# How to get new concentration from percent transmittance?

Apr 20, 2017

Transmittance and absorbance are related as follows.

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

Evidently, percent transmittance is simply some fraction of $T$. Absorbance is known to be related to concentration:

$A = \epsilon b c$

where $\epsilon$ is the molar absorptivity, $b$ is the path length, and $c$ is the concentration in $\text{M}$. $\epsilon$ and $b$ are CONSTANTS for the same substance.

Therefore:

$- \log \left({T}_{1}\right) = \epsilon b {c}_{1}$

$- \log \left({T}_{2}\right) = \epsilon b {c}_{2}$

And we have:

c_2/c_1 = log(T_2)/(log(T_1)

Thus, the new concentration is:

$\textcolor{b l u e}{{c}_{2}} = {c}_{1} \log \frac{{T}_{2}}{\log} \left({T}_{1}\right)$

$= \left(\text{26 mg/L}\right) \log \frac{0.617}{\log} \left(0.528\right)$

$=$ $\textcolor{b l u e}{\text{19.66 mg/L}}$

This should make sense... if it transmits more, it absorbs less. It absorbs less if less particles are in solution that block the path of the incoming light.