Consider a titration of 50.00 mL of 0.1005 M HCl with 0.1012 M NaOH.

To determine the result of the titration, we can use the concept of stoichiometry and the balanced chemical equation. In this case, the balanced chemical equation for the reaction between HCl and NaOH is:

HCl + NaOH → NaCl + H2O

The stoichiometry of the reaction tells us that 1 mole of HCl reacts with 1 mole of NaOH, producing 1 mole of NaCl and 1 mole of water.

Given that the volume of the HCl solution is 50.00 mL, and the concentration of HCl is 0.1005 M, we can calculate the number of moles of HCl using the formula:

Moles of HCl = Volume of HCl (in liters) x Concentration of HCl

Converting 50.00 mL to liters:

Volume of HCl = 50.00 mL / 1000 mL/L = 0.0500 L

Moles of HCl = 0.0500 L x 0.1005 mol/L = 0.005025 mol

Since the stoichiometry of the reaction is 1:1, the number of moles of NaOH that reacts with HCl will also be 0.005025 mol.

Now, let's determine the volume of NaOH solution required to neutralize the HCl. We can use the formula:

Moles of NaOH = Moles of HCl

Concentration of NaOH = 0.1012 M

Volume of NaOH = Moles of NaOH / Concentration of NaOH

Volume of NaOH = 0.005025 mol / 0.1012 mol/L = 0.0497 L

Converting 0.0497 L to milliliters:

Volume of NaOH = 0.0497 L x 1000 mL/L = 49.7 mL

Therefore, approximately 49.7 mL of the 0.1012 M NaOH solution is required to neutralize 50.00 mL of 0.1005 M HCl.