Question

What is the concentration of SCN- in a solution prepared by adding 3.0 mL of 0.0020M SCN- to a test tube and diluting to a total volume of 10.0 mL?

Answer 1

To determine the concentration of SCN- in the final solution, we can use the dilution formula:

C1V1 = C2V2

Where:

C1 = Initial concentration of the solution

V1 = Initial volume of the solution

C2 = Final concentration of the solution

V2 = Final volume of the solution

In this case, the initial volume of SCN- solution (V1) is 3.0 mL, the initial concentration (C1) is 0.0020 M, and the final volume (V2) is 10.0 mL.

Now we can plug in these values into the dilution formula and solve for C2:

(0.0020 M)(3.0 mL) = C2(10.0 mL)

Simplifying the equation:

0.0060 mol/mL = C2(10.0 mL)

Dividing both sides by 10.0 mL:

C2 = 0.0060 mol/mL / 10.0 mL

C2 = 0.00060 M

Therefore, the concentration of SCN- in the final solution is 0.00060 M.