By titration it is found that 12.1 mL of 0.130 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.
NaOH + HCl ==> NaCl + H2O
mols NaOH = M x L = ?
mols HCL = mols NaOH since 1 mol NaOH reacts with 1 mols NaOH.
Then M HCl = mols HCl/L HCl = ?
To calculate the concentration of the HCl solution, we can use the concept of stoichiometry and the equation:
NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)
First, let's determine the number of moles of NaOH used in the titration:
Number of moles of NaOH = Molarity of NaOH * Volume of NaOH used (in liters)
= 0.130 M * 0.0121 L
= 0.001573 moles
From the balanced equation, we can see that the mole ratio between NaOH and HCl is 1:1. Therefore, the number of moles of HCl in the solution is also 0.001573 moles.
Now, we can calculate the concentration of the HCl solution:
Concentration of HCl = Number of moles of HCl / Volume of HCl used (in liters)
= 0.001573 moles / 0.0250 L
= 0.0629 M
Therefore, the concentration of the HCl solution is 0.0629 M.