Calculating the pH at the Equivalence Point in the Titration of a Weak Acid by a Strong Base Calculate the pH of the solution at the equivalence point when 25 mL of 0.10 M nicotinic acid is titrated by 0.10 M sodium hydroxide. Ka for nicotinic acid equals 1.4 × 10^-5. PROBLEM STRATEGY At the

Question

Calculating the pH at the Equivalence Point in the Titration of a Weak Acid by a Strong Base

Calculate the [latex]\mathrm{pH}[/latex] of the solution at the equivalence point when [latex]25 \mathrm{~mL}[/latex] of [latex]0.10 \mathrm{M}[/latex] nicotinic acid is titrated by [latex]0.10 \mathrm{M}[/latex] sodium hydroxide. [latex]K_{a}[/latex] for nicotinic acid equals [latex]1.4 imes 10^{-5}[/latex].

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 [latex]\mathrm{pH}[/latex] of this solution (hydrolysis problem).

Accepted Answer

CONCENTRATION OF NICOTINATE ION Assume that the reaction of the base with the acid is complete. In this case, [latex]25 \mathrm{~mL}[/latex] of [latex]0.10 \mathrm{M}[/latex] sodium hydroxide is needed to react with [latex]25 \mathrm{~mL}[/latex] of [latex]0.10 \mathrm{M}[/latex] nicotinic acid. The molar amount of nicotinate ion formed equals the initial molar amount of nicotinic acid.

[latex]25 imes 10^{-3} \cancel{\mathrm{~L} ext { soln }} imes \frac{0.10 \mathrm{~mol} ext { nicotinate ion }}{1 \cancel{ ext { L soln }}}=2.5 imes 10^{-3} ext { mol nicotinate ion }[/latex]

The total volume of solution is [latex]50 \mathrm{~mL}[/latex] ( [latex]25 \mathrm{~mL} \mathrm{NaOH}[/latex] solution plus [latex]25 \mathrm{~mL}[/latex] 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.

[latex] ext { Molar concentration }=\frac{2.5 imes 10^{-3} \mathrm{~mol}}{50 imes 10^{-3} \mathrm{~L}}=0.050 \mathrm{M}[/latex]

HYDROLYSIS OF NICOTINATE ION This portion of the calculation follows the method given in Example 17.8. You find that [latex]K_{b}[/latex] for nicotinate ion is [latex]7.1 imes 10^{-10}[/latex] and that the concentration of hydroxide ion is [latex]6.0 imes 10^{-6} \mathrm{M}[/latex]. The [latex]\mathrm{pH}[/latex] is 8.78 .

Question 17.14: Calculating the pH at the Equivalence Point in the Titration...

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