What is the concentration of a sulphuric acid solution if 25.0 mL of the 36.0% w/v acid is taken by pipette and added to 500. mL of water?

Apr 2, 2017

1.71%

Explanation:

The idea here is that you should expect the concentration of the acid to decrease as a result of the dilution because the amount of acid remains unchanged, but the volume of the solvent increases.

When you dilute a solution, the increase in volume that results from adding solvent must match the decrease in concentration. In other words, the concentration of the solution decreases by the same factor as its volume increases.

color(darkblue)(ul(color(black)(overbrace(" DF ")^(color(blue)("the dilution factor")) = c_"concentrated"/c_"diluted" = V_"diluted"/V_"concentrated")))

Now, you know that after you add the acid to the water, and keep in mind that you must never add water to acid, the total volume of the solution will be

${V}_{\text{diluted" = "25.0 mL" + "500. mL}}$

${V}_{\text{diluted" = "525 mL}}$

This means that the volume of the solution decreased by a factor of

"DF" = (525 color(red)(cancel(color(black)("mL"))))/(25.0color(red)(cancel(color(black)("mL")))) = color(blue)(21)

You can thus say that the concentration of the solution must decrease by a factor of $\textcolor{b l u e}{21}$.

Since

$\text{DF" = "% concentrated"/"% diluted" implies "% diluted" = "% concentrated"/"DF}$

you will have

"% diluted" = "36.0 %"/color(blue)(21) = color(darkgreen)(ul(color(black)(1.71%)))

The answer is rounded to three sig figs.

You can double-check the result by using the solution's mass by volume percent concentration to figure out the mass of sulfuric acid present in the initial sample

25.0 color(red)(cancel(color(black)("mL solution"))) * overbrace(("36.0 H H"_2"SO"_4)/(100color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 36.0% w/v")) = "9.00 g H"_2"SO"_4

After the dilution, the mass of sulfuric acid present in '100 mL" of the diluted solution will be

100color(red)(cancel(color(black)("mL solution"))) * ("9.00 g H"_2"SO"_4)/(525color(red)(cancel(color(black)("mL solution")))) = "1.7143 g H"_2"SO"_4

Therefore, you will once again have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% H"_2"SO"_4 = "1.71% w/v}}}}$