The titration of HCl with NaOH is represented by the equation

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

What volume of 0.100 M HCl is required to titrate 50.0 mL of 0.500 M NaOH?

Any volume can titrate an acid with a base; you probably mean titrate to the equivalence point. For a 1:1 acid/base system such as NaOH/HCl you may use

mLHCl x MHCl = mLNaOH x MNaOH

To solve this problem, we will use the concept of stoichiometry and the balanced chemical equation to determine the volume of HCl required.

First, let's write down the balanced chemical equation:
HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)

From the balanced equation, we can see that the stoichiometric ratio between HCl and NaOH is 1:1. This means that for every 1 mole of NaOH, we need 1 mole of HCl to react completely.

Step 1: Calculate the moles of NaOH in 50.0 mL of 0.500 M NaOH.
Moles of NaOH = Volume (in liters) x Concentration
Moles of NaOH = 0.050 L x 0.500 mol/L
Moles of NaOH = 0.025 mol

Step 2: Determine the volume of 0.100 M HCl required.
Since the stoichiometric ratio is 1:1, the number of moles of HCl required is also 0.025 mol.

Using the formula:
Moles = Volume (in liters) x Concentration

we can rearrange it to calculate the volume:
Volume (in liters) = Moles / Concentration

Volume of HCl = 0.025 mol / 0.100 mol/L
Volume of HCl = 0.25 L

Step 3: Convert the volume of HCl from liters to milliliters.
Volume of HCl = 0.25 L x 1000 mL/L
Volume of HCl = 250 mL

Therefore, 250 mL of 0.100 M HCl is required to titrate 50.0 mL of 0.500 M NaOH.

To determine the volume of 0.100 M HCl required to titrate 50.0 mL of 0.500 M NaOH, we can use the concept of stoichiometry.

First, let's write the balanced equation for the reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

From the balanced equation, we can see that the mole ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, we need 1 mole of NaOH to react completely.

Given that the concentration of NaOH is 0.500 M and the volume is 50.0 mL, we can calculate the number of moles of NaOH:
moles of NaOH = concentration × volume
moles of NaOH = 0.500 M × 50.0 mL
moles of NaOH = 0.500 mol/L × 0.050 L (converting mL to L)
moles of NaOH = 0.025 mol

Since the mole ratio is 1:1 between HCl and NaOH, we know that we need 0.025 moles of HCl to react completely with 0.025 moles of NaOH.

Now, let's calculate the volume of 0.100 M HCl required:
volume of HCl = moles of HCl / concentration of HCl
volume of HCl = 0.025 mol / 0.100 mol/L
volume of HCl = 0.250 L or 250 mL

Therefore, 250 mL of 0.100 M HCl is required to titrate 50.0 mL of 0.500 M NaOH.