What volume of 2.5% (m/v) KOH can be prepared from 125 mL of a 5.0% KOH solution?

1 Answer
Dec 25, 2015

Answer:

#"250 mL"#

Explanation:

The short answer is that the volume of the target solution will be equal to #"250 mL"#.

Think about it like this - you want the concentration of the target solution to be half of the concentration of the stock solution.

Since you're keeping the amount of solute unchanged, you can say that doubling the volume of the solution will result in half the initial concentration #-># you're essentially diluting the initial solution by a factor of #2#.

Now for the long answer.

So, you're dealing with #"125 mL"# of a #"5.0% m/v"# potassium hydroxide stock solution, and are interested in finding out what volume of #"2.5% m/v"# potassium hydroxide solution can be prepared using this starting solution.

A mass by volume percent concentration is defined as mass of solute, usually expressed in grams, divided by volume of solution, usually expressed in milliliters, and multiplied by #100#

#color(blue)("%m/v" = "mass of solute"/"volume of solution" xx 100)#

In your case, the stock solution will contain

#"5.0% g/mL" = m_(KOH)/"125 mL" xx 100#

#m_(KOH) = (5.0"g"/color(red)(cancel(color(black)("mL"))) * 125 color(red)(cancel(color(black)("mL"))))/100 = "6.25 g"#

So, your stock solution contains #"6.25 g"# of potassium hydroxide. This is exactly how much potassium hydroxide the #"2.5%"# target solution must contain.

This means that you have

#"2.5% g/mL" = "6.25 g"/V_"sol" * 100#

#V_"sol" = (6.25 color(red)(cancel(color(black)("g"))))/(2.5color(red)(cancel(color(black)("g")))/"mL") * 100 = color(green)("250 mL")#