Question

Question #a034f

Answer 1

Here's what I got.

Explanation:

The `"pH"` of the nitric acid solution will tell you its molarity. As you know, nitric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations and nitrate anions.

`"HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO"_ (3(aq))^(-)`

This means that a nitric acid solution has

`["HNO"_3] = ["H"_3"O"^(+)]`

The `"pH"` of the solution is defined as

`color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))`

This means that you have

`["H"_3"O"^(+)] = 10^(-"pH")`

which, in your case, will get you

`["H"_3"O"^(+)] = 10^(-0.29) = "0.513 M"`

Now, notice that you have

`"Na"_ 2"CO"_ (3(aq)) + color(red)(2)"HNO"_ (3(aq)) -> "CO"_ (2(g)) uarr + "H"_ 2"O"_ ((l)) + 2"NaNO" _(3(aq))`

The `1:color(red)(2)` mole ratio that exists between the sodium carbonate and the nitric acid tells you that the reaction consumes `color(red)(2)` moles of nitric acid for every `1` mole of sodium carbonate that takes part in the reaction.

In your case, the reaction consumed

`25.4 color(red)(cancel(color(black)("cm"^3))) * ("0.10 moles Na"_2"CO"_3)/(10^3color(red)(cancel(color(black)("cm"^3)))) = "0.00254 moles Na"_2"CO"_3`

This implies that it also consumed

`0.00254 color(red)(cancel(color(black)("moles Na"_2"CO"_3))) * (color(red)(2)color(white)(.)"moles HNO"_3)/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3)))) = "0.00508 moles HNO"_3`

If you know the concentration of the nitric acid solution

`["HNO"_3] = "0.513 M"`

you can find the volume used in the titration

`0.00508 color(red)(cancel(color(black)("moles HNO"_3))) * (10^3color(white)(.)"cm"^3)/(0.513 color(red)(cancel(color(black)("moles HNO"_3)))) = "9.9 cm"^3`

Therefore, you can say that you titrated `"25.4 cm"^3` of `"0.10 mol dm"^(-3)` sodium carbonate solution to the equivalence point by using `"9.9 cm"^3` of `"0.513 M"` nitric acid solution.