Answer:
97.9 g/mol
Explanation:
There is some info missing. I think this is the original question.
An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.0600 M NaOH solutions. When the titration reaches the equivalence point, the chemist finds he has added 95.9 mL of NaOH solution.
Calculate the molar mass of the unknown acid.
Let's consider the neutralization between a generic triprotic acid and NaOH.
H₃X + 3 NaOH → Na₃X + 3 H₂O
The moles of NaOH that reacted are:
0.0600 mol/L × 0.0959 L = 0.00575 mol
The molar ratio of NaOH to H₃X is 3:1. The moles of H₃X are 1/3 × 0.00575 mol = 0.00192 mol.
The molar mass of the acid is:
0.188 g / 0.00192 mol = 97.9 g/mol
