Question

When measured with a 1.00-cm cell, a 7.50 3 1025 M solution of species A exhibited absorbances of 0.155 and 0.755 at 475 and 700 nm, respectively. A 4.25 3 1025 M solution of species B gave absorbances of 0.702 and 0.091 under the same circumstances. Calculate the concentrations of A and B in solutions that yielded the following absorbance data in a 2.50-cm cell: (a) 0.439 at 475 nm and 1.025 at 700 nm; (b) 0.662 at 475 nm and 0.815 at 700 nm

Short answer

Part(a) $$\begin{gathered} c_{\text{A}=}3.95\times {10}^{-5}\text{M} \\ {c}_{\text{B}=}5.69\times {10}^{-6}\text{M} \end{gathered}$$

Part(b)$$\begin{gathered} c_{\text{A}=}2.98\times {10}^{-5}\text{M} \\ {c}_{\text{B}=}1.23\times {10}^{-6}\text{M} \end{gathered}$$

Step-by-step solution

Part (a) Step1. Given information

For species A:-

$$\begin{gathered} \text{Absrobance}\left(A\right)=0.155 \\ c\left(A\right)=7.5\times {10}^{-5}\text{M} \\ l\left(A\right)=1\mathrm{cm} \end{gathered}$$

For species B:-

$$\begin{gathered} \text{Absrobance}\left(B\right)=0.702 \\ c\left(B\right)=4.25\times {10}^{-5}\text{M} \\ l\left(B\right)=1\mathrm{cm} \end{gathered}$$

Part (a) Step2. Beer's Lambert Law

$A=\epsilon cl$

Here,

A – absorbance

$\epsilon$- molar absorptivity

$l$– length of the solution light passes through (cm)

$c$– concentration of solution (mol/L)

Part (a) Step3. Calculate molar absorptivity for A and B by using the formula given in step2 at 475 nm and 700 nm.

At 475 nm:-

The absorptivity of A is as follows:-

$$\begin{gathered} {\epsilon }_A=\frac{A}{cl} \\ =\frac{0.155}{7.5\times {10}^{-5}\text{M×1cm}} \\ =2066.67{\text{M}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

The absorptivity of B is as follows:-

$$\begin{gathered} {\epsilon }_{\text{B}}=\frac{A}{cl} \\ =\frac{0.702}{4.25\times {10}^{-5}\text{M×1cm}} \\ =16517.6{\text{M}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

At 700nm:-

The absorptivity of A is as follows:-

$$\begin{gathered} {\epsilon }_A=\frac{A}{cl} \\ =\frac{0.755}{7.5\times {10}^{-5}\text{M×1cm}} \\ =10066.7{\text{M}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

The absorptivity of B is as follows:-

$$\begin{gathered} {\epsilon }_{\text{B}}=\frac{A}{cl} \\ =\frac{0.091}{4.25\times {10}^{-5}\text{M×1cm}} \\ =2141.2{\text{M}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

Part (a) Step4. Calculate concentration of species A and B

$$\begin{gathered} A_{475}=0.439=2067\times 2.5\times c_{\text{A}}+16518\times 2.5\times c_B \\ 0.1756=2067\times c_{\text{A}}+16518\times c_B.............\left(1\right) \\ {A}_{700}=1.025=10067\times 2.5\times c_{\text{A}}+2141\times 2.5\times c_B \\ 0.41=10067\times c_{\text{A}}+2141\times c_B............\left(2\right) \end{gathered}$$

$\left(1\right)\times 2141-\left(2\right)\times 16518$

$$\begin{gathered} 375.96-6772.38=4425447c_{\text{A}}-166286706c_{\text{A}} \\ 6396.42=161861259c_{\text{A}} \\ {c}_{\text{A}}=3.95\times {10}^{-5}\text{mol/L} \end{gathered}$$

Substitute$3.95\times {10}^{-5}\text{mol/L}$for$c_{\text{A}}$in equation (1) as follows:-

role="math" localid="1646899242169" $$\begin{gathered} 0.1756=2067\times 3.9\times {10}^{-5}+16518\times c_B \\ 0.1756=0.0816+16518c_B \\ {c}_B=5.69\times {10}^{-6}{\text{molL}}^{-1} \end{gathered}$$

Part(b) Step1 .Calculate the concentrations of A and B in solutions that yielded the following absorbance data in a 2.50-cm cell: 0.662 at 475 nm and 0.815 at 700 nm 

$$\begin{gathered} A_{475}=0.662=2067\times 2.5\times c_{\text{A}}+16518\times 2.5\times c_B \\ 0.2648=2067\times c_{\text{A}}+16518\times c_B.............\left(1\right) \\ {A}_{700}=0.815=10067\times 2.5\times c_{\text{A}}+2141\times 2.5\times c_B \\ 0.326=10067\times c_{\text{A}}+2141\times c_B............\left(2\right) \end{gathered}$$

$\left(1\right)\times 2141-\left(2\right)\times 16518$

role="math" localid="1646899197941" $$\begin{gathered} 566.9368-5384.868=4425447c_{\text{A}}-166286706c_{\text{A}} \\ 4817.7492=161861259c_{\text{A}} \\ {c}_{\text{A}}=2.98\times {10}^{-5}\text{mol/L} \end{gathered}$$

Substitute $2.98\times {10}^{-5}\text{mol/L}$for $c_B$in equation (1) as follows:-

$$\begin{gathered} 0.2648=2067\times 2.98\times {10}^{-5}+16518\times c_B \\ 0.2648=0.0616+16518c_B \\ {c}_B=1.23\times {10}^{-6}{\text{molL}}^{-1} \end{gathered}$$