What is the {\rm HCl} concentration if 54.2 mL of 0.100 M {\rm NaOH} is required to titrate a 30.0 mL sample of the acid?
To find the concentration of the hydrochloric acid (HCl), we can use the concept of stoichiometry in a titration.
In a titration, a known volume and concentration of one solution (the titrant) is added to another solution until the reaction between the two is complete. The point at which the reaction is complete is called the equivalence point.
In this case, 54.2 mL of 0.100 M sodium hydroxide (NaOH) is required to fully react with a 30.0 mL sample of hydrochloric acid (HCl).
First, we can write the balanced equation to determine the stoichiometry of the reaction:
HCl + NaOH → NaCl + H₂O
From the equation, we can see that the molar ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, 1 mole of NaOH is required for complete neutralization.
Now, let's use the concept of stoichiometry to calculate the moles of NaOH used in the reaction.
moles of NaOH = concentration of NaOH x volume of NaOH (in L)
= 0.100 M x 0.0542 L
= 0.00542 moles
Since the molar ratio between HCl and NaOH is 1:1, the moles of HCl used in the reaction will also be 0.00542 moles.
Next, we need to calculate the concentration of HCl in the original solution using the volume of the acid and the moles of acid.
concentration of HCl = moles of HCl / volume of HCl (in L)
= 0.00542 moles / 0.0300 L
= 0.1807 M
Therefore, the concentration of HCl is 0.1807 M.