An analytical chemist weighs out 0.281g of an unknown diprotic acid into a 250mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.0700M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 48.3mL of NaOH solution. Calculate the molar mass of the unknown acid.

166 g/mol

Explanation:

Step 1: Write the neutralization reaction

H₂A + 2 NaOH ⇒ Na₂A + 2 H₂O

Step 2: Calculate the reacting moles of NaOH

48.3 mL of 0.0700 M NaOH react.

0.0483 L × 0.0700 mol/L = 3.38 × 10⁻³ mol

Step 3: Calculate the reacting moles of H₂A

The molar ratio of H₂A to NaOH is 1:2. The reacting moles of H₂A are 1/2 × 3.38 × 10⁻³ mol = 1.69 × 10⁻³ mol.

Step 4: Calculate the molar mass of H₂A

1.69 × 10⁻³ moles of H₂A have a mass of 0.281 g. The molar mass of H₂A is:

M = 0.281 g / 1.69 × 10⁻³ mol = 166 g/mol

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'