Question

In the titration of 50.0 mL of HCl of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26.5 mL of 0.130 M Ca(OH)2 is added. Show the calculation of the molarity of the HCI.

Answer 1

The molarity of HCl is 0.138 M

Step-by-step explanation:

The titration reaction is as follows:

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:

At the endpoint:

moles Ca(OH)₂ = moles HCl / 2

Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:

mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂

mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂

The number of moles of HCl will be:

mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl

This number of moles was present in 50.0 ml, then, in 1000 ml:

mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol

Then:

Molarity HCl = 0.138 M