Question

Determine the pH during the titration of 22.7 mL of 0.134 M HCIO, by 0.134 M NaOH at the following points: (a) Before the addition of any NaOH (b) After the addition of 11.4 mL of NaOH (c) At the equivalence point (d) After adding 28.6 mL of NaOH

Determine the pH during the titration of 22.7 mL of 0.134 M HClO4 by 0.134 M NaOH at the following points

Answer 1

NaOH Titration of Strong acid with strong base Titration of 22.7 mL of 0.134 M HCIO4 with 0.134 M answer-a Before addition of NaOH Strong acid dissociates completely reaction HCIO4 + H+ + A concentration(M) 0.134 0.134 Concentration of HCIO4 = Concentration of Ht Concentration of H = 0.134 M pH = -log[H+] = -log[ 0.134 ] = 0.873 answer-b After addition of 11.4 ml NaOH number of moles – Molarity Volume number of moles of HCIO4 = 0.134 x 22.7 3.04 mmol number of moles of NaOH = 0.134 * 11.4 = 1.53 mmol moles left of HClO4 3.04 - 1.53 1.51 mmol Total Volume 22.7 + 11.40 34.1 mL moles 1.51 Concentration of HCIO4 = - = 0.044 M Volume 34.1 Concentration of H+ = 0.044 M pH = -log[H+] = -log[ 0.044 ] = 1.353 answer-cat equivalnece point all the acid is completely neutralized by base. So, pH = 7 answer-d After addition of 28.6 ml NaOH number of moles = Molarity * Volume number of moles of HCIO4 = 0.134 x 22.7 3.04 mmol number of moles of NaOH = 0.134 % 28.6 = 3.83 mmol moles left of NaOH = 3.83 - 3.04 = 0.79 mmol Total Volume = 22.7 + 28.60 = 51.3 ml moles 0.79 Concentration of NaOH = = 0.015 M Volume 51.3 Concentration of OH = 0.015 M. pOH = -log[H0] = -log[ 0.015 ] = 1.812 p H = 14 - pOH = 14 - 1.812 = 12.188