Question

53.3 mL of 1.69 M perchloric acid is added to 38.5 mL of potassium hydroxide, and the resulting solution is found to be acidic. 19.8 mL of 1.62 M calcium hydroxide is required to reach neutrality. What is the molarity of the original potassium hydroxide solution?

Answer 1

HClO4 + KOH ------------> KClO4 + H2O

moles of HClO4 = 53.3 x 1.69 /1000 = 0.09

HClO4 + KOH ------------> KClO4 + H2O

0.09    x 0

0.09 -x    0                            x

since acidic; KOH is completely reacted..and some amount of HClO4 left

given.19.8 ml of 1.62 M Ca[OH]2 is required to neutralise the left amount of HClO4

therefore

moles of Ca(OH)2 = 19.8 x 1.62 /1000 = 0.0321

moles of HClO4 = 2 x 0.0321 = 0.0642 mol

so

amount of HClO4 reacted = 0.09 - 0.0642

                                         = 0.0258 moles

0.0258 = 38.5 x M

M = 0.671

Molarity of original KOH = 0.671 M