What is the molarity of an HCl solution if 14.0mL HCl solution is titrated with 22.6mL of 0.155 M NaOH solution?
HCl + NaOH ==> NaCl + H2O
mols NaOH = M x L = ?
mols HCl = mols NaOH (from the coefficients in the balanced equation.)
MHCl = mols HCl/L HCl
To find the molarity of the HCl solution, we can use the concept of stoichiometry and the balanced chemical equation of the reaction between HCl and NaOH.
The balanced chemical equation 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 one mole of HCl reacts with one mole of NaOH.
To begin, we need to determine the number of moles of NaOH used in the titration. The concentration of the NaOH solution is given as 0.155 M, and the volume used is 22.6 mL. To convert the volume to liters, we divide by 1000:
22.6 mL / 1000 = 0.0226 L
Now we can calculate the number of moles of NaOH:
moles of NaOH = concentration x volume
moles of NaOH = 0.155 M x 0.0226 L
Next, since the stoichiometric ratio between HCl and NaOH is 1:1, the moles of HCl used in the titration would be equal to the moles of NaOH used:
moles of HCl = moles of NaOH
Now we need to find the molarity of the HCl solution. The number of moles of HCl is equal to the number of moles of NaOH from the titration. The volume of the HCl solution used in the titration is 14.0 mL, which is equivalent to 0.0140 L:
molarity of HCl = moles of HCl / volume of HCl in liters
molarity of HCl = moles of HCl / 0.0140 L
Now we can substitute the moles of HCl with the moles of NaOH:
molarity of HCl = moles of NaOH / 0.0140 L
By substituting the values we calculated earlier:
molarity of HCl = (0.155 M x 0.0226 L) / 0.0140 L
Performing the calculation:
molarity of HCl = 0.2519 M
Therefore, the molarity of the HCl solution is 0.2519 M.