Question 17.14: Calculating the pH at the Equivalence Point in the Titration...
Calculating the pH at the Equivalence Point in the Titration of a Weak Acid by a Strong Base
Calculate the \mathrm{pH} of the solution at the equivalence point when 25 \mathrm{~mL} of 0.10 \mathrm{M} nicotinic acid is titrated by 0.10 \mathrm{M} sodium hydroxide. K_{a} for nicotinic acid equals 1.4 \times 10^{-5}.
PROBLEM STRATEGY
At the equivalence point, equal molar amounts of nicotinic acid and sodium hydroxide react to give a solution of sodium nicotinate. You first calculate the con- centration of nicotinate ion (stoichiometry problem). Then you find the \mathrm{pH} of this solution (hydrolysis problem).
CONCENTRATION OF NICOTINATE ION Assume that the reaction of the base with the acid is complete. In this case, 25 \mathrm{~mL} of 0.10 \mathrm{M} sodium hydroxide is needed to react with 25 \mathrm{~mL} of 0.10 \mathrm{M} nicotinic acid. The molar amount of nicotinate ion formed equals the initial molar amount of nicotinic acid.
25 \times 10^{-3} \cancel{\mathrm{~L} \text { soln }} \times \frac{0.10 \mathrm{~mol} \text { nicotinate ion }}{1 \cancel{\text { L soln }}}=2.5 \times 10^{-3} \text { mol nicotinate ion }
The total volume of solution is 50 \mathrm{~mL} ( 25 \mathrm{~mL} \mathrm{NaOH} solution plus 25 \mathrm{~mL} of nicotinic acid solution, assuming there is no change in volume on mixing). Dividing the molar amount of nicotinate ion by the volume of solution in liters gives the molar concentration of nicotinate ion.
\text { Molar concentration }=\frac{2.5 \times 10^{-3} \mathrm{~mol}}{50 \times 10^{-3} \mathrm{~L}}=0.050 \mathrm{M}
HYDROLYSIS OF NICOTINATE ION This portion of the calculation follows the method given in Example 17.8. You find that K_{b} for nicotinate ion is 7.1 \times 10^{-10} and that the concentration of hydroxide ion is 6.0 \times 10^{-6} \mathrm{M}. The \mathrm{pH} is 8.78 .