Question

28.7 mL of KOH solution required 21.3 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was in the original sample? (MW KOH = 56 g/mole)

Express your answer using 3 significant figures

Answer 1

KOH + HCl --------------------> KCl + H2O

moles of KOH = moles of HCl

moles of KOH = moles of HCl   = 21.3 x 0.118 / 1000

                                                = 2.51 x 10^-3

mass = moles x molar mass

          = 2.51 x 10^-3 x 56

          = 0.141 g

mass of KOH required = 0.141 g