Find the equivalence point (pH and mL of added acid) for the titration of 50ml of .1M Ca(OH)2 with .5M HCl. Assume that calcium hydroxide is completely soluble at this concentration.

Answer 1

First, always write a correctly balanced equation for the reaction:

2HCl + Ca(OH)₂ ==> CaCl₂ + 2H₂O

Next, find the moles of Ca(OH)₂ present:

50 ml x 1 L/1000 ml x 0.1 mol/L = 0.005 moles Ca(OH)₂

To neutralize 1/2 (0.0025 moles) how many moles of HCl do you need?:

0.0025 moles Ca(OH)₂ x 2 mol HCl/1 mol Ca(OH)₂ = 0.005 moles HCl needed

Find the volume of HCl needed:

(x L)(0.5 mol/L) = 0.005 moles and x = 0.01 L = 10 ml of HCl needed

Now, for the pH...

At 1/2 neutralization, you still have 0.0025 moles Ca(OH)₂ left in a total volume of (50 +10) 60 ml (0.06 L)

Molarity of Ca(OH)₂ = 0.0025 mol/0.06 L = 0.0416 M

Molarity of OH^- = 2 x 0.0416 M = 0.083 M because Ca(OH)₂ ==> Ca²⁺ + 2OH^-

pOH = -log OH- = -log 0.083 = 1.08

pH = 14 - pOH

pH = 14 - 1.08

pH = 12.9

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