Example 4.14: Titration Analysis

Titration Analysis

The end point in a titration of a 50.00 mL sample of aqueous \(\ce{HCl}\) was reached by addition of 35.23 mL of 0.250 M \(\ce{NaOH}\) titrant. The titration reaction is:

\(\ce{HCl(aq)}\)\(\ce{ + }\)\(\ce{NaOH(aq)}\)\(\ce{->}\)\(\ce{NaCl(aq)}\)\(\ce{ + }\)\(\ce{H2O(l)}\)\(\ce{ }\)

What is the concentration of the \(\ce{HCl}\)?

Solution

\(V_{\mathrm{\ce{HCl}}}\) \(= 50.00\ \mathrm{mL}\)


\(c_{\mathrm{\ce{HCl}}}\) = ?


\(c_{\mathrm{\ce{NaOH}}}\) \(= 0.250\ \mathrm{M}\)


\(V_{\mathrm{\ce{NaOH}}}\) \(= 35.23\ \mathrm{mL}\)


At the end point of the titration, the reactants are present at stochiometric ratio

\(\dfrac{n_{\mathrm{\ce{HCl}}}}{n_{\mathrm{\ce{NaOH}}}} = \dfrac{1}{1}\)     

As for all reaction stoichiometry calculations, the key issue is the relation between the chemical amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous modules of this chapter is followed, with additional considerations required, since the amounts of reactants provided and requested are expressed as solution concentrations.
For this exercise, the calculation will follow the following outlined steps:

\(n_{\mathrm{\ce{NaOH}}}\) \(= c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH}}}\)

\(\ \ \ =0.250\ \mathrm{M} \cdot 35.23\ \mathrm{mL}\)

\(\ \ \ =8.81\times 10^{-3}\ \mathrm{mol}\)


\(n_{\mathrm{\ce{HCl}}}\) \(= n_{\mathrm{\ce{NaOH}}} \cdot 1\)

\(\ \ \ =8.81\times 10^{-3}\ \mathrm{mol} \cdot 1\)

\(\ \ \ =8.81\times 10^{-3}\ \mathrm{mol}\)


\(c_{\mathrm{\ce{HCl}}}\) \(= \dfrac{n_{\mathrm{\ce{HCl}}}}{V_{\mathrm{\ce{HCl}}}}\)

\(\ \ \ =\dfrac{8.81\times 10^{-3}\ \mathrm{mol}}{50.00\ \mathrm{mL}}\)

\(\ \ \ =0.1761\ \mathrm{M}\)