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>
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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>
