A 0.5435 gram sample of an unknown acid, known to be monoprotic, was dissolved in water and titrated with 0.1960 M NaOH solution. 22.34 mL of the NaOH solution was required to reach the endpoint. What is the molecular weight of the unknown acid?

Question

jfraser · Answer

knowing the acid is monoprotic guarantees that the molar ratio of acid to base will be 1:1.
use the concentration and volume of base to find moles of base used. That moles of base must equal moles of acid that were originally present in the acid before the neutralization.
You know the mass of acid that was neutralized, so divide mass by moles to find molar mass.
(I get about 124g/mol)

anonymous · Answer

do you mind showing me a step by step process on how to solve this because I really am lost

jfraser · Answer

because the acid is monoprotic makes the chemical reaction:$$HA + NaOH \rightarrow H_2O + NaA$$ where "A" is the other part of the acid (doesn't matter what it's formula is, it's just a double replacement reaction so it stays whole).
the volume of base times concentration of base gets us moles of base used to neutralize the solution of acid
$$0.02234L NaOH * \frac{0.1960 mol}{1L NaOH} = 0.00438mol NaOH$$Because the acid is monoprotic, the SAME NUMBER OF MOLES OF ACID MUST HAVE BEEN USED ALSO. 
The molar mass of the acid is found using this moles and the mass of acid measured earlier.$$\frac{0.5435g HA}{0.00438mol HA} = \frac{124g}{1mol}$$

anonymous · Answer

The L represents?

jfraser · Answer

Liters, the concentration of a solution give in moles/Liter