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kipiarov [429]
2 years ago
12

n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

enolphthalein end point. Calculate the molarity of the acetic acid solution, and the percent (by weight) of acetic acid in the solution (assuming its density to be 1.00 g/mL).
Chemistry
1 answer:
olasank [31]2 years ago
8 0

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

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

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

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

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>

<em></em>

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>

Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>

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