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melisa1 [442]
3 years ago
12

A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5

mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119
Chemistry
1 answer:
matrenka [14]3 years ago
3 0

Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

<em>1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.</em>

<em />

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

<em>0.263M of CH₃COOH is the concentration of the solution</em>

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Weight of Chloroform : = 2.862 kg

<h3>Further explanation</h3>

Given

Density 1.483 g/ml

Volume = 1.93 L

Required

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Solution

Density is a quantity derived from the mass and volume  

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