Question

Question #80271

Answer 1

Here's what I got.

Explanation:

You know that you're dealing with a triprotic acid, so start by writing the balanced chemical equation that describes its reaction with sodium hydroxide

`"H"_ 3"A"_ ((aq)) + color(blue)(3)"NaOH"_ ((aq)) -> "Na"_ 3"A"_ ((aq)) + 3"H"_ 2"O"_ ((l))`

Notice that a complete neutralization requires `color(blue)(3)` moles of sodium hydroxide for every `1` mole of acid.

Use the molarity and volume of the sodium hydroxide solution to calculate the number of moles of solute it contained

`23.36 color(red)(cancel(color(black)("mL solution"))) * "0.107 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0024995 moles NaOH"`

This means that the acid solution used in the titration contained

`0.0024995 color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole H"_3"A")/(color(blue)(3)color(red)(cancel(color(black)("moles NaOH")))) = "0.0008332 moles H"_3"A"`

Now, you know that this solution was made by taking a `"20-mL"` sample of the original solution, the one that contained `"2 g"` of acid in `"250 mL"` of solution.

You can say that the `"20-mL"` sample contained

`20 color(red)(cancel(color(black)("mL solution"))) * ("2 g H"_3"A")/(250color(red)(cancel(color(black)("mL solution")))) = "0.16 g H"_3"A"`

Since you know that `0.0008332` moles of acid have a mass of `"0.16 g"`, you can say that the molar mass of the acid is equal to

`"molar mass" = "0.16 g"/"0.0008332 moles" = "192 g mol"^(-1)`

The relative molecular mass of the acid, `A_r`, is a unitless quantity calculated by dividing the molar mass by `"1 g mol"^(-1)`.

`A_r = (192 color(red)(cancel(color(black)("g mol"^(-1)))))/(1color(red)(cancel(color(black)("g mol"^(-1))))) = color(darkgreen)(ul(color(black)(190)))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of the acid present in the original solution.

This means that the answer should actually be given as

`A_r = 200`