Titration: Back Titration

Question

44.2 mL of 1.27 M hydroiodic acid is added to 22.6 mL of calcium hydroxide, and the resulting solution is found to be acidic.

28.2 mL of 1.33 M sodium hydroxide is required to reach neutrality.

What is the molarity of the original calcium hydroxide solution?

Answer 1

There are actually two separate reactions to consider. In the first, hydroiodic acid reacts with calcium hydroxide to completion, according to the balanced equation.

hydroiodic acid + calcium hydroxide ----> calcium iodide + water

But there is excess hydroiodic acid, so it is left in the solution, until sodium hydroxide (not calcium hydroxide) is added to the solution.

hydroiodic acid + sodium hydroxide ----> sodium iodide + water

Only enough sodium hydroxide is added to consume the remaing moles of hydroiodic acid. Therefore, the total number of moles of hydroiodic acid that reacted is equal to the sum of the moles used up in each of the two reactions:

1. Since you have both the concentration and total volume of the hydroiodic acid used, you can determine the total number of moles consumed in both reactions.
2. By using the given concentration and volume of sodium hydroxide added, and the balanced chemical equation for the second reaction, you can calculate the number of moles of hydroiodic acid consumed.
3. Subtracting that amount from the total number of moles of hydroiodic acid, you'll determine how many moles of hydroiodic acid was present for the first reaction with the calcium hydroxide.
4. Using the balanced chemical equation for the first reaction, determine the number of moles of calcium hydroxide that reacted.
5. Using the volume of the calcium hydroxide solution and moles of calcium hydroxide reacted, you can determine its concentration.