A sample of solid calcium hydroxide, Ca(OH)₂is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00×10^-2M HCl requires 36.6 mL of the acid to reach the end point.

\({\rm{Ca(OH}}{{\rm{)}}_{2({\rm{aq}})}}{\rm{ + 2HC}}{{\rm{l}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ CaC}}{{\rm{l}}_{2({\rm{aq}})}}{\rm{ + 2}}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)

What is the molarity?

According to the question, it is given that

The concentration of hydrochloric acid is \({\rm{5}}{\rm{.00 }} \times {\rm{ 1}}{{\rm{0}}^{ - 2}}{\rm{ M}}\)and the volume is 36.6 ml.

The volume of calcium hydroxide is 75 ml and the concentration is unknown.

Moreover, 1 mole of calcium hydroxide reacts with 2 moles of hydrochloric acid.

Thus, to calculate the concentration or molarity of calcium hydroxide, we can use:

\({{\rm{n}}_{\rm{H}}}{{\rm{M}}_{\rm{H}}}{{\rm{V}}_{\rm{H}}}\,{\rm{ = }}{{\rm{n}}_{\rm{c}}}{{\rm{M}}_{\rm{c}}}{{\rm{V}}_{\rm{c}}}\)

Here, \({{\rm{n}}_{\rm{H}}}{\rm{, }}{{\rm{M}}_{\rm{H}}}{\rm{, }}{{\rm{V}}_{\rm{H}}}\,\)is moles, molarity and volume of HCl whereas \({{\rm{n}}_{\rm{c}}}{\rm{, }}{{\rm{M}}_{\rm{c}}}{\rm{, }}{{\rm{V}}_{\rm{c}}}\) is moles, molarity and volume of calcium hydroxide.

\({\rm{1 }} \times {\rm{ 5 }} \times {\rm{ }}{10^{ - 2}}{\rm{ }} \times {\rm{ 36}}{\rm{.6 }}\,{\rm{ = 2 }} \times {\rm{ }}{{\rm{M}}_{\rm{c}}}{\rm{ }} \times {\rm{ }}75\)

\({{\rm{M}}_{\rm{c}}}{\rm{ = 0}}{\rm{.0122 M}}\)