Question

Determine the pH during the titration of 25.0 mL of 0.322 M nitrous acid (K4 = 4.5x104) by 0.350 M KOH at the following points. (a) Before the addition of any KOH (b) After the addition of 5.80 mL of KOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 34.5 mL of KOH

Answer 1

Titration of Weak acid with strong base Titration of 25.0mL of 0.322 M HNO2 with 0.350 M KOH answer-1 Before addition of KOH Construct ICE table to obtain their equilibrium concentrations: reaction HNO2 + H+ + OH Initial(M) 0.322 Change(M) -X +X Equilibrium(M) 0.322-X ka ] = - [H+ OH- [ HNO2 ] 4.5*104 = 4.5x104 _[x][x] = 0.32-X 0.011815 M By solving the above equation, we get x = * = pH = -log[H+] = -log[ 0.011815 ]= 1.93 H2O + answer-2 After addition of 5.80 mL KOH pka = -log[Ka] = -log[ 4.5x10-4 ]= 3.35 HNO2 + KOH KNO2 + number of moles - Molarity * Volume number of moles of HNO2 = 0.322 25.0 = number of moles of KOH = 0.350x 5.80 = moles left of HNO2 = 8.05 - 2.03 = number of moles of KNO2 = 2.03 mmol According to Henderson-Hasselbalch equation pH = pka + log[salt/acid] = 3.35 + log [ 2.03 / 6.02 ] pH = 2.87 8.05 mmol 2.03 mmol 6.02 mmol

answer-3 At halfway point pH = pka = 3.35 answer-4 at equivalnece point only salt remains the number of moles acid = base = salt = 8.05 mmol M1 V1 = M2 x V2 + 0.322 25.0 = 0.35 x V2 Volume of Base V2 = 23.00 total volume = 25.0 + 23.0 = 48.0 mL moles 8.05 Concentration of KNO2 Volume 48.0 = 0.17 M 14 + 3.35 + log 0.17 ] pH = [14+ pka +logc ]/2 = pH = 2 8.29 answer-5 After addition of 34.5 mL KOH number of moles - Molarity Volume number of moles of HNO2 = 0.322 X 25.0 = 8.05 mmol number of moles of KOH = 0.350 x 34.5 = 12.08 mmol moles left of KOH = 12.08 - 8.05 = 4.03 mmol Total Volume = 25.0 + 34.50 = 59.5 ml moles 4.03 Concentration of KOH - = 0.068 M Volume 59.5 Concentration of OH = 0.068 M pOH = -log[H0] = -log[ 0.068] = 1.17 pH = 14 - pOH = 14 - 1.17 = 12.83