In a titration experiment involving an unknown concentration of HCl and a 0.100M solution of NaOH(aq), it was determined that it took 12.5mL of NaOH to neutralize 20.0mL of the HCl(aq). Write a balanced chemical equation for this reaction and determine the concentration of the HCl(aq).

HCl + NaOH ==> NaCl + H2O

mols NaOH = M x L = ?
mols HCl = mols NaOH and you know that from the 1:1 ratio in the balanced equation.
Then M HCl = mols HCl/L HCl.

To write a balanced chemical equation for the reaction between HCl and NaOH, we need to remember that acids, like HCl, react with bases, like NaOH, to form a salt and water. The balanced chemical equation can be written as follows:

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

Now, let's determine the concentration of the HCl(aq):

First, we need to calculate the number of moles of NaOH used in the reaction. We can use the equation:

moles of NaOH = concentration of NaOH × volume of NaOH (in liters)

Given that the concentration of NaOH is 0.100M and the volume of NaOH used is 12.5mL (which is 0.0125L), we can calculate the moles of NaOH:

moles of NaOH = 0.100M × 0.0125L = 0.00125 moles

Since the balanced chemical equation shows a 1:1 ratio between NaOH and HCl, we know that the moles of HCl used in the reaction is also 0.00125 moles.

Next, we need to determine the concentration of the HCl(aq). We can rearrange the equation for moles of a substance:

moles of a substance = concentration of the substance × volume of the substance (in liters)

In this case, we want to solve for the concentration of HCl:

concentration of HCl = moles of HCl / volume of HCl (in liters)

Given that the volume of HCl used is 20.0mL (which is 0.0200L) and we found that the moles of HCl used is 0.00125 moles, we can calculate the concentration:

concentration of HCl = 0.00125 moles / 0.0200L = 0.0625M

Therefore, the concentration of the HCl(aq) solution is 0.0625M.