Question

Determine how to dilute a 0.001M stock solution of cyanine molecules so that it has an absorbance between 0.2 and 0.8 when run through UV-Vis spectroscopy. The extinction coefficient for cyanine is about (1-7) x 10^5 M^-1cm^-1

Answer 1

Answer:

Absorbance = A = c l

where, = extinction coefficient = 1 x 10⁵ M^-1cm^-1 to 7 x 10⁵ M^-1cm^-1

c = concentration

l = length = 1 cm

So, c = (A/ l)

for A = 0.2

c = 0.2 / (7 x 10⁵ M^-1cm^-1 x 1 cm) M to (0.2 / (1 x 10⁵ M^-1cm^-1 x 1cm)) M = 2.9 x 10^-7 M to 20 x 10^-7 M

or for A = 0.8

c = 0.8 / (7 x 10⁵ M^-1cm^-1 x 1 cm) M to (0.8 / (1 x 10⁵ M^-1cm^-1 x 1cm)) M = 11.6 x 10^-7 M to 80 x 10^-7 M

it means,

the concentration between 11.6 x10^-7 M to 80 x 10^-7 M will give us absorbance in range of 0.2 to 0.8.

How to make such dilution?

stock solution = 0.001 M

lets say we want to make 1000mL of final solution of molarity 11.6 x 10^-7 M

then

M1V1 = M2V2     ...(1)

M1 = 0.001         V1 = x

M2 = 11.6 x 10^-7 M                   V2 = 1000 mL

now putting in (1) and solving for x gives

x = 1.16 mL

So if we dilute 1.16mL of stock solution to add enough water to get final volume = 1000 mL, then we will get absorbance of the solution nearly 0.2

similarly,

lets say we want to make 1000mL of final solution of molarity 80 x 10^-7 M

then

M1 = 0.001         V1 = x

M2 = 80 x 10^-7 M                   V2 = 1000 mL

now putting in (1) and solving for x gives

x = 8 mL

So if we dilute 8 mL of stock solution to add enough water to get final volume = 1000 mL, then we will get absorbance of the solution nearly 0.8.