Question

Suppose a student titrates a 10.00-mL aliquot of saturated Ca(OH)2 solution to the equivalence point with 13.70 mL of 0.0234 M HCl. What was the initial [OH − ]?

Answer 1

Outline to solve the problem:

The balanced equation: Ca(OH)₂ + 2HCl CaCl₂ + 2H₂O

i.e. At the equivalence point, no. of moles of HCl = 2*no. of moles of Ca(OH)₂

The no. of moles of HCl = no. of moles of H⁺ = 0.0234 mmol/mL * 13.7 mL = 0.32058 mmol (since 1 M = 1 mmol/mL)

At the equivalence point, The no. of moles of Ca(OH)₂ = 0.32058/2 = 0.16029 mmol

Therefore, the concentration of Ca(OH)₂ = 0.16029 mmol / 10 mL = 0.016029 mmol/mL

i.e. The intitial concentration of OH^- = 2*0.016029 M = 0.032058 M

Note: 1 mole of Ca(OH)₂ releases 2 moles of OH^-.