How many mL of 1.211 Molarity sulfuric acid is required to completely neutralize 10.0 mL of 3.210 Molarity calcium hydroxide?

0.01L x 3.20 mol Ca(OH)2/1L x 1000mL/1.211mol H2SO4 =

To find out how many mL of sulfuric acid is required to neutralize calcium hydroxide, we need to use the concept of stoichiometry and the balanced equation of the reaction.

The balanced equation for the reaction between sulfuric acid (H2SO4) and calcium hydroxide (Ca(OH)2) is:

H2SO4 + 2 Ca(OH)2 -> CaSO4 + 2 H2O

From the balanced equation, we can see that one mole of sulfuric acid reacts with two moles of calcium hydroxide. This means the stoichiometric ratio between sulfuric acid and calcium hydroxide is 1:2.

Now, let's calculate the moles of calcium hydroxide:

Moles of calcium hydroxide = Molarity of calcium hydroxide * Volume of calcium hydroxide
= 3.210 mol/L * 0.010 L
= 0.0321 mol

Since the stoichiometric ratio is 1:2, we need double the moles of sulfuric acid to completely neutralize the calcium hydroxide.

Moles of sulfuric acid = 2 * Moles of calcium hydroxide
= 2 * 0.0321 mol
= 0.0642 mol

Next, let's calculate the volume of sulfuric acid needed.

Volume of sulfuric acid = Moles of sulfuric acid / Molarity of sulfuric acid
= 0.0642 mol / 1.211 mol/L
≈ 0.053 mL (rounded to three decimal places)

Therefore, approximately 0.053 mL of 1.211 Molarity sulfuric acid is required to completely neutralize 10.0 mL of 3.210 Molarity calcium hydroxide.