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

To find the molarity of the original Ca(OH)₂, we will need to find the moles of Ca(OH)₂ present. Then since we know the volume (22.6 ml), we'll be able to find moles/liter which is the definition of molarity.

So how do we find the moles of Ca(OH)₂ that are present? Look at the reaction with hydroiodic acid (HI).

2HI + Ca(OH)₂ ==> CaI₂ + 2H₂O and notice it takes 2 moles of HI for each 1 mole of Ca(OH)₂.

We know that we used 44.2 ml of 1.27 M Hl and from this we can find the moles of HI that we used.

-- 44.2 ml x 1 L/1000 ml x 1.27 mol/L = 0.05613 moles HI used. This was enough to react with ALL of the Ca(OH)₂ and still have some HI left over (we know this because it tells us the resulting soln is acidic)

How much HI is left over?

HI + NaOH ==> NaI + H2O

moles NaOH used = 28.2 ml x 1 L/1000 ml x 1.33 mol/L = 0.03751 moles NaOH = 0.03751 moles HI left over

So, moles of HI needed to neutralize the Ca(OH)₂ = 0.05613 mol - 0.03751 mol = 0.01862 mol HI used to neutralize the Ca(OH)₂, BUT remember that it takes 2 moles HI to neutralize 1 mole Ca(OH)₂.

So, moles Ca(OH)₂ originally present = 0.01862 /2 = 0.00931 moles Ca(OH)₂

Molarity of Ca(OH)₂ = 0.00931 moles / 0.0226 L = 0.412 M