number of moles of Pd(II) in 10.00 mL of solution:

data-custom-editor="chemistry" $\frac{1\mathrm{mol}}{2\mathrm{mol}}\times 7.59\times {10}^{-7}\mathrm{mol}=3.79\times {10}^{-7}\mathrm{mol}$

Concentration of Pd(II):

$\frac{3.79\times {10}^{-7}\mathrm{mol}}{10\mathrm{mL}}\times 1000\mathrm{mL}=3.79\times {10}^{-7}\mathrm{mol}/\mathrm{L}$