Question #11605

1 Answer
Aug 25, 2017


Here's what I got.


The trick here is to realize that when performing a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution is equal to the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

So for any dilution calculation, you know that you must have

#"DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock" -># the dilution factor

Now, your starting solution has a volume of #"250 mL"# and a concentration of #"0.136 M"#.

As you know, solutions are homogeneous mixtures, i.e. they have the same composition throughout, so the #"25-mL"# sample of this starting solution will have a concentration of #"0.136 M"# as well.

This #"25-mL"#, #"0.136-M"# sample represents your stock solution.

After you dilute this solution, its final volume is equal to #"100 mL"#. This means that the dilution factor is equal to

#"DF" = (100 color(red)(cancel(color(black)("mL"))))/(25color(red)(cancel(color(black)("mL")))) = color(blue)(4)#

The volume of the solution increased by a factor of #color(blue)(4)#, which can only mean that its concentration decreased by a factor of #color(blue)(4)#.

#"DF" = c_"stock"/c_"diluted" implies c_"diluted" = c_"stock"/"DF"#

Therefore, you will have

#c_"diluted" = "0.136 M"/color(blue)(4) = color(darkgreen)(ul(color(black)("0.034 M")))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the final volume of the diluted solution.

As an important note, your stock solution of hydrochloric acid is not that concentrated to begin with, but you should always remember that when diluting concentrated acids, you must always add acid to water and never water to acid!