Acid-Base Titrations

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?


Step 1:  Write a Balanced Equation

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

Step 2: Determine mols HCl added

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

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

Step 3: Stochiometrically calculate mols of Ca(OH)2 in sample

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

Step 4: Calculate Molarity

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