Question

An unknown acid is titrated with 0.1 M NaOH following a reaction. Calculate the Molecular Weight and identify what the compound could possibly be if the amount of titrant used was 5.2 mL and the amount of unknown acid is 1.632 grams. (I believe the compound is benzoic acid or close to it; I am mainly looking for the molecular weight).

Answer 1

0.1 M NaOH = 0.1 moles/ litre = 0.1 mol/ L

Volume of tritrant (NaOH) used = 5.2 ml = 5.2/1000 = 0.0052 L

Moles of tritrant (NaOH) used = 0.0052 L x 0.1 mol /L = 0.00052 mol

Note: The moles of titrant used = Moles of acid present

Therefore,

Moles of acid present = 0.00052 mol

Amount of acid used = 1.632 grams

We know that,

Moles = Weight / Molecular Weight

Therefore,

Molecular weight of acid = Weight (acid) / Moles (acid)

Molecular weight of acid = 1.632 g/ 0.00052 mol = 3138.46 g/mol

Now thinking practically, these high molecular weights compounds are not the ones that are used for titration.

Conclusion: There is some mistake in any one of the value out of the following:

1. The value of normality is incorrect

2. The value of the weight of acid is incorrect

3. The value of the titrant (in ml) is incorrect

You can see the error that has occurred, and use the above methodology to arrive at the correct answer.

Thank you.

Have a great day.