Question

Constant-boiling HCl can be used as a primary standard for acid-base titrations. A 50.00 mL sample of constant-boiling HCl with a concentration of 0.1225 M was collected and titrated to an end point with 35.60 mL of Ba(OH)2 solution. What is the molarity of the Ba(OH)2 solution?

Answer 1

Balanced neutralization reaction is

2HCl + Ba(OH)2  → 2H2O + BaCl2

no. of moles = molarity X volume of solution in liter

molarity of HCl = 0.1225 M

volume of HCl = 0.050 L ( 1 ml = 0.001 L then 50 ml = 50 X 0.001 = 0.050 L )

moles of HCl = 0.1225 X 0.050 = 0.006125 moles

According to balanced neutralization reaction 2 mole HCl react with 1 mole of Ba(OH)2 molar ratio between HCl to Ba(OH)2 is 2:1 thefore to react with 0.006125 moles of HCl required Ba(OH)2 = 0.006125 / 2 = 0.0030625 moles

moles of Ba(OH)2 = 0.0030625 moles

Volume of Ba(OH)2 = 0.0356 L ( 1 ml = 0.001 L then 35.60 ml = 35.60 X 0.001 = 0.0356 L )

Molarity = no.of moles / volume of solution in liter

molarity of Ba(OH)2 = 0.0030625 / 0.0356 = 0.086 M

molarity of Ba(OH)2 solution  = 0.086 M