what volume of 1.00M ca(oh)2 is needed to neutralize 250.0mL of 0.100M HCL?
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To determine the volume of 1.00M Ca(OH)2 needed to neutralize 250.0mL of 0.100M HCl, we can use the concept of the balanced chemical equation and stoichiometry.
The balanced chemical equation representing the neutralization reaction between Ca(OH)2 and HCl is:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
From the equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, the mole ratio between Ca(OH)2 and HCl is 1:2.
First, we need to calculate the number of moles of HCl present in 250.0mL (0.2500L) of 0.100M HCl using the formula:
moles = volume (L) x concentration (M)
moles of HCl = 0.2500L x 0.100M = 0.0250 moles of HCl
Since the mole ratio is 1:2 (Ca(OH)2:HCl), we need twice the number of moles of Ca(OH)2 to neutralize the given amount of HCl.
moles of Ca(OH)2 = 2 x 0.0250 moles of HCl = 0.0500 moles of Ca(OH)2
Now, with the number of moles of Ca(OH)2, we can calculate the volume of 1.00M Ca(OH)2 required using the formula:
volume (L) = moles / concentration (M)
volume = 0.0500 moles / 1.00M = 0.050 L
However, the volume is usually expressed in milliliters (mL), so we need to convert it back:
volume = 0.050 L x 1000 mL/1 L = 50.0 mL
Therefore, to neutralize 250.0mL of 0.100M HCl, you will need 50.0mL of 1.00M Ca(OH)2.