40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 m naoh. after 20.0 ml of the base solution has been added, the ph in the titration flask is 5.10. what was the concentration of the original acetic acid solution? [ka(ch3cooh) = 1.8 x 10â5]

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

elements make up everything, and they combine to make compounds.

the answer is c