Question

a student titrated a 15.00-ml sample of a solution containing a monoprotic weak acid with naoh. if the titration required 13.36 ml of 0.1015 m naoh to reach the equivalence point, calculate: a. molarity of acid in the sample b. mass (in g) of acid in the sample if the molecular mass of the acid is 60.00 g/mol

Answer 1

The molarity of the acid in the sample is calculated to be 0.090416 M, and the mass of the acid in the sample is found to be 0.0813744 g.

To calculate the molarity of the acid in the sample, we use the concept of stoichiometry and the fact that the reaction between a monoprotic acid and NaOH is a 1:1 reaction. Begin by calculating the moles of NaOH titrated.

Moles of NaOH = Volume (L) × Molarity (M)

= 0.01336 L × 0.1015 M

= 0.00135624 mol of NaOH

Since the acid is monoprotic, we know one mole of acid reacts with one mole of NaOH. Therefore, there are also 0.00135624 mol of acid in the 15.00 mL sample.

a) Molarity of the acid = Moles of Acid / Volume of Acid (L)

= 0.00135624 mol / 0.015 L

= 0.090416 M

b) For the mass of the acid in the sample:

Mass = Moles × Molecular Mass

= 0.00135624 mol × 60.00 g/mol

= 0.0813744 g