What volume of 0.23M HCl is required to completely neutralize 25.0mL of a 0.45M Ca(OH)2 solution?
To determine the volume of HCl needed to neutralize the Ca(OH)2 solution, we need to use the following balanced chemical equation:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
From the balanced equation, we can see that it takes 2 moles of HCl to neutralize 1 mole of Ca(OH)2. Therefore, the mole ratio is 2:1.
First, let's determine the moles of Ca(OH)2 in the 25.0mL of 0.45M solution:
Moles of Ca(OH)2 = (0.45 mol/L) * (25.0 mL / 1000 mL) = 0.01125 mol
From the mole ratio, we can determine the moles of HCl needed to neutralize the Ca(OH)2:
Moles of HCl needed = 2 * 0.01125 mol = 0.0225 mol
Now, we can determine the volume of 0.23M HCl needed to provide 0.0225 moles:
Volume of HCl = moles / Molarity
Volume of HCl = 0.0225 mol / 0.23 mol/L = 0.0978 L = 97.8 mL
Therefore, 97.8mL of 0.23M HCl is required to completely neutralize 25.0mL of a 0.45M Ca(OH)2 solution.