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Answer:
7.92 mL
Explanation:
There is some info missing. I think this is the original question.
<em>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.
</em>
<em>
Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.</em>
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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
What do you need help with
Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:
1). Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water. So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease. If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.
So you would have to choose some other fluid besides water ... one in
which rock salt doesn't dissolve. I don't know right now what that could
be. You'd have to shop around and find one.
2). Whatever fluid you did choose, it would also have to be less dense
than rock salt. If it's more dense, then the rock salt just floats in it, and
never goes all the way under. If that happens, then you have a tough
time measuring the total volume of the lump.
So the displacement method could perhaps be used, in principle, but
it would not be easy.