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

1 Answer
Apr 28, 2017

Answer:

(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"#