Question

Consider the titration of 25.0 mL of 0.050 0 MHClO_4 with 0.100 M KOH. Find the pH of the flask solution after 13.00 mL of titrant have been added. (a) 1.47 (b) 2.88 (c) 7.00 (d) 11.12 (e) 12.53

Answer 1

the reaction is

HClO₄ + KOH KClO₄ + H₂O

No. of moles of Hclo_4 = 0.025 lit times 0.05 M = 0.00125 no. of moles of KOH = 0.013 lit times 0.1 M = 0.0013 n_OH^- > n_H^+ Therefore no. of moles of OH^- left = 0.0013 - 0.00125 = 0.00005 p^OH = -log [OH^-] [OH^-] = no. of moles/total volume = 0.00005 moles/0.038 lit = 0.00131 Rightarrow p^OH = -log [0.00131] = 2.88 p^H = 14 - p^OH = 14 - 2.88 = 11.12.

1 mole of HClO4 reacts with 1 mole of KOH