# A 2.4 x 10^-4 M solution of a copper complex has transmittance of 58 % when measured in a cell of path length 1.5 cm. Calculate the transmittance of solution if: (a) the concentration is doubled, (b) concentration is made half?

Jun 5, 2016

(a)" "27%

(b)" "72.1%

#### Explanation:

The Beer - Lambert Law states:

$A = \epsilon c l$

$A$ is the absorbance as measured by the spectrometer

$c$ is the molar concentration

$l$ is the path length of the cell in $\text{cm}$

The absorbance $A$ can be written in terms of the transmittance $I$ as:

$A = \log \left[\frac{{I}_{0}}{I}\right]$

${I}_{0}$ is set to 100 so this becomes:

$A = \log \left[\frac{100}{I}\right] = 2 - \log I$

The first part of the question involves finding $\epsilon$. Then this can be used to calculate $\left(a\right)$ and $\left(b\right)$.

$2 - \log I = \epsilon c l$

$\therefore \epsilon = \frac{2 - \log I}{c l} = \frac{2 - 1.763}{2 \times {10}^{- 4} \times 1.5} = 790 {\text{ ""mol"^(-1)."l"."cm}}^{- 1}$

$\textcolor{b l u e}{\left(a\right)}$

The concentration is doubled:

$2 - \log I = 790 \times 4.8 \times {10}^{- 4} \times 1.5 = 0.5688$

$\therefore \log I = 2 - 0.5688 = 1.4312$

:.color(red)(I=27%)

This seems a reasonable answer as you would expect less light to be transmitted if the solution is more concentrated.

$\textcolor{b l u e}{\left(b\right)}$

The concentration is halved:

$\therefore 2 - \log I = 790 \times 1.2 \times {10}^{- 4} \times 1.5 = 0.1422$

$\therefore \log I = 2 - 0.1422 = 1.8578$

color(red)(I=72.1%)

Again, this seems a reasonable answer as we have diluted the solution so you would expect more light to pass through.