What volume (in L) of a 3.95 M potassium chloride, #KCl#, solution would be needed to make 325 mL of a 2.76 M solution by dilution?
1 Answer
Explanation:
One possible strategy to use here is to find the dilution factor first, then use it to find the volume of the stock solution that is needed in order to make your target solution.
The dilution factor essentially tells you how many times more concentrated the stock solution was compared with the target solution.
#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_"stock"/c_"target"color(white)(a/a)|)))#
In your case, the dilution factor is equal to
#"D.F." = (3.95 color(red)(cancel(color(black)("M"))))/(2.76color(red)(cancel(color(black)("M")))) = 1.43#
Now, because a dilution implies that the number of moles of solute remains constant, the dilution factor also tells you the ratio that exists between the volume of the target solution, i.e. the diluted solution, and the volume of the stock solution, i. e. the concentrated solution.
#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"target"/V_"stock"color(white)(a/a)|)))#
In your case, the volume of the stock solution needed to make the target solution would be
#V_"stock" = V_"target"/"D.F."#
#V_"stock" = "325 mL"/1.43 = "227 mL"#
Expressed in liters, the answer will be
#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.227 L")color(white)(a/a)|))) -># rounded to three sig figs
So, in order to make
Notice that the concentration of the solution decreases by the same ratio by which the volume increases. This is what the dilution factor is useful for.
