Quantitative Titration of Unknown Solid Acid: Find Molar Mass & Formula

AI Thread Summary
To determine the molar mass and formula of an unknown monoprotic solid acid, the mass of the acid (0.7 g) and the volume of NaOH used in titration (29.5 mL at 0.095 mol/L) can be utilized. The stoichiometry of the neutralization reaction allows for the calculation of moles of the acid based on the moles of NaOH reacted. With the number of moles and the mass of the acid, the molar mass can be easily calculated. The acid dissociation constant (Ka) is not necessary for this calculation, as the stoichiometric relationship provides sufficient information. This method effectively identifies the molar mass and formula of the acid.
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If you were given a sample unknown solid acid that had a mass of 0.7 g and it was being dissolved in 50 mL of water and was being titrated with NaOH, is it possible to find the molar mass or the formula of the acid?

Given information
Mass of solid acid: 0.7 g
Volume of acid solution: 50 mL = 0.05 L
Volume of base used: 29.5 mL = 0.0295 L
[NaOH] = 0.095 mol/L

I was also told that this is monoprotic.
HX (aq) + NaOH (aq) --> H2O (l) + NaX (aq)
 
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Hi,
I would try to get the Acid dissociation constant with this little formula:
Ka=( [A-]*[H+])/[HA]
How many H+ were in the solution?
How many mol are 0,7 g H+ ?= c (O)
HA = c0 H+ - [H+].
Then you just have to search for the specific the ka in and you have you acid.
Greetings Firelion
 
No need for acid dissociation constant, in fact it is completely unrelated to problem.

If acid is monoprotic you can easily calculate number of moles from the neutralization stoichiometry. Then you know mass of the sample and number of moles in the sample - that's enough to calculate molar mass.

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oh sorry.
We leraned in our exams to calculate very different types of data with the Ka so I thought this task is similar.
 
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