Question

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

Answer 1

`"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.