Question

A buffer that contains 0.35 M of a base, B and 0.28 M of its conjugate acid BH+, has a pH of 9.71. What is the pH after 0.01 mol of HCl are added to 0.71 L of the solution? (Answer: 9.67)

Calculate the pH during the titration of 30.00 mL of 0.1000 M LiOH with 0.1000 M HClO4 after 30.5 mL of the acid have been added. (Answer: 3.08)

Calculate the pH during the titration of 20.00 mL of 0.1000 M HClO(aq) with 0.1000 M NaOH(aq) after 5 mL of the base have been added. Ka of hypochlorous acid = 3.0 x 10-8 (Answer: 7.05)

Answer 1

Given, pH of buffer = 9.71

               [H⁺] = 1.95 * 10^-10 M

         and [OH^-] = (1.0*10^-14 ) / [H⁺]

                          = 5.13*10^-5 M

             B (aq) + H₂O (l) <-----> BH⁺ (aq) + OH^- (aq)

I(M)      0.35 0.28                0

C(M) -5.13*10^-5 +5.13*10^-5 +5.13*10^-5

         -------------------------------------------------------

E(M)     0.35 0.28   5.13*10^-5

                 K_b   = [BH⁺][OH^-] / [B]

                         = 0.28 * 5.13*10^-5 / 0.35

                         = 4.104*10^-5

When 0.01 mol HCl is added to 0.71 L of this solution, concentration is 0.01 mol / 0.71 L

                                                                                                                  = 0.01408 M

             B + HCl   ------> BH⁺   + Cl^-

I(M)      0.35 0.01408 0.28

C(M) -0.01408 ^- 0.01408 +0.01408

          ------------------------------------

E(M) 0.336 0                   0.294

                 K_b   = [BH⁺][OH^-] / [B]

         4.104*10^-5= 0.294 *  [OH^-] / 0.336

              [OH-] = 4.69*10^-5 M

   

Then, [H⁺] = 2.13*10^-10 M

      so pH = 9.67