Question

Question #c8cd9

Answer 1

Here's what I got.

Explanation:

Start by writing the balanced chemical equation that describes this neutralization reaction

`color(red)(2)"NaOH"_ ((aq)) + "H"_ 2"SO"_ (4(aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l))`

Notice that the reaction consumes `color(red)(2)` moles of sodium hydroxide for every mole of sulfuric acid that takes part in the reaction.

You can thus use the molarity and concentration of the sulfuric acid solution to figure out how many moles of acid were needed to neutralize the moles of sodium hydroxide

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))`

In your case, you have

`n_("H"_ 2"SO"_ 4) = "0.050 mol" color(red)(cancel(color(black)("dm"^3))) * overbrace(20.0 * 10^(-3)color(red)(cancel(color(black)("dm"^(-3)))))^(color(blue)("volume in dm"^3))`

`n_("H"_ 2"SO"_ 4) = "0.00100 moles H"_ 2"SO"_4`

According to the aforementioned mole ratio, this many moles of acid neutralized

`0.00100 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(red)(2)color(white)(a)"moles NaOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.00200 moles NaOH"`

You know that this many moles of sodium hydroxide were present in the `"25.0 cm"^3` sample of the diluted solution. Use this information to find the concentration of the sample

`c = "0.00200 moles"/(25.0 * 10^(-3)"dm"^3) = "0.0800 mol dm"^(-3)`

Now, the important thing to realize here is that the concentration of the `"25.0 cm"^3` sample is equal to the concentration of the diluted `"250 cm"^3` solution.

That is the case because the `"25.0 cm"^3` sample is taken directly from the `"250 cm"^3` sample, without any dilution involved.

You can thus say that the concentration of the diluted solution is equal to `"0.0800 mol dm"^(-3)`.

To find the concentration of the original, undiluted solution, use the fact that when a dilution is performed, the concentration of the solution is decreased because

• the number of moles of solute remains constant
• the volume of the solution is increased

You can thus use the dilution factor, which gives you the ratio that exists between the final volume of the solution, i.e. the volume of the diluted solution, and the initial volume of the solution, i.e. the volume of the concentrated solution, to find the initial concentration of the solution

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

In your case, you have

`"DF" = (250 color(red)(cancel(color(black)("cm"^3))))/(40.0color(red)(cancel(color(black)("cm"^3)))) = color(blue)(6.25)`

This tells you that the original solution was `color(blue)(6.25)` times more concentrated than the diluted solution

`c_"concentrated" = color(blue)(6.25) * "0.0800 mol dm"^(-3) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.50 mol dm"^(-3))color(white)(a/a)|)))`

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the diluted solution, i.e. for `"250 cm"^3`.