Question

Calculate the concentration of the solutions prepared as described: Show your work!

a. A 25.00 mL aliquot of 1.00 x10^-3 M methylene blue (MB) was added to a 100.00 mL volumetric flask, and diluted with water to the mark. This solution was labelled MB-1. What is the concentration of methylene blue in the new solution?

b. A second solution labelled MB-2 was prepared by adding a 10.00 mL aliquot of solution MB-1 and 10.00 mL of ethanol (0.10M). Distilled water was used to complete the solution in a 250.0 mL volumetric flask. What is the concentration of methylene blue and ethanol in MB-2 solution?

Answer 1

a. Volume of methylene blue added (V₁)= 25mL

Molarity of methylene blue(M₁) = 1.00 * 10^-3 M

Let Molarity of new methylene blue be M₂

Volume of new methylene blue solution (V₂)= 100.00mL

Now, M₁V₁ = M₂V₂

1.00 * 10^-3 M * 25.00 mL =M₂ *  100.00mL

M₂ = 2.5 * 10^-4 M

the concentration of methylene blue in the new solution =  2.5 * 10^-4 M

b) the concentration of methylene blue in the new solution (M₁)=  2.5 * 10^-4 M

volume of methylene blue taken (V₁) = 10.00 mL

Molarity of ethanol (M₂) = 0.10 M

Volume of ethanol (V₂) = 10.00 mL

Volume of new solution(V₃) = 250 mL

Let oncentration of methylene blue and ethanol in MB-2 solution be M₃

NOw, by molarity equation,

M₁V₁ + M₂V₂ = M₃V₃

( 2.5 * 10^-4 M * 10.00mL ) + ( 10mL * .10 M) = M₃ * 250 mL

(0.0025 mol + 1 mol)/ 250 mL = M₃

M₃ = 0.0040M

thus, the concentration of methylene blue and ethanol in MB-2 solution = 0.0040 M