Answer:
7.92 mL
Explanation:
There is some info missing. I think this is the original question.
A chemistry student weighs out 0.0617g of acetic acid (HCH₃CO₂) into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1300 M NaOH solution.
Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.
First, we will calculate the moles of acetic acid (MW 60.05 g/mol).
0.0617 g × (1 mol/60.05 g) = 1.03 × 10⁻³ mol
Let's consider the neutralization reaction between HCH₃CO₂ and NaOH.
HCH₃CO₂ + NaOH ⇄ NaCH₃CO₂ + H₂O
The molar ratio of HCH₃CO₂ to NaOH is 1:1. The moles of NaOH are 1.03 × 10⁻³ moles.
The volume of 0.1300 M NaOH that contains 1.03 × 10⁻³ mol is:
1.03 × 10⁻³ mol × 1000 mL/0.1300 mol = 7.92 mL
