Question

A portable photometer with a linear response to radiation registered $56.3$µA with the solvent in the light path. The photometer was set to zero with no light striking the detector. Replacement of the solvent with an absorbing solution yielded a response of $36.7$µA. Calculate

(a) the percent transmittance of the sample solution.

(b) the absorbance of the sample solution.

(c) the transmittance to be expected for a solution in which the concentration of the absorber is one third that of the original sample solution.

(d) the transmittance to be expected for a solution that has twice the concentration of the sample solution.

Short answer

Part (a) The percentage transmittance is $65.2\%$

Part (b) The absorbance is $0.186$

Part (c) The transmittance is localid="1651341612991" $0.867$

Part (d) The transmittance is$0.425$

Step-by-step solution

Step 1. Given information

Absorbing yield is $36.7\mu A$and linear response is $56.3\mu A$

And the formulas are as,

localid="1645506860954" $T=\frac{P_{\text{solution}}}{P_{\text{solvent}}}$and

$\%T=\frac{P}{P_{\circ }}\times 100\%$

The relationship between transmittance and absorbance is

$A=-\log T$

Part (a) Step 1. percent transmittance

Use the formula of percent transmittance for the given values in the question as,

$$\begin{gathered} \%T=\left(\frac{36.7\mu \mathrm{A}}{56.3\mu \mathrm{A}}\right)(100\%) \\ =65.2\% \end{gathered}$$

Part (b) Step 1. absorbance

Now apply absorbance formula for the obtained transmittance value as,

$$\begin{gathered} A=-{\log }_{10}\left(\frac{65.2}{100}\right) \\ =0.186 \end{gathered}$$

Part (c) Step 1. expected transmittance

For absorbance the concentration is now one third of original,

$A=\frac{0.186}{3}=0.062$

Apply relation between transmittance and absorbance,

localid="1651168444684" $T={10}^{(-0.062)}=0.867$

Part (d) Step 1. expected transmittance

For absorbance concentration is now twice of the original value,

$$\begin{gathered} A=(0.186)\left(2\right) \\ =0.372 \end{gathered}$$

Apply relation between transmittance and absorbance as,

$$\begin{gathered} T={10}^{(-0.372)} \\ =0.425 \end{gathered}$$