Question

millimoles of NaOH titrant

Chemical Equilibria: K, of a Weak Monoprotic Acid Report Form Name: Partner's Name: (if any) Lab Section MWITTHM-TH (Circle) Data and Calculations 1. Measurement of pH and titration of acetic acid solution Concentration of standardized NaOH titrant 0.1054 Mass concentration of acetic acid 2.40 - mol/L Trial 3 Measured pH of the acetic acid solution Mass of acetic acid solution taken for titration Trial 1 3.09 20.16 0 11.4 Trial 2 3.32 30.06 Initial buret reading of NaOH titrant ml ml Final buret reading of NaOH titrant 23.0 mL mL Net volume of NaOH Millimoles of NaOH to end point of titration mmol mmol mmol Millimoles of acetic acid in sample mmol mmol m mol Molar concentration of acetic acid solution mol/L mol/L mol/L Calculated molar mass of acetic acid /mol g/mol /mol Calculation of the lonization constant for acetic acid from the measured pll 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: 12 mol/L Average pH 3.205 Calculate the corresponding [H,O'l, compute the concentration of A and HA, and calculate the dissociation constant, K. MH,O MA MHA K HOA. HA : pK, 2. Measurement of the pk, of acetic acid by the half-neutralization method Concentration of standardized NaOH titrant 1024 m ol/L Trial 3 Mass of acetic acid solution Volume of NaOH added to half-neutralize the acetic acid Trial 1 30.03 5.1 4.12 & mL Trial 2 30-02 5.8 4.72 8 ml 11. Measured pH of half-neutralized solution

Answer 1

conc. of Standardized NaOH = 0.1054 mol/L volume of NaOH = 11.6 mL ( Trial 1) in concento arion = no.of moles Volume (L) moles of NQOH = conc. X volume = 0.1054 X 0.0116 = 0.00122 mol - 11.22 mmol | Nol of NaOH = 11.4 MOL ( Trial 2) :: moles of NaOH = 0.1054 X 0.0114 L = 0.00120 mol 1.20 mmol/