Question

Suppose you start with a solution of red dye #40 that is 3.1 ✕ 10−5 M. If you do four successive volumetric dilutions pipetting 1.00 mL of solution and diluting with water in a 40.00 mL volumetric flask, what is the molarity of the final dilution?

Answer 1

1.3 × 10⁻¹¹ M

Step-by-step explanation:

We are going to do 4 successive dilutions. In each dilution, we will apply the dilution rule.

C₁.V₁=C₂.V₂

where,

C₁ and V₁ are concentration and volume of the initial state

C₂ and V₂ are concentration and volume of the final state

First dilution

C₁ = 3.1 × 10⁻⁵ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

`C_(2)=(C_(1).V_(1))/(V_(2)) =(3.1* 10^(-5)M * 1.00mL )/(40.00mL) =7.8 * 10^(-7)M`

Second dilution

C₁ = 7.8 × 10⁻⁷ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

`C_(2)=(C_(1).V_(1))/(V_(2)) =(7.8 * 10^(-7)M * 1.00mL )/(40.00mL) =2.0 * 10^(-8)M`

Third dilution

C₁ = 2.0 × 10⁻⁸ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

`C_(2)=(C_(1).V_(1))/(V_(2)) =(2.0 * 10^(-8)M * 1.00mL )/(40.00mL) =5.0 * 10^(-10)M`

Fourth dilution

C₁ = 5.0 × 10⁻¹⁰ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

`C_(2)=(C_(1).V_(1))/(V_(2)) =(5.0 * 10^(-10)M * 1.00mL )/(40.00mL) =1.3 * 10^(-11)M`