Question 14.7: When 50.00 mL of 1.000 M HCl is titrated with 0.7450 M NaOH ...

When 50.00 mL of 1.000 M HCl is titrated with 0.7450 M NaOH (Figure 14.9), the pH increases.

a How many milliliters of NaOH are required to reach the equivalence point and a pH of 7.00?

ANALYSIS

Information given:                                       HCl: V (50.00 mL); M (1.000)                                                                                                                  NaOH: M (0.7450)
Information implied:                                     acid-base reaction
Asked for:                                      volume of NaOH required to reach the equivalence point

STRATEGY

1. Recall the stoichiometry of acid-base reactions discussed in Chapter 4 and write the reaction.
2. Find mol HCl.
3. Follow the plan below to find the mol of NaOH and the volume of NaOH needed.
\text{mol}\> HCl \xrightarrow[\text{ratio}]{\text{atomic}} \text{mol}\> H^+ \xrightarrow[\bigwedge_{ratio}^{}]{\text{stoichiometric}} \text{mol}\> OH^- \xrightarrow[\text{ratio}]{\text{atomic}} \text{mol}\> NaOH \xrightarrow{M} V\> NaOH

b Find the pH when the volume of NaOH added is 0.02 mL less than the volume required to reach the equivalence point.

ANALYSIS

Information given:                              HCl: V (0.05000 L); M (1.000)
from part (a): mol H^+ (0.05000); V NaOH (67.11 mL)
volume NaOH in the titration (67.11 – 0.02 = 67.09 mL)
Asked for:                            pH of the solution after NaOH is added

STRATEGY

1. Find mol OH^-.
2. Fill in the following stoichiometric table

\begin{array}{ll}\hline & H^+&&OH^- \\ \hline \text{Mol before reaction}\\ \text{Change}\\ \text{Mol}\> \text{after}\> \text{reaction}\\ \text{Volume}\\ \hline \end{array}

This table almost looks like the equilibrium table in Chapter 12.
3. Find [excess reactant]
[excess reactant] = \frac{\text{mol excess reactant}}{(\text{volume}\> H^+)  + (\text{volume}\> OH^-)}
4. Find pH

c Find the pH when the volume of NaOH added is 0.02 mL more than the volume required to reach the equivalence point.

ANALYSIS

Information given:                          HCl: V (0.05000 L); M (1.000)                                                                              from part (a): mol H^+ (0.05000); V NaOH (67.11 mL)                                                                                           volume NaOH in the titration (67.11 + 0.02 = 67.13 mL)
Asked for:                   pH of the solution after NaOH is added

STRATEGY

1. Find mol OH^-.
2. Fill in a stoichiometric table as in part (b).
3. Find [excess reactant] as in part (b).
4. Find pH.