Question

A student is asked to standardize a solution of barium hydroxide. He weighs out 0.912 g potassium hydrogen phthalate (KHC₈H₄O₄, treat this as a monoprotic acid).

It requires 15.2 mL of barium hydroxide to reach the endpoint.

A. What is the molarity of the barium hydroxide solution?  M

This barium hydroxide solution is then used to titrate an unknown solution of perchloric acid.

B. If 12.6 mL of the barium hydroxide solution is required to neutralize 27.7 mL of perchloric acid, what is the molarity of the perchloric acid solution?  M

Answer 1

Part A

Number of moles = amount in g / molar mass

= 0.912 g potassium hydrogen phthalate/ 204.22 g/mol

=0.00447 moles potassium hydrogen phthalate

The balance the equationof KPH and Ba(OH)2 is as follows:

Ba(OH)2 + 2KHP ==> 2H2O + BaP + K2P

Moles of Ba(OH)2 = 0.00447 moles potassium hydrogen phthalate *1/2

= 0.002235 Mole Ba(OH)2

Molarity = number of moles / volume in L

= 0.002235 Mole Ba(OH)2/15.2 mL of barium hydroxide *1 L/1000

= 0.147 M

PART b

Ba(OH)₂ + 2HClO₄ = Ba(ClO₄)₂ + 2H₂O

Number of moles = molarity * volume in L

=0.147 M *12.6 ml*1 L/1000 ml

= 0.0018522 moles Ba(OH)₂

Moles of perchloric acid = 0.0018522 moles Ba(OH)₂ *2/1

= 0.0037044 mole HClO4

Molarity = number of moles / volume in L

= 0.0037044 mole HClO4/27.7 ml*1 L/1000 ML

= 0.134 M