A0.229-g sample of an unknown monoprotic acid is titrated with 0.112 m naoh. the resulting titration curve is shown here. part a determine the molar mass of the acid. express your answer using two significant figures.

81.8 g/mol

Explanation:

We’re being asked to calculate the molar mass of an unknown acid based on our interpretation of the titration curve.  

Recall that at the equivalence point of a titration:

We can use this equation to calculate the moles of the unknown acid. Since the unknown acid only has a single Ka value, it is a monoprotic acid, which means there will be one equivalence point for the titration.

Mole-to-mole comparison: A mole-to-mole comparison will then show that 1 mole of the unknown acid reacts with 1 mole of Na(OH).

Molarity of NaOH: The molarity of NaOH (the base) is given as 0.112 M or 0.112 mol/L NaOH.

As for the volume of NaOH (the base) added, we can interpret this from the graph:  

we can see that at the equivalence point, 25 mL of the vase was added.

Vbase is more or less 25.00 mL

Recall:  

1 mL = 10-3 L  

Solving for MMacid:  

Again, at the equivalence point of a titration:

Molar mass is defined as the mass in grams of one mole of a substance.  

The units of molar mass are grams per mole, abbreviated as g/mol and uses the formula below:

molar mass acid = 81.8 g/mol

The molar mass of the unknown acid is 81.8 g/mol.