# Molecular weight of unknown acid through titration

• Chemistry

## Homework Statement

The molecular weight of a monoprotic acid HX was to be determined. A sample of 15.126 grams of HX was dissolved in distilled water and the volume were brought to exactly 250mL in a volumetric flask. Several 50mL portions of this solution were titrated against NaOH solution, requiring an average of 38.21mL of NaOH.

The NaOH solution was standardized against oxalic acid dihydratem H2C2O4(2H2O) (molecular weight: 126.066 grams per mol). The volume of NaOH solution required to neutralize 1.2596 grams of oxalic acid dihydrate was 42.21mL

a) Calculate the molarity of the NaOH solution

## Homework Equations

pH = pKa + log (A/HA)

That's what I think is going on here, no?

## The Attempt at a Solution

I'm stuck at the second paragraph of the problem, am I supposed to assume that oxalic acid dihydrate completely dissociate into the hydromium ions and the conjugate base? If it is then this turns into a simple problem.

But what if not all of the oxalic acid dihydrate dissociate completely? How am I supposed to calculate the concentration of the NaOH if I don't have the Ka of the oxalic acid dihydrate?

epenguin
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Gold Member

## Homework Statement

The molecular weight of a monoprotic acid HX was to be determined. A sample of 15.126 grams of HX was dissolved in distilled water and the volume were brought to exactly 250mL in a volumetric flask. Several 50mL portions of this solution were titrated against NaOH solution, requiring an average of 38.21mL of NaOH.

The NaOH solution was standardized against oxalic acid dihydratem H2C2O4(2H2O) (molecular weight: 126.066 grams per mol). The volume of NaOH solution required to neutralize 1.2596 grams of oxalic acid dihydrate was 42.21mL

a) Calculate the molarity of the NaOH solution

## Homework Equations

pH = pKa + log (A/HA)

That's what I think is going on here, no?

## The Attempt at a Solution

I'm stuck at the second paragraph of the problem, am I supposed to assume that oxalic acid dihydrate completely dissociate into the hydromium ions and the conjugate base? If it is then this turns into a simple problem.

But what if not all of the oxalic acid dihydrate dissociate completely? How am I supposed to calculate the concentration of the NaOH if I don't have the Ka of the oxalic acid dihydrate?

It is a simple problem.

'But what if not all of the oxalic acid dihydrate dissociate completely?'

It will have by the time you have titrated it with NaOH! Any serious acid that you can titrate will. All you need to know is that both oxalic acid groups are fairly strong acids. Revise titration curves.

Why do they use oxalic acid? By memory because it forms nice pure crystals of defined composition. Therefore you can weigh it out to good precision and make a solution of precisely defined concentration and (calculated) molarity. As opposed to messy hygroscopic stuff like NaOH where you would know the concentration only somewhat approximately by weighing. So the oxalic acid it's called a primary standard. Then after you have measured your [NaOH] by titration, it's standardised (then called secondary standard I seem to remember) and you can use it to titrate your HX.

But why am I telling you? I am doing it, as I said, from distant memory as I haven't got a book on that, while you probably do have a book or a class handout which you seem to be not using (well OK, you could have got confused about what parts are relevant, but I'm sure titration curves are in your book) and it's you who is in the better position to tell me!

Last edited:
symbolipoint
Homework Helper
Gold Member
The two dissociation constants for oxalic acid are relatively less weak compared to other weak acids so the "weak acid" quality does not make any complications for such a titration as you have done.

symbolipoint
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