Problem PageQuestion A chemistry student weighs out of acetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.

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

43.452

step-by-step explanation:

284 times .153 = 43.452

it is a solid

explanation: