A 100.0 ml sample of .50 M HCl (aq) is titrated with a .10 M NaOH. What volume of the NaOH solution is required to reach the endpoint of the titration?

I don't know about the end point because I don't know what indicator you are using; however, I can tell you about the equivalence point (you must be careful with those two words.)

mols HCl = M x L = ?
mols NaOH = mols HCl
M HCl = moles HCl/L HCl. You know mols and M, solve for L

To find the volume of NaOH solution required to reach the endpoint of the titration, we need to use the balanced chemical equation and the concept of stoichiometry.

The balanced chemical equation for the reaction between HCl and NaOH is:

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

From the balanced equation, we see that the stoichiometric ratio between HCl and NaOH is 1:1, meaning that for every 1 mole of HCl, we need 1 mole of NaOH.

Given:
- Volume of HCl solution = 100.0 ml
- Concentration of HCl solution = 0.50 M
- Concentration of NaOH solution = 0.10 M

To calculate the volume of NaOH solution required, we can use the following equation:

(Concentration of HCl) × (Volume of HCl) = (Concentration of NaOH) × (Volume of NaOH)

However, we need to convert the volume of HCl from milliliters (ml) to liters (L) to match the units of concentration. Therefore:

Volume of HCl = 100.0 ml = 100.0 ml × (1 L/1000 ml) = 0.100 L

Now we can plug in the values into the equation:

(0.50 M) × (0.100 L) = (0.10 M) × (Volume of NaOH)

Solving for the volume of NaOH:

Volume of NaOH = (0.50 M) × (0.100 L) / (0.10 M) = 0.50 L

Therefore, 0.50 liters (or 500.0 ml) of NaOH solution is required to reach the endpoint of the titration.