A solution of weak acid was titrated with base NaOH . The equivalence point was reached when 36.12 ml of 0.1 M − NaOH have been added. Now, 18.06 ml 0.1 M − HCl was added to the titrated solution, the

Question

A solution of weak acid was titrated with base NaOH . The equivalence point was reached when 36.12 ml of 0.1 M − NaOH have been added. Now, 18.06 ml 0.1 M − HCl was added to the titrated solution, the pH was found to be 5.0. The pK a ​ of acid is

TestHub · Accepted Answer

At equivalence: moles acid = moles base Added HCl = half of NaOH used → forms buffer (acid = base) In buffer, pH = pKa Given pH = 5 Answer: pKa = 5

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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