What volume of NaOH was required to have neutralized the HCl.
HCl + NaOH --> H2O + NaCl
There no other information given in the problem except in later questions like the NaOH had 0.1 M and that there was 50.00 mL of HCl.
To determine the volume of NaOH required to neutralize HCl, you will need to perform a stoichiometric calculation using the balanced equation given:
HCl + NaOH --> H2O + NaCl
Since the stoichiometry of the reaction is 1:1 between HCl and NaOH, it means that one mole of HCl reacts with one mole of NaOH.
To solve this problem, you will need to know the molarity (M) and volume (V) of one of the substances involved. In this case, you mentioned that the NaOH concentration is 0.1 M and that there is 50.00 mL of HCl.
First, let's convert the given volume of HCl from milliliters to liters:
V(HCl) = 50.00 mL = 50.00 mL * (1 L / 1000 mL) = 0.0500 L
Next, we can use the equation: Molarity (M) = moles / volume (L)
We can rearrange this equation to solve for moles:
moles(HCl) = Molarity(HCl) * volume(HCl)
Now we can calculate the moles of HCl:
moles(HCl) = 0.0500 L * 0.10 M = 0.005 moles HCl
Since the stoichiometric ratio between HCl and NaOH is 1:1, this means that the number of moles of NaOH required to neutralize the HCl is also 0.005 moles.
Finally, using the definition of molarity, we can determine the volume of NaOH required:
V(NaOH) = moles(NaOH) / Molarity(NaOH) = 0.005 moles / 0.10 M = 0.050 L
Therefore, the volume of NaOH required to neutralize the HCl is 0.050 liters, or equivalently, 50.00 mL.