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 $$pH > 7$$.