# A student prepares a solution of hydrochloric acid that is approximately 0.1 M and wanted to determine its precise concentration. A 25.00 mL portion of the HCl solution is transferred to a flask, and after a few drops of indicator are added, the HCl ?

Mar 28, 2015

The molarity of the hydrochloric acid solution is $\text{0.095 M}$.

You're performing a titration of the $\text{HCl}$ solution using sodium hydroxide. The balanced chemical equation for the neutralization reaction that takes place is

$H C {l}_{\left(a q\right)} + N a O {H}_{\left(a q\right)} \to N a C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

Notice the $\text{1:1}$ mole ratio that exists between sodium hydroxide and hydrochloric acid - at the equivalence point, the number of moles of the former will be equal to the number of moles of the latter.

The number of moles of added sodium hydroxide is

$C = \frac{n}{V} \implies n = C \cdot V$

n_("NaOH") = 38.71 * 10^(-3)"L" * "0.0615 M" = "0.002381 moles" $\text{NaOH}$

The number of moles of hydrochloric acid will be equal to

n_("HCl") = n_("NaOH") = "0.002381"

This means that the molarity of the hydrochloric acid solution will be

$C = \frac{n}{V} = \text{0.002381 moles"/(25.00 * 10^(-3)"L") = "0.095 M}$