Question

Zinc(II) and the ligand L form a 1:1 complex that absorbs strongly at $600$nm. As long as the molar concentration of L exceeds that of zinc(II) by a factor of $5$, the absorbance depends only on the cation concentration. Neither zinc(II) nor L absorbs at 600 nm. A solution that is $1.59\times {10}^{-4}\mathrm{M}$in zinc(II) and $1.00\times {10}^{-3}\mathrm{M}$in L has an absorbance of $0.352$in a $1.00$-cm cell at $600$nm. Calculate (a) the percent transmittance of this solution.

(b) the percent transmittance of this solution in a $2.50$-cm cell.

(c) the molar absorptivity of the complex.

Short answer

Part (a) The percent transmittance is $44.5\%$

Part (b) The percent transmittance is $13.2\%$

Part (c) The molar absorptivity is$=2.21\times {10}^3\mathrm{L}{\mathrm{cm}}^{-1}{\mathrm{mol}}^{-1}$

Step-by-step solution

Part (a) Step 1. percent transmittance

The formula of absorbance is

$A=-\log T$

Substitute the value of absorbance,

$$\begin{gathered} A=-\log T \\ 0.352=-\log T \\ T={10}^{-0.362} \\ T=0.445 \end{gathered}$$

Now percent transmittance is,

$$\begin{gathered} \%T=0.445\times 100\% \\ \%T=44.5\% \end{gathered}$$

Part (b) Step 1. percent transmittance

The following is the mathematical formula for the relationship between absorbance, molar absorptivity, cell path length, and concentration:

$A=\epsilon bc......\left(1\right)$

The negative logarithm of transmittance is known as absorbance, and it is stated mathematically as:

$A=-\log T.....\left(2\right)$

From equation $\left(1\right)$and $\left(2\right)$,

$-\log T=\epsilon bc.....\left(3\right)$

Now substitute the values given in the question as,

$$\begin{gathered} \frac{-\log T}{-\log 0.445}=\frac{(2.50cm)}{(1.00\mathrm{cm})} \\ \frac{-\log T}{0.352}=2.50 \\ -\log T=0.880 \\ T=0.132 \end{gathered}$$

Now the percent transmittance is,

$$\begin{gathered} \%T=0.132\times 100\% \\ =13.2\% \end{gathered}$$

Part (c) Step 1. molar absorptivity

The following is the mathematical formula for the relationship between absorbance, molar absorptivity, cell path length, and concentration:

$A=\epsilon bc$

Rearrange the above equation,

role="math" localid="1645511687812" $\epsilon =\frac{A}{cb}......\left(4\right)$

The solution has an absorbance of $0.352$in a $1.00cm$cell.

To calculate the molar absorptivity of complex, calculate the concentration complex.

Part (c) Step 2. Conclusion

A 1:1 combination formed by zinc (II) and the ligand $L$absorbs significantly at $600nm$. When the molar concentration of $L$surpasses that of zinc (II) by a factor of $5$, only the cation concentration determines the absorbance.

In solution, a $1.59\times {10}^{-4}\mathrm{M}$complex is generated because Zinc (II) and the ligand $L$form a $1:1$complex. As a result, the concentration of the complex with an absorbance of $0.352$is $1.59\times {10}^{-4}\mathrm{M}$

Substitute the values in equation $\left(4\right)$,

$$\begin{gathered} \epsilon =\frac{A}{cb} \\ =\frac{0.352}{(1.00\mathrm{cm})\times \left(1.59\times {10}^{-4}\mathrm{mol}/\mathrm{L}\right)} \\ =2.21\times {10}^3\mathrm{L}{\mathrm{cm}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$