KEY2CHEM

Partial Neutralization Reactions 

 

In the titration of a strong acid with a strong base, the moles of acid initially present = moles of base added at the equivalence point. Before a stoichiometric quantity of strong base has been added, there are varying amounts of chemical species in the solution.

Prior to the equivalence point (when \([\require{mhchem}\ce{H+}] > [\require{mhchem}\ce{OH-}]\)), the pH is determined by the excess acid present. A strong acid is one that dissociates completely: \(\require{mhchem}\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A^{-}(aq)}\) so the remaining \([\require{mhchem}\ce{HA}] = [\require{mhchem}\ce{H3O+}]\). Remember that \(pH = -log[\require{mhchem}\ce{H3O+}]\).

Past the equivalence point, the pH is determined by the excess \([\require{mhchem}\ce{OH-}]\) added.

 


Example 1.

When \(50.0 \text{ mL}\) of\( 0.1 \text{ M } \require{mhchem}\ce{HCl}\) is titrated with\( 0.2 \text{ M } \require{mhchem}\ce{NaOH}\), what is the pH after \(20.0 \text{ mL}\) of \(\require{mhchem}\ce{NaOH}\) has been added?

 

A. 7.00

B. 1.85

C. 5.15

 

 

Solution

 

 

B. 1.85

 

 

Calculate the moles of acid \((\require{mhchem}\ce{HCl}) \) initially present.

\(50.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times \frac{0.1 \text{ mol}}{1 \text{ L}} = 0.005 \text{ mol HCl}\)

 

Now calculate the moles of base  \((\require{mhchem}\ce{NaOH})\) added.

\(20.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times \frac{0.2 \text{ mol}}{1 \text{ L}} = 0.004 \text{ mol NaOH}\)

 

The acid is present in excess, with the amount remaining calculated as \(0.005 \text{ mol} – 0.004 \text{ mol} = 0.001 \text{ mol acid remaining}\)

 

The pH of the solution is determined by the excess acid present. The concentration of remaining acid (which is equal to the \(\require{mhchem}\ce{H3O+}\) present since the acid is a strong acid and dissociates completely) is calculated using the remaining amount and the total current volume of the solution.

\([\require{mhchem}\ce{HCl}] = \frac{0.001 \text{ mol }}{0.0700 \text{ L}} = 0.0143 \text{ M}\)

 

Since \([\require{mhchem}\ce{HCl}] = [\require{mhchem}\ce{H3O+}] = 0.0143 \text{ M}\), the pH is calculated as:

\(pH = -log[\require{mhchem}\ce{H3O+}] = -log(0.0143 \text{ M}) = 1.85\)

 


Example 2.

At the equivalence point in the titration of a strong acid (\(\require{mhchem}\ce{HA}\)) with a strong base, which species is present in the largest concentration?

 

 

A. \(\require{mhchem}\ce{HA}\)

B. \(\require{mhchem}\ce{H3O+}\)

C. \(\require{mhchem}\ce{A-}\)

 

 

Solution

 

 

C. \(\require{mhchem}\ce{A-}\)

Strong acid dissociate completely (undergo complete proton transfer to water), so the concentration of undissociated \(\require{mhchem}\ce{HA}\) is very small. At the equivalence point, the \(\require{mhchem}\ce{H3O+}\) has reacted with strong base to form a neutral solution. As such, the species in the largest concentration is \(\require{mhchem}\ce{A-}\).


Example 3.

When \(10.0 \text{ mL}\) of \(0.25 \text{ M } \require{mhchem}\ce{HBr}\) and \(25.0 \text{ mL}\) of \(0.15 \text{ M } \require{mhchem}\ce{KOH}\), what is the pH of the resulting solution?

 

 

A. \(pH = 7\)

B. \(pH > 7\)

C. \( pH < 7\)

 

 

 

 

 

 

Solution

 

 

B. \(pH > 7\)

 

Since \([\require{mhchem}\ce{OH-}] > [\require{mhchem}\ce{H3O+}]\), the solution is basic with \(pH > 7\).