Question

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

Answer 1

(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:

`color(blue)(bar(ul(|color(white)(a/a)A = epsiloncl color(white)(a/a)|)))" "`

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 = (color(red)(cancel(color(black)(epsilon)))c_2color(red)(cancel(color(black)(l))))/(color(red)(cancel(color(black)(epsilon)))c_1color(red)(cancel(color(black)(l)))) = c_2/c_1`

We can rearrange this formula to give

`c_2 = c_1 × A_2/A_1`

In your problem,

`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 = epsiloncl`

`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`.

Start with

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