How much 1.0 M NaOH would be required to completely neutralize 40.0 mL of 0.60 M HCl?

To determine the amount of 1.0 M NaOH required to neutralize 40.0 mL of 0.60 M HCl, we need to use the concept of stoichiometry.

First, let's write down the balanced chemical equation for the reaction between NaOH and HCl:

NaOH + HCl → NaCl + H2O

From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl. This means the stoichiometric ratio between NaOH and HCl is 1:1.

Next, we can calculate the number of moles of HCl in the given volume (40.0 mL) and concentration (0.60 M). To do this, we use the formula:

moles of solute = volume (in liters) × concentration (in M)

Converting the volume of HCl to liters:

40.0 mL = 40.0 mL × (1 L / 1000 mL) = 0.040 L

Calculating the moles of HCl:

moles of HCl = 0.040 L × 0.60 M = 0.024 moles

Since the stoichiometric ratio between NaOH and HCl is 1:1, we know that we need 0.024 moles of NaOH to neutralize 0.024 moles of HCl.

Now, let's calculate the volume of 1.0 M NaOH solution needed to provide 0.024 moles of NaOH. We'll again use the formula:

volume (in liters) = moles of solute / concentration (in M)

volume of NaOH = 0.024 moles / 1.0 M = 0.024 L = 24.0 mL

Therefore, 24.0 mL of 1.0 M NaOH would be required to completely neutralize 40.0 mL of 0.60 M HCl.