Question

An aqueous solution of hydroiodic acid is standardized by titration with a 0.133 M solution of calcium hydroxide.

If 13.6 mL of base are required to neutralize 26.5 mL of the acid, what is the molarity of the hydroiodic acid solution?

M hydroiodic acid

Answer 1

The reaction we are analyzing is

2 HI + Ca(OH)₂ ==== CaI₂ + 2 H₂O

The base Ca(OH)₂ has a molarity of 0.133 M and you have a volume of 13.6 ml so calculate the moles of base

moles = molarity * volume

moles = 0.133 * 0.0136 = 0.0018 moles of base available

since you require 2 moles acid for every mole of base then calculate the moles of acid

moles of acid = 2 * 0.0018 = 0.0036 moles of HI available

since you have 26.5 ml of acid then calculate the molarity using the equation

Molarity = moles / volume

Molarity = 0.0036 / 0.0265 L = 0.1358 M

the concentration of the acid is 0.1358 M or 0.136 M

*hope it helps =)