Question

A 69.0 mL portion of a 1.60 M solution is diluted to a total volume of 218 mL. A 109 mL portion of that solution is diluted by adding 187 mL of water. What is the final concentration? Assume the volumes are additive.

Answer 1

To calculate the final concentration, dilution steps are evaluated sequentially using the formula M1V1 = M2V2. After two sets of dilutions, the final concentration of the solution is determined to be 0.185 M.

Step-by-step explanation:

To determine the final concentration after a sequence of dilutions, we use the dilution equation M1V1 = M2V2, where M1 and V1 are the molarity and volume of the initial solution, respectively, and M2 and V2 are the molarity and volume of the final solution, respectively.

The student's question involves two dilution steps. In the first step, the original 1.60 M solution is diluted from 69.0 mL to 218 mL. The molarity after this dilution can be calculated:

M1V1 = M2V2

For the first dilution, M1 is 1.60 M and V1 is 69.0 mL. V2 is 218 mL. We solve for M2, which will be used as the M1 for the second dilution.

1.60 M * 69.0 mL = M2 * 218 mL

M2 = (1.60 M * 69.0 mL) / 218 mL = 0.50 M

In the second dilution, a 109 mL portion of this new 0.50 M solution is further diluted by adding 187 mL of water. Using the dilution equation again:

0.50 M * 109 mL = Mf * (109 mL + 187 mL)

Mf = (0.50 M * 109 mL) / (109 mL + 187 mL)

Mf = (0.50 M * 109 mL) / 296 mL = 0.185 M

The final concentration of the solution is 0.185 M.