# What is the Importance of Choosing the Correct Endpoint in a Titration?

• Valenti
In summary, the conversation discusses a problem with a lab report where the molar mass of an unknown diprotic acid needs to be determined by comparing it to a list of given acids. The individual is having trouble with the calculations, but after being given experimental information and equations, they are able to correctly find the formula weight for H2X. It is also noted that the endpoint used for the calculation depends on the speed of pH change and the stoichiometry of the neutralization needs to be taken into consideration.
Valenti
Hey, I've been having some trouble with my Lab report. After doing a titration of an unknown diprotic acid we had to find the molar mass of the acid and then find out which acid it is by comparing it to a list of given acids with molar masses. But when I run through the calculations my answer is off by quite a bit to list (lowest molar mass there being 104)

1. Homework Statement

Determine the molar mass of H2X based on the number of moles of NaOH used in the titration.

Concentration of NaOH = 0.2M
Mass of unknown = 0.0987 g
Volume used (first equivalence point) = 10.49 mL
Volume used (second equivalence point) = 20.31 mL

Endpoints were found via 1st and 2nd derivative
The 2nd equivalence point is steeper so it will be used for the calculation

n=m/M
n=cv

## The Attempt at a Solution

n=cv
Moles of NaOH = 0.2(20.31/1000)
= 0.004 mols
H2X : NaOH = 1:2

Mols of H2x = 0.002 mols

M=m/n
0.0987/0.002
M= 49.35g

Dissolved 0.0987 grams of H2X and titrated to second equivalence point volume 20.31 mL of 0.2 M NaOH

0.2(moles/liter)*0.02031*liters=0.004062 moles of NaOH
which means that
0.0987 grams of H2X contains 0.004062 moles of H+1 .

Can you use this to find the formula weight for H2X ?

Last edited:
symbolipoint said:
Dissolved 0.0987 grams of H2X and titrated to second equivalence point volume 20.31 mL of 0.2 M NaOH

0.2(moles/liter)*0.02031*liters=0.04062 moles of NaOH
which means that
0.0987 grams of H2X contains 0.04062 moles of H+1 .

Can you use this to find the formula weight for H2X ?

using M=m/n I get formula weight to be 24.29g/mol

I got 48.6 g/mol (without rounding intermediate results, which you should never do). Your number is in a correct ballpark (that is, correctly calculated from the given data). No idea what and where went wrong, but it is not the calculation to blame.

symbolipoint
Borek said:
I got 48.6 g/mol (without rounding intermediate results, which you should never do). Your number is in a correct ballpark (that is, correctly calculated from the given data). No idea what and where went wrong, but it is not the calculation to blame.

When doing the same calculation for the first endpoint (10.24 mL used) I get a molar mass of 98g/mol. Should it matter which endpoint I use for the calculation?

Valenti said:
When doing the same calculation for the first endpoint (10.24 mL used) I get a molar mass of 98g/mol.

At the first endpoint what you have neutralized was a monoprotic acid, not a diprotic.

First, or Second endpoint to use - depends on how fast pH changes with change in titrant volume. You still need to be sure titration is complete enough so you can decide if you have one or two (or maybe more) endpoints.

symbolipoint said:
First, or Second endpoint to use - depends on how fast pH changes

Yes, one should choose the endpoint which is easier to detect, at the same time they can't be treated the same way as they mean different stoichiometry of the neutralization, which is apparently what OP missed.

## What is the molar mass of diprotic?

The molar mass of diprotic refers to the mass of one mole of a diprotic compound, which is a compound that contains two hydrogen atoms bonded to an element or molecule.

## How do you calculate the molar mass of diprotic?

To calculate the molar mass of diprotic, you need to determine the atomic mass of each element in the compound and then multiply it by the number of atoms present. Then, add all the masses together to get the total molar mass.

## Why is the molar mass of diprotic important?

The molar mass of diprotic is important because it is used to convert between the mass of a substance and the number of moles present. It is also used in stoichiometric calculations and determining the concentration of a solution.

## Can the molar mass of diprotic change?

The molar mass of diprotic is a constant value for a specific compound and does not change. However, if the compound is a mixture of different isotopes, the molar mass can vary slightly due to the different masses of the isotopes.

## What is the difference between molar mass and atomic mass?

Molar mass refers to the mass of one mole of a substance, while atomic mass refers to the mass of one atom of an element. Molar mass takes into account the number of atoms present in a compound, while atomic mass is the average mass of all the isotopes of an element.

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