91影视

Beer's law states that through the sample and the concentration of the absorbing species, the absorbance is proportional to the path length.

$\mathrm{A}=\mathrm{蔚产颁}$

is the absorbance,$蔚$ is the molar absorptivity, b is the length of light path, C is the concentration

Find the absorbance atfor,

$\begin{array}{r}{\mathrm{A}}_{\mathrm{A}}=鈭�\log 鈦�\left(\mathrm{T}\right)=鈭�\log 鈦�(0.310)=0.50\\ \mathrm{Use}\mathrm{Beer}鈥�\mathrm{s}\mathrm{law}\mathrm{to}{\mathrm{肠补濒肠耻濒补迟别蔚}}_{\mathrm{A}},\\ {\stackrel{藱}{\mathrm{鈭�}}}_{\mathrm{A}}=\frac{\mathrm{A}}{\mathrm{b}鈰�\mathrm{c}}=\frac{0.5086}{0.005\mathrm{cm}鈰�0.01\mathrm{M}}\\ {\stackrel{藱}{\mathrm{鈭�}}}_{\mathrm{A}}=10172{\mathrm{M}}^{鈭�1}{\mathrm{cm}}^{鈭�1}\end{array}$

Thus, absorbance for B,

${\mathrm{A}}_{\mathrm{B}}=鈭�\log 鈦�\left(T\right)=鈭�\log 鈦�\left(0.974\right)=0$

Calculate:${\mathrm{蔚}}_{\mathrm{B}}$

$\begin{array}{r}{\stackrel{藱}{o}}_{\mathrm{B}}=\frac{\mathrm{A}}{\mathrm{b}鈰�\mathrm{c}}=\frac{0.0114}{0.005\mathrm{cm}鈰�0.01\mathrm{M}}\\ {\stackrel{藱}{o}}_{\mathrm{B}}=228{\mathrm{M}}^{鈭�1}{\mathrm{cm}}^{鈭�1}\end{array}$

Evaluate the absorbance at 1993${\mathrm{cm}}^{-1}$ for A

$A_A=鈭�\log 鈦�\left(T\right)=鈭�\log 鈦�\left(0.797\right)=0.0985$

Calculate:$蔚{\text{'}}_A$

$\begin{array}{r}{\stackrel{藱}{o}}_{\mathrm{A}}^{\mathrm{鈥�}}=\frac{\mathrm{A}}{\mathrm{b}鈰�\mathrm{c}}=\frac{0.0985}{0.005\mathrm{cm}鈰�0.01\mathrm{M}}\\ {\stackrel{藱}{o}}_{\mathrm{A}}^{\mathrm{鈥�}}=197{\mathrm{OM}}^{鈭�1}{\mathrm{cm}}^{鈭�1}\end{array}$

Hence, absorbance for B

$A_B=鈭�\log 鈦�\left(T\right)=鈭�\log 鈦�\left(0.20\right)=0.69897$

Calculate:

$\begin{array}{r}{\stackrel{藱}{\mathrm{鈭�}}}_{\mathrm{B}}^{\mathrm{鈥�}}=\frac{\mathrm{A}}{\mathrm{b}鈰�\mathrm{c}}=\frac{0.69897}{0.005\mathrm{cm}鈰�0.01}\\ {\stackrel{藱}{\mathrm{鈭�}}}^{\mathrm{鈥�}}=13979.4{\mathrm{M}}^{鈭�1}{\mathrm{cm}}^{鈭�1}\end{array}$

Hence, the mixture absorbance at 2022 ${\mathrm{cm}}^{-1}$is,

$\begin{array}{r}{\mathrm{A}}_1=鈭�\log 鈦�\left(\mathrm{T}\right)=-\log (0.34)\\ {\mathrm{A}}_1=0.4685\end{array}$

Thus, the mixture absorbance at $1993{\mathrm{cm}}^{-1}$is,

$\begin{array}{r}{\mathrm{A}}_2=鈭�\log 鈦�\left(\mathrm{T}\right)=鈭�\log 鈦�\left(0.383\right)\\ {\mathrm{A}}_2=0.4168\end{array}$

Use the formula to find the concentration of

$\begin{array}{r}\left[\mathrm{A}\right]=\frac{\left|\begin{array}{ll}{\mathrm{A}}_1& {\stackrel{藱}{o}}_{\mathrm{B}}鈰�\mathrm{b}\\ {\mathrm{A}}_2& {\stackrel{藱}{o}}_{\mathrm{B}}^{\mathrm{鈥�}}鈰�\mathrm{b}\end{array}\right|}{\left|\begin{array}{cc}{\stackrel{藱}{o}}_{\mathrm{A}}鈰�\mathrm{b}& {\stackrel{藱}{o}}_{\mathrm{B}}鈰�\mathrm{b}\\ {\stackrel{藱}{o}}_{\mathrm{A}}^{\mathrm{鈥�}}鈰�\mathrm{b}& {\stackrel{藱}{o}}_{\mathrm{B}}^{\mathrm{鈥�}}鈰�\mathrm{b}\end{array}\right|}\\ \left[A\right]=\frac{\left|\begin{array}{cc}0.4685& 228鈰�0.005\\ 0.4168& 13979.4鈰�0.005\end{array}\right|}{\left|\begin{array}{cc}10172鈰�0.005& 228鈰�0.005\\ 1970鈰�0.005& 13979.4鈰�0.005\end{array}\right|}\end{array}$

$\begin{array}{r}\left[\mathrm{A}\right]=\frac{\left|\begin{array}{cc}0.4685& 1.14\\ 0.4168& 69.897\end{array}\right|}{\left|\begin{array}{cc}50.86& 1.14\\ 9.85& 69.897\end{array}\right|}\\ \left[\mathrm{A}\right]=\frac{(0.4685鈰�69.897)鈭�(1.14鈰�0.4168)}{(50.86鈰�69.897)鈭�(1.14鈰�9.85)}\\ \left[\mathrm{A}\right]=9.11鈰�{10}^{鈭�3}\mathrm{M}\end{array}$

Evaluate the concentration for B,

$\begin{array}{r}\left[\mathrm{B}\right]=\frac{\left|\begin{array}{cc}{\stackrel{藱}{o}}_{\mathrm{A}}鈰�\mathrm{b}& {\mathrm{A}}_1\\ {\stackrel{藱}{o}}_{\mathrm{A}}^{\mathrm{鈥�}}鈰�\mathrm{b}& {\mathrm{A}}_2\end{array}\right|}{\left|\begin{array}{ll}{\stackrel{藱}{o}}_{\mathrm{A}}鈰�\mathrm{b}& {\stackrel{藱}{o}}_{\mathrm{B}}鈰�\mathrm{b}\\ {\widehat{o}}_{\mathrm{A}}^{\mathrm{鈥�}}鈰�\mathrm{b}& {\stackrel{藱}{o}}_{\mathrm{B}}^{\mathrm{鈥�}}鈰�\mathrm{b}\end{array}\right|}\\ \left[B\right]=\frac{\left|\begin{array}{cc}10172鈰�0.005& 0.4685\\ 1970鈰�0.005& 0.4168\end{array}\right|}{\left|\begin{array}{cc}10172鈰�0.005& 228鈰�0.005\\ 1970鈰�0.005& 13979.4鈰�0.005\end{array}\right|}\\ \left[B\right]=\frac{\left|\begin{array}{cc}50.86& 0.4685\\ 9.85& 0.4168\end{array}\right|}{\left|\begin{array}{cc}50.86& 1.14\\ 9.85& 69.897\end{array}\right|}\\ \left[B\right]=\frac{(50.86鈰�0.4168)鈭�(9.85鈰�0.4685)}{(50.86鈰�69.897)鈭�(1.14鈰�9.85)}\\ \left[B\right]=4.68鈰�{10}^{鈭�3}M\end{array}$

Therefore, the concentrations of A and B are $\left[\mathrm{A}\right]=9.11鈰�{10}^{鈭�3}\mathrm{M}\text{and}\left[\mathrm{B}\right]=4.68鈰�{10}^{鈭�3}\mathrm{M}$.