Question

In the reaction `H_3A(s) + 3NaOH(aq) -> 3H_2O(l) + Na_3A(aq)`, if it takes 38.22 mL of .1665 M `NaOH` to neutralize .4750 g of the acid, what is the molar mass of the acid?

Answer 1

Here's what I got.

Explanation:

Start by writing down the balanced chemical equation for this neutralization reaction

`"H"_3"A"_text((aq]) + color(red)(2)"NaOH"_text((aq]) -> 3"H"_2"O"_text((l]) + "Na"_3"A"_text((aq])`

The key to this problem is the `1:color(red)(3)` mole ratio that exists between the triprotic acid `"H"_3"A"` and sodium hydroxide, `"NaOH"`.

This mole ratio will allow you to find the number of moles of acid present in solution.

Simply put, if you know how many moles of sodium hydroxide were needed to neutralize the acid, you can use this mole ratio to figure out how many moles of acid were present in solution.

So, you know the molarity and volume of the sodium hydroxide solution used in the reaction. The number of moles of sodium hydroxide will be equal to

`color(blue)(c = n/V implies n = c * V)`

`n_(NaOH) = "0.1665 M" * 38.22 * 10^(-3)"L" = "0.0063636 moles NaOH"`

This means that the solution must have contained

`0.0063636 color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole H"_3"A")/(color(red)(3)color(red)(cancel(color(black)("moles NaOH")))) = "0.0021212 moles H"_3"A"`

As you know, the molar mass of a substance tells you what the exact mass of one mole of that substance is. In your case, you know that a `"0.4750-g"` sample of this acid contained `0.0021212` moles, which means that one mole will be equivalent to

`1 color(red)(cancel(color(black)("mole H"_3"A"))) * "0.4750 g"/(0.0021212color(red)(cancel(color(black)("moles H"_3"A")))) = "223.93 g"`

The molar mass of the acid will thus be

`M_M = color(green)("223.9 g/mol")`