Question

16.4 mL of 0.135 M NaOH is used to neutralize 46.9 mL of HCl. The concentration of the HCl solution is ____ M (3 decimal places).

Answer 1

The concentration of the HCl solution is 0.047 M.

Explanation:

Data given about acid and base:

volume of acid Vacid = 46.9 ml

molarity of acid =?

volume of the base (NaOH) = 16.4 ml

molarity of the base = 0.135 M

To know the concentration of the acid in this reaction, the formula used in titration  is used. It is

Macid X Vacid = Mbase X Vbase

the formula is rewritten as:

Macid = $\frac{Mbase X Vbase}{Vacid}$

putting the values in the equation:

Macid = $\frac{0.135 X 16.4}{46.9}$

          = 0.047 M

the concentration of the acid i.e HCl in the solution is of 0.047 M.