Titration of 12.50 mL of HCl solution requires 22.50 mL of 0.1250 M NaOH solution. What is the molarity of the HCl solution?

HCl + NaOH ==> NaCl + H2O

mols NaOH = M x L = ?
mols HCl = mols NaOH (look at the coefficients; 1:1)
Then M HCl = mols HCl/L HCl.

To find the molarity of the HCl solution, we can use the concept of stoichiometry in a titration reaction.

In a titration, a measured volume of one solution with a known concentration, called the titrant, is added to a known volume of another solution with an unknown concentration, called the analyte, until the reaction between the two is complete.

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 can see that the stoichiometric ratio between HCl and NaOH is 1:1. This means that for every 1 mole of NaOH, there is 1 mole of HCl.

First, let's calculate the number of moles of NaOH used in the titration:

moles of NaOH = volume of NaOH solution (in L) × molarity of NaOH solution

Given that the volume of NaOH solution used is 22.50 mL (which is 0.02250 L) and the molarity of NaOH solution is 0.1250 M, we can substitute these values into the formula:

moles of NaOH = 0.02250 L × 0.1250 M = 0.0028125 moles

Since the stoichiometric ratio between NaOH and HCl is 1:1, the number of moles of HCl is also 0.0028125 moles.

Now, let's calculate the molarity of the HCl solution:

molarity of HCl solution = moles of HCl / volume of HCl solution (in L)

Given that the volume of HCl solution used is 12.50 mL (which is 0.01250 L), we can substitute these values into the formula:

molarity of HCl solution = 0.0028125 moles / 0.01250 L = 0.225 M

Therefore, the molarity of the HCl solution is 0.225 M.