# 51.0 mL of a 1.40 M solution is diluted to a total volume of 208 mL. A 104-mL portion of that...

## Question:

51.0 mL of a 1.40 M solution is diluted to a total volume of 208 mL. A 104-mL portion of that solution is diluted by adding 141 mL of water. What is the final concentration? Assume the volumes are additive.

## Dilution:

In doing dilution, a certain volume of the solvent is added to an aliquot or to a part of the concentrated solution. This is often used in the preparation of reagents from a stock solution. This is also done in certain analyses such as in spectroscopy to allow the concentration of the analyte to be at the level supported by Beer's law.

The equation that will be used is:

{eq}\rm C1V1 = C2V2 {/eq}

where:

• C1 - initial concentration
• V1 -initial volume
• C2 - final concentration
• V2 - final volume

There are two dilutions performed in the problem which is why there are also two calculations that will be done..

First Dilution

{eq}\rm 1.40\ M \times 51.0\ mL = C2 \times 208\ mL\\[3ex] C2 = \dfrac{1.40\ M \times 51.0\ mL}{208\ mL} = 0.343\ M {/eq}

Second Dilution

The concentration calculated in the first dilution will be used in the second dilution. Here, the total final volume is not yet given which is why it will be calculated first.

{eq}\rm V2 = 104\ mL + 141\ mL = 245\ mL {/eq}

This will then be used to calculate the final concentration of the solution.

{eq}\rm 0.343\ M \times 104\ mL = C2 \times 245\ mL\\[3ex] C2 = \dfrac{0.343\ M \times 104\ mL}{245\ mL} = \boxed{\mathbf{0.146\ M}} {/eq}