Question

How do I calculate the ionization constant for acetic acid from the measured pH of the acedic acid samples?

Data and Calculations 1. Measurement of pH and fitration of acetfic acid solution Concentration of standardized NaOH titrant Mass concentration of acetic acid-aao Trial 3 Trial 2 Trial 1 Measured pht of the acetic acid solution Mass of acetic acid solution taken for titration. Initial buret reading of NaOH titrant Final buret reading of NaOH titrant Net volume of NaOH Millimoles of NaOH to end point of titration Millimoles of acetic acid in sample Molar concentration of acetic acid solution Calculated molar mass of acetic acid 2.91 3.00 20.00 30. g . -oor ru ml mmol mmol mol/L mol/L mol/L g/mol /mol Calculation of the ionization constant for acetic acid from the measured pH of the acetic acid samples. (Average the pH and molar concentration values for the samples of acetic acid you titrated: use the average values in the calculation of the ionization constant.) Average molar concentration of acetic acid for the samples you titrated mol/L Average plH Calculate the corresponding [Hy0'1, compute the concentration of A" and HA, and calculate the dissociation constant, K 2. Measurement of the pK, of acetic acid by the half-neutralization method Concentration of standardized NaOH titrant mol/L Trial 3 Trial 2 Trial 1 Mass of acetic acid solution U.gmL mł. | -Uフ G-S Volume of NaOH added to half-neutralize the acetic acid ml. mL. Measured pH of half-neutralized solution4.0

Answer 1

Hello, you have not mentioned molarity of NaOH. I am talking it 0.1 M you have to change it with your molarity.

Mmoles of NaOH	1.3	1.3	1.3
Mmoles of acetic acid	1.3	1.3	1.3
Molar Concentration of Acetic acid M	0.104	0.1037	0.104
Molar mass of acetic acid Some part is missing

Average molar concentration of acetic acid = 0.1039 M

Average pH = 2.99

[H3O+] = 10^(-2.99) = 0.00102 M,

[A-]= [H3O+] = 0.00102 M

[HA] = 0.1039 M

Ka = 1.007 × 10^-5 , pKa = 4.99

Half method average pH at half equivalent point = 4.716 =pka

Ka = 1.92 × 10^-5