What is the molarity of an HCl solution if 38.5 mL of it is titrated to the equivalence point with 41.6 mL of a 1.41 M NaOH solution?
HCl + NaOH ==> NaCl + H2O
mols NaOH = M x L = ?
mols HCl = mols NaOH (See the coefficientgs in the balanced equation.)
M HCl = mols HCl/L HCl.
To determine the molarity of an HCl solution, you can use the equation for the titration:
M₁V₁ = M₂V₂
Where:
M₁ is the molarity of HCl solution
V₁ is the volume of HCl solution (in liters)
M₂ is the molarity of NaOH solution
V₂ is the volume of NaOH solution (in liters)
First, convert the volumes from milliliters (mL) to liters (L):
V₁ = 38.5 mL = 38.5/1000 L = 0.0385 L
V₂ = 41.6 mL = 41.6/1000 L = 0.0416 L
Substitute the values into the equation:
M₁ * 0.0385 = 1.41 * 0.0416
Now, rearrange the equation to solve for M₁:
M₁ = (1.41 * 0.0416) / 0.0385
Calculating the expression:
M₁ = 0.0586 mol/L
Therefore, the molarity of the HCl solution is 0.0586 M.