Question

A 0.288 g sample of an unknown monoprotic acid is dissolved in water and titrated with a 0.115 M NaOH solution. After the addition of 17.84 mL of base, a pH of 4.92 is recorded. The equivalence point is reached when a total of 33.83 mL of NaOH is added. What is the molar mass of the acid

Answer 1

Answer:

74.0 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

At the equivalence point, 33.83 mL of 0.115 M NaOH react.

0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol

Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.

Step 4: Calculate the molar mass of the acid

3.89 × 10⁻³ moles of HA have a mass of 0.288 g.

M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol