Question

titration:

2. If 45.0 mL of 1.50 M Ca(OH)2 are needed to neutralize 25.0 mL of HI of unknown concentration, what is the molarity of the HI?

Answer 1

2.

Given molarity of Ca(OH)₂ = 1.50 M

Given volume of Ca(OH)₂ = 45.0 mL = 0.045 L       ($\because$ 1 mL = 0.001 L)

Hence, moles of Ca(OH)₂ used = molarity of Ca(OH)₂ x liter of Ca(OH)₂

                                                   = 1.50 M x 0.045 L

                                                   = 0.0675 mol

Given, the volume HI titrated = 25.0 mL = 0.025 L

Let us say that the molarity of HI = z M

Hence, moles of HI titrated = molarity of HI x liter of HI

                                              = z M x 0.025 L

                                              = 0.025z mol

Ca(OH)₂ + 2HI $\rightarrow$ CaI₂ + 2H₂O

Now, from the balanced equation above, it is clear that 1 mol of Ca(OH)₂ is required to neutralize 2 mol of HI .

Thus, equating the moles of Ca(OH)₂ and HI to get

0.025z = 2 x 0.0675

or, z = 5.40 M

Hence, the molarity of HI = 5.40 M