Question

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 ?

Answer 1

The molarity of the hydrochloric acid solution is `"0.095 M"`.

You're performing a titration of the `"HCl"` solution using sodium hydroxide. The balanced chemical equation for the neutralization reaction that takes place is

`HCl_((aq)) + NaOH_((aq)) -> NaCl_((aq)) + H_2O_((l))`

Notice the `"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 = n/V => n = C * V`

`n_("NaOH") = 38.71 * 10^(-3)"L" * "0.0615 M" = "0.002381 moles"` `"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 = n/V = "0.002381 moles"/(25.00 * 10^(-3)"L") = "0.095 M"`