Question

solve for Ka of acid please

Data: Experiment 2; Determination of an Acid lonization Constant A. Determination of the Molar Mass of an Unknown Acid Trial 3 Trial 2 Trial 1 400 405 405 Sample Mass 0100m O.Sm) O.1D Initial reading NaOH buret 3H.3m 34ml 34 3ml OH buret Final reading o.0342 0-034 0-0343 Volume NaOH used O.0034 Moles NaOH used TLTOO O.0034 Moles H in sample O.0034 601 220.5glmal 23.lginmal 19.4glmol Unknown Acid Molar Mass (MM) B. Determination of K, for the Unknown Acid Trial 3 Trial 2 Trial 1 NaOH volume at half way -0.5mL e o w.0 pH of this solution NaOH volume at half way la o le.o pH of this solution NaOH volume at half way +0.5mL pH of this solution Le.O la.0 K, of acid

Answer 1

For a weak acid, at half equivalence point i.e. at half volume of NaOH, pH = pKa

Now, given in B., pH = 6.0

As this is pH at half equivalence point, pH = pKa,

pKa = 6.0

pKa = -logKa

6.0 = -logKa

Ka = 10^-6.0

So, Ka = 1 x 10^-6