How to calculate the concentration of the acid solutions?

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All the help is appreciated!! Information is in the picture.

1 Answer
Sep 14, 2016

Answer:

#"0.0800 mol dm"^(-3)#

Explanation:

All you have to do here is look at the titration graph and find the equivalence point for the two titrations.

The cool thing to notice here, and the problem actually provides this information, is that because both acids are monoprotic and their solutions have equal concentrations, they will have the same equivalence point.

Keep in mind that you can base that statement on the fact that both acids are being titrated using the same #"0.100 mol dm"^(-3)# sodium hydroxide solution.

In other words, the same equivalence point means that complete neutralization occurs when you're adding the same volume of sodium hydroxide solution.

For hydrochloric acid, #"HCl"#, you have

#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

Here #1# mole of base is needed to neutralize #1# mole of acid.

Moreover, you know that because you're titrating a strong acid with a strong base, the pH at the equivalence point, i.e. at complete neutralization, must be equal to #7#.

As you can see from the graph, the titration curve for hydrochloric acid shows that #"20.0 cm"^(-3)# of sodium hydroxide are added when the pH is equal to #7#.

In other words, #"20 cm"^(-3)# of #"0.100 mol dm"^(-3)# sodium hydroxide solution contains the same numer of moles of hydroxide anions, #"OH"^(-)#, as #"25.0 cm"^(-3)# of hydrochloric acid solution contains hydronium cations, #"H"_3"O"^(+)#.

You thus have

#20.0color(red)(cancel(color(black)("cm"^(-3)))) * (1color(red)(cancel(color(black)("dm"^(-3)))))/(10^3color(red)(cancel(color(black)("cm"^(-3))))) * "0.100 moles OH"^(-)/(1color(red)(cancel(color(black)("dm"^(-3))))) = "0.00200 moles OH"^(-)#

This implies that the #"25.0 cm"^(-3)# aliquot of hydrochloric acid contains #0.00200# moles of #"H"_3"O"^(+)#, and so the molarity of the hydrochloric acid solution will be

#["HCl"] = "0.00200 moles"/(25.0 * 10^(-3)"dm"^(-3)) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.0800 mol dm"^(-3))color(white)(a/a)|)))#

Since you know that the two acid solutions have equal concentrations, you can say that

#["CH"_3"COOH"] = color(green)(bar(ul(|color(white)(a/a)color(black)("0.0800 mol dm"^(-3))color(white)(a/a)|)))#

The answers are rounded to three sig figs.

Notice that despite the fact that both solutions have the same equivalence point, their pH at complete neutralization is not equal.

At equivalence point, the weak acid solution will have #"pH" > 7# because the conjugate base of ethanoic acid, the ethanoate anion, #"CH"_3"COO"^(-)#, acts as a base in the resulting solution, i.e. it produces hydroxide anions, #"OH"^(-)#.

#"CH"_ 3"COOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

At equivalence point, you have

#"CH"_ 3"COO"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"COOH"_ ((aq)) + "OH"_ ((aq))^(-)#