4.00 grams of an unknown monoprotic acid is titrated with 0.75 m naoh. it takes 88.81 ml of naoh to completely neutralize the acid. what is the molecular weight of the acid?

The molecular weight of the acid is 60.05 g/mol

Explanation:

Let's state the balanced chemical equation to represent the neutralization reaction:

NaOH + HAc → NaAc + H2O

where HAc is the representation of the monoprotic acid. As we can see, the relationship between the base and the acid is 1:1, that is 1 mole of NaOH reacts with 1 mole of the monoprotic acid HAc. So, let's calculate the moles of NaOH that where present in the 88.81 mL aliquot used to neutralize the acid:

1000 mL 0.75 moles of NaOH

88.81 mL --- x = (88.81 mL × 0.75 moles)/1000 mL = 0.0666075 moles NaOH

As we stated before, 1 mole of NaOH will react with 1 mole of HAc, so 0.0666075 moles of NaOH will reacted with 0.0666075 moles of the acid. Having said that, because we already know the mass of the acid, we are able to determine the molecular weight of it:

0.0666075 moles of HAc 4.00 g

1 mole of HAc x = (1 mole × 4.00 g)/0.0666075 moles = 60.05 g/mole