Question

A student titrates 0.1719 g of an unknown monoprotic acid to the equivalence point with 21.85 mL of 0.1062 M NaOH. Which of the following is most likely to be the unknown acid? propionic acid (MM 74.08 g/mol) nitrous acid (MM 47.01 g/mol) chlorous acid (MM 68.46 g/mol) benzoic acid (MM 122.12 g/mol) lactic acid (MM: 90,08 gimol)​

Answer 1

Propionic acid

Step-by-step explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

21.85 mL of 0.1062 M NaOH react.

0.02185 L × (0.1062 mol/L) = 2.320 × 10⁻³ mol

Step 3: Calculate the reacting moles of HA

The molar ratio of HA to NaOH is 1:1. The moles of HA are 1/1 × 2.320 × 10⁻³ mol = 2.320 × 10⁻³ mol.

Step 4: Calculate the molar mass of HA

0.1719 g of HA correspond to 2.320 × 10⁻³ moles.

MM = 0.1719 g/2.320 × 10⁻³ mol = 74.09 g/mol

With this molar mass, the most likely acid is propionic acid.