Question

4. An monoprotic acid was isolated from an organic chemistry reaction. Titration of a sample of pure acid (1.2131 g) with NaOH (0.0944 M) to a phenolphthalein endpoint required 41.21 mL of base. What is the molecular weight of the unknown acid?

Answer 1

Assume the monoprotic acid is HA.

Mass of HA = 1.2131 g

Concentration of NaOH = 0.0944 M

The volume of NaOH required to reach endpoint = 41.21 mL x ( 1L/1000 mL) = 0.04121 L

The balanced reaction between an acid (HA) and base(NaOH) is as follows:

HA(aq) + NaOH(aq) $\rightarrow$ NaA(aq) + H₂O(l)

Determine the number of moles of NaOH from the given concentration and the volume of NaOH as follows:

The formula to calculate molarity is as follows:

Molarity = Number of moles / L of solution

Rearrange the formula for number of moles as follows:

Number of moles = Molarity x L of solution

Number of moles of NaOH = 0.0944 M x 0.04121 L

Number of moles of NaOH = 0.00389 mol NaOH

Use the moles of NaOH and mole ratio from the balanced chemical reaction, calculate the number of moles of acid reacted as follows:

= 0.00389 mol NaOH x ( 1 mol HA / 1 mol NaOH)

= 0.00389 mol HA

The molecular weight of the unknown acid is calculated as follows:

Molar mass = grams of acid / Moles

Molar mass = 1.2132 g / 0.00389 mol

Molar mass = 312 g/mol

Thus, the molecular weight of the unknown acid is 312 g/mol