Question

16. Exactly 25.00 ml of 0.0685 M HClO₄ solution is placed in a flask and titrated with standardized 0.050M NaOH.

        A. Determine the endpoint of the titration

        B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution:

                       a.) Initially, 0.00 ml NaOH.                   b.) 5.00 ml                                c.) 15.00 ml

                       d.) 25.00 ml                                              e.) 34.00 ml                             f.) At the endpoint of the titration      

                      g.) 34.35 ml                                              h.) 45.00 ml                             i.) 50.00 ml

    

17. Exactly 25.00 ml of a 0.0685 M pentanoic acid solution,(C₄H₈O₂), is placed in a flask and titrated with standardized 0.050 M NaOH. (the pKa of this acid is 4.84).

        A. Determine the endpoint of the titration

        B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution:

                       a.) Initially, 0.00 ml NaOH.                   b.) 5.00 ml                                c.) 15.00 ml

                       d.) 25.00 ml                                              e.) 34.00 ml                             f.) At the endpoint of the titration      

                      g.) 34.35 ml                                              h.) 45.00 ml                             i.) 50.00 ml

18. Make a labeled rough sketch of pH vs. ml of NaOH for the titrations performed in questions #16 and #17

Answer 1

16. too, is a strong feid. too. is moles de redone - Moleauty & roleme (2) | -0.0685 * 25x10² = 1.1125 x103 ndes. As redou is a monoprotic Acid it needs equal moles des Naot for Complete neutralization. in no. de moles de laote needed - 1:7125x10 indes. no volume moles. Molarity (mole ) 161125 xro? - 31.25x103 L - 31.25 m) A To reach endpoint we need 34.25 min Naot toleich is 0.050M a) om so daoth, When 'om of Acid dissociates. Naot is added entire

reclou z tt + cou [++] = {reclow] - 0.0685 prl = 10g [1] = 10g (0.0685). = 1.164 6) smides c.05M NaOH moles et daotha Molasity &volume (4) 0.05 mxsxioll - 0.25 milli ndes. But we have 107125 milli moles de neid So, 0.28 milli moles ils neid neutralizes and remaining dissociates. . (H} - 107125 – 0.25 milli moles. 10 4625 x10 males moles ut 1.4625 x Molarity of H+ = 1 vol. Ctolal) (25 +5 x (Acid) (Base) - 0.04875 pH = 109 [tet) = -log Co.ou8r5) = 1.312

c) In the Similar way. moles As laore - Molarity xvol. = 0.05 x 15 x = 0.15x10 moles. moes it tot = 1.7125 x0 – 0.75*10 - v. 9625 x 10 moles. Molarity of tot – moles vol 0.9625 x 40 x 0.024) pH = -log (0.0241) = 1.6187
NOTE: As per HomeworkLib policy we are allowed to do only 4 subparts hope you understand. Thank you.