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Acid-Base Titrations

A 100ml Sample of a Ca(OH)2 solution is titrated until neutralized with 23mL 1.00M HCL.

What was the molarity of the Ca(OH)2 solution?

$CaOH_2(aq)$ + $2HCl(aq)$ → $CaCl_2$ + $H_2O$

$M=\dfrac{mols}{volume}$ $mols = Molarity \cdot Volume$

$0.023\,L \left( 1.0 \right) = 0.02\,{\it mols}$

$0.01150000000\,{\it mol}\,{\it Ca} \left( {\it OH} \right) _{{2}}= 0.115\,{\it mol}\,{\it CaOH}_{{2}}$

Molarity = ${\frac {{\it mols}}{{\it Volume}\left( L \right) }}= 0.1150000000\,{ \frac {{\it mol}\,{\it CaOH}_{{2}}}{0.100 L}}$ = 0.115 M

Answer: 0.115 M ${\it CaOH}_{{2}}$ Solution