# Question #c8cd9

##### 1 Answer

#### Answer:

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 **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 **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 **equal** to the concentration of the *diluted*

That is the case because the **directly from** the

You can thus say that the concentration of the **diluted** solution is equal to

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

thenumber of molesof solute remainsconstantthevolumeof the solution isincreased

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 **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