Answer:
a. 0.393M CH₃COOH.
b. 2.360% of acetic acid in the solution
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺
<em>That means 1 mole of acid reacts per mole of NaOH.</em>
Moles of NaOH to reach the equivalence point are:
35.75mL = 0.03575L × (0.2750mol / L) = <em>9.831x10⁻³ moles of NaOH</em>
As 1 mole of acid reacts per mole of NaOH, moles of CH₃COOH in the acid solution are 9.831x10⁻³ moles.
a. As the volume of the acetic acid solution is 25.00mL = 0.02500L, the molarity of the solution is:
9.831x10⁻³ moles / 0.02500L =
<h3>0.393M CH₃COOH</h3>
b. Molar mass of acetic acid is 60g/mol. The mass of 9.831x10⁻³ moles is:
9.831x10⁻³ moles ₓ (60g / mol) = <em>0.590g of CH₃COOH</em>.
As volume of the solution is 25.00mL, the percentage of acetic acid is:
(0.590g CH₃COOH / 25.00mL) ₓ 100 =
<h3>2.360% of acetic acid in the solution</h3>
