Question

Calculate the pH when 54.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

Answer 1

Titration of Weak acid with strong base molarity of HBrO = 0.300 M volume of HBrO = 20.0 mL = 0.0200 L molarity of KOH = 0.150 M volume of KOH = 54.0 mL = 0.0540 L after addition of KOH Calculating initial number of moles: moles = Molarity Volume number of moles of HBrO = 0.300 x 0.0200 = 0.0060 mol number of moles of KOH = 0.150 0.0540 = 0.0081 mol total volume = 0.0200 + 0.0540 = 0.0740 L HBrO + KOH + HCOOK + H2O initial moles 0.0060 0.0081 change in moles 0.0060 0.0060 final moles 0.0000 0.0021 left moles of KOH = 0.0021 mol Calculating Concentration : concentration = moles / Volume concentration of KOH = 0.0021 / 0.0740 = 0.0284 M concentration of KOH = Concentration of OH Concentration of OH = 0.0284 M. Calculating pOH: [OH] = 0.0284 M pOH = -log(OH] pOH = -log[ 0.0284 ] pOH = 1.55 Calculating pH: pH = 14 - POH pH = 14 -1.55 pH = 12.45