# Question aa2dd

Dec 15, 2017

$\text{1400 mL}$

#### Explanation:

The thing to remember about dilutions is that if you divide the concentration of the stock solution by the concentration of the diluted solution you will end up with the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

This ratio is called the dilution factor

$\text{DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock}$

In your case, you know that the stock solution is being diluted by a dilution factor equal to

"DF" = (2.8 color(red)(cancel(color(black)("M"))))/(1.0color(red)(cancel(color(black)("M")))) = color(blue)(2.8)#

This means that the volume of the diluted solution must be $\textcolor{b l u e}{2.8}$ times that of the stock solution.

${V}_{\text{diluted" = "DF" * V_"stock}}$

In your case, the volume of the diluted solution will be equal to

${V}_{\text{diluted" = color(blue)(2.8) * "750 mL}}$

${V}_{\text{diluted" = "2100 mL}}$

This means that the diluted solution will contain

${m}_{\text{water" = "2100 mL " - " 750 mL}}$

${m}_{\text{water" = "1400 mL}}$

of water. The answer is rounded to two sig figs.

Keep in mind that you're dealing with a strong acid, so you should not add water to the acid, but the other way around. So remember, when dealing with strong acids, you never add water to the acid, you always add the acid to the water.

In this case, you would simply add the $\text{750-mL}$ solution to enough water to make the total volume of the solution equal to $\text{2100 mL}$. This will get your solution from an initial concentration of $\text{2.8 M}$ to a final concentration of $\text{1 M}$.