58.0 mL of a 1.30 M solution is diluted to a total volume 248 mL. A 124-mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive

1 Answer
Jul 8, 2016

Answer:

#"0.130 M"#

Explanation:

You can solve this problem by using the dilution factor, #"DF"#, which essentially tells you the factor by which the concentration of an initial solution decreased upon dilution.

The dilution factor is calculated by using either volumes or molarities

#color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"color(white)(a/a)|)))#

So, the dilution factor for the first dilution

#"58.0 mL " -> " 248 mL"#

will be

#"DF"_1 = (248 color(red)(cancel(color(black)("mL"))))/(58.0color(red)(cancel(color(black)("mL")))) = 4.276#

The first solution was diluted by a factor of #4.276#, which means that its concentration decreased by a factor of #4.276#. Therefore, the concentration of the diluted solution will be

#c_"diluted" = c_"concentrated"/"DF"_ 1#

#c_"diluted" = "1.30 M"/4.276 = "0.304 M"#

Now, you take a sample of #"124 mL"# of this diluted solution and add another #"165 mL"# of water. The final volume of the solution will be

#V_"final" = "124 mL" + "165 mL" = "289 mL"#

The concentration of the #"124 mL"# sample is equal to the concentration of the first diluted solution, i.e. #"0.304 M"#.

The dilution factor for the second dilution will be

#"DF"_ 2 = V_"final"/V_"sample"#

#"DF"_ 2 = (289 color(red)(cancel(color(black)("mL"))))/(124color(red)(cancel(color(black)("mL")))) = 2.331#

Therefore, the concentration of the final solution will be

#c_"final" = c_"sample"/"DF"_ 2#

#c_"final" = "0.304 M"/2.331 = color(green)(|bar(ul(color(white)(a/a)color(black)("0.130 M")color(white)(a/a)|)))#

The answer is rounded to three sig figs.