Question

Calculate pH for a weak acid/strong base titration. Determine the pH during the titration of 60.9 mL of 0.396 M hypochlorous acid (K4 = 3.5x10-8) by 0.396 M KOH at the following points. (a) Before the addition of any KOH (b) After the addition of 15.0 mL of KOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 91.4 mL of KOH

Answer 1

Titration of Weak acid with strong base Titration of 60.9 mL of 0.396 M HCIO with 0.396 M KOH answer-a Before addition of KOH Construct ICE table to obtain their equilibrium concentrations: reaction HClO H H+ + Clo Initial(M) 0.396 Change(M) +x Equilibrium(M) 0.396-X Equilibrium expression: Ka = [H+][ CIO- 1/ HCIO ] 3.5x108 = (x)/( 0.396 -x) Calculating x Value: x2 + 3.5x10-8 X - 0.000000013860 = 0 By solving the above quadratic equation, we get x = * = 0.000118 M Calculating pH: pH = -log[H+ = -log[ 0.000118 ] = 3.93

KOH + H2O X answer-b After addition of 15.00 mL pKa = -log[Ka] = -log[ 3.5x10-8 ] 7.46 HCIO + KOHH KCI number of moles = Molarity x Volume number of moles of HCIO = 0.396 x 60.9 number of moles of KOH = 0.396 X 15.0 moles left of HCIO = 24.1 - 5.9 number of moles of KCl = 5.9 mmol According to Henderson-Hasselbalch equation pH = pka + log[salt/acid] = 7.46 + log [ 5.9 / 18.2 pH = 6.97 = = = 24.1 mmol 5.9 mmol 18.2 mmol X ]

answer-c At halfequivalence point pH = pka = 7.46 answer-d at equivalnece point only salt remains the number of moles acid = base = salt = 24.1 mmol M1 x V1 = M2 * V2 = 0.396 * 60.9 = 0.396 * V2 Volume of Base V2 60.9 total volume = 60.9 + 60.9 = 121.8 mL moles 24.1 Concentration of KCl = - = 0.20 M Volume 121.8 14 + 7.46 + log 0.21 pH = [14+ pka +logo ]/2 = pH = 10.38

answer-e After addition of 91.4 mL KOH number of moles – Molarity * Volume number of moles of HCIO = 0.396 x 60.9 = number of moles of KOH = 0.396 X 91.4 = moles left of KOH = 36.2 - 24.1 = Total Volume 60.9 + 91.40 moles 12.1 Concentration of KOH - = Volume 152.3 Concentration of OH = 0.079 M pOH = -log[H0] = -log[ 0.079 ] = 1.10 pH = 14 - pOH = 14 - 1.10 = 12.90 24.1 mmol 36.2 mmol 12.1 mmol 152.3 mL + 11 0.079 M