Question

A chemist makes 0.940 L of mercury(Il) iodide (HgI_2) working solution by adding distilled water to 0.260 L of a 0.0000155 mol/L stock solution of mercury (II) iodine in water
Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer 1

To calculate the concentration of the working solution, we need to first determine the number of moles of mercury (II) iodide in the stock solution.

Given:
Volume of stock solution (V1) = 0.260 L
Concentration of stock solution (C1) = 0.0000155 mol/L

Number of moles of mercury (II) iodide in stock solution = C1 * V1
Number of moles = 0.0000155 mol/L * 0.260 L
Number of moles = 0.00000403 mol

Next, we need to determine the concentration of the working solution. The total volume of the working solution is 0.940 L, which includes the volume of the stock solution and the distilled water.

Volume of working solution (V2) = 0.940 L
Volume of distilled water = V2 - V1
Volume of distilled water = 0.940 L - 0.260 L
Volume of distilled water = 0.680 L

Since mercury (II) iodide is not being diluted when distilled water is added, the number of moles of mercury (II) iodide in the working solution remains the same as in the stock solution.

Concentration of working solution (C2) = Number of moles / Volume of working solution
Concentration = 0.00000403 mol / 0.940 L
Concentration ≈ 0.00000429 mol/L

Therefore, the concentration of the chemist's working solution is approximately 0.00000429 mol/L, rounded to 3 significant digits.