# Calculating an initial Cr2O72-(aq) concentration of 1.0 x 10-3 M and an initial C2H5OH(aq) concentration of 0.500M?

Apr 28, 2017

(d) At 1.50 min, ["Cr"_2"O"_7^"2-"] = 7.1 × 10^"-4"color(white)(l)"mol/L".
(e) Use an initial ["Cr"_2"O"_7^"2-"] = 5.0 × 10^"-4" color(white)(l)"mol/L".

#### Explanation:

(c) Calculating concentration

The formula for absorbance $A$ is given by Beer's Law:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} A = \epsilon c l \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

$\epsilon =$ the molar absorptivity of the sample
$c =$ the concentration of the sample
$l =$ the path length through the cuvette

It states that the absorbance is directly proportional to the concentration of the sample. Thus,

${A}_{2} / {A}_{1} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\epsilon}}} {c}_{2} \textcolor{red}{\cancel{\textcolor{b l a c k}{l}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\epsilon}}} {c}_{1} \textcolor{red}{\cancel{\textcolor{b l a c k}{l}}}} = {c}_{2} / {c}_{1}$

We can rearrange this formula to give

c_2 = c_1 × A_2/A_1

c_1 = 1.0 ×10^"-3"color(white)(l) "mol/L"; A_1 = 0.782
c_2 = ?; color(white)(mmmmmmml)A_2 = 0.553

c_2 = 1.0 × 10^"-3"color(white)(l) "mol/L" × 0.553/0.782 = 7.1 × 10^"-4"color(white)(l)"mol/L"

(d) Determining the new setup

You know that

$A = \epsilon c l$

$\epsilon$ is a constant that is characteristic of the sample.

You want to keep ${A}_{0}$ constant at 0.782.

Thus, if you double the path length $l$, you must halve the concentration $c$.

c_0 = 1/2 × 1.0 × 10^"-3"color(white)(l) "mol/L" = 5.0 × 10^"-4"color(white)(l)"mol/L"