Answer: Concentration of original the acetic acid = 0.11M
Explanation:
This neutralization reaction takes place between acetic acid, CH 3 COOH , a weak acid, and sodium hydroxide, NaOH , a strong base. Since the neutralization of the acid is not complete as indicated by the pH of the resulting solution, a buffer solution is formed that will contain acetic acid and its conjugate base, the acetate anion.
The balanced ionic equation of the reaction is as follows:
CH₃COOH + OH⁻ ----> CH₃COO⁻ + H₂O
Since the pH of the solution and Ka of the acid is given, the conjugate acid-base ratio is to be found using the Henderson-Hasselbalch equation:
pH = pKa + log ([conjugate base] / [conjugate acid])
where pH = 5.10, pKa= -log Ka
pKa = -log(1.8 x 10⁻⁵)
pKa = 4.74
log([conjugate base] / [conjugate acid]) = pH - pKa
log([conjugate base] / [conjugate acid]) = 5.10 - 4.74 = 0.36
[conjugate base] / [conjugate acid] = antilog(0.36) = 2.3
The moles of base reacted = no of moles of conjugate base
no of moles of NaOH reacted = molarity * vol in liters
no of moles of NaOH reacted= 0.100M * (20mL/1000mL*1L) = 0.002moles
[conjugate base] = no of moles / volume in liters
total volume of solution = (40 + 20)mL = 60mL
[conjugate base] = 0.002/ 60mL/1000mL*1L) = 0.033M
[conjugate acid] = 2.3 * [conjugate base] = 2.3 *0.033M
[conjugate base] = 0.076M
Concentration of original the acetic acid = [conjugate base] +[conjugate acid]
Concentration of original the acetic acid = 0.076M + 0.033M = 0.11M