Question

An acid-base titration was performed. Calcium hydroxide was used to titrate HCl. It took only 9.16 mL of the base, Ca(OH)₂, to titrate 25.00 mL of 0.985 M HCl, the acid. What was the Molarity of the Ca(OH)₂?

Answer 1

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

$V(Ca(OH)_2) = 9.16 mL \\ C(HCl) = 0.985 \;M \\ V(HCl) = 25.0\; mL$

Proportion 1 (HCl):

0.985 – 1000 mL

n(HCl) – 25.0 mL

$n(HCl) = \frac{0.985 \times 25}{1000} = 0.024625 \;mol$

According to the reaction

n(Ca(OH)₂)$= \frac{1}{2}n(HCl) = 0.0123 \; mol$

Proportion 2 (Ca(OH)₂):

0.0123 mol – 9.16 mL

C – 1000 mL

C(Ca(OH)₂) $= \frac{0.0123 \times 1000}{9.16} = 1.34 \;M$

Answer: 1.34 M