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Chapter 17: Problem 43

Explain why the volume of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to reach the equivalence point in the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HA is the same regardless of whether HA is a strong or a weak acid, yet the \(\mathrm{pH}\) at the equivalence point is not the same.

Short Answer

Expert verified
The volume of NaOH required to reach the equivalence point is the same whether HA is a strong or weak acid because this volume is determined by the number of H+ ions in solution, which are present in equal quantities for a given concentration and volume of any acid. However, the pH at the equivalence point varies because a strong acid will have all its H+ ions neutralized by NaOH, giving a neutral pH of 7, while a weak acid will have residual unmolested acid molecules that yield H+ ions, making the solution acidic and hence the pH at the equivalence point will be less than 7.

Step by step solution

01

Understanding Titration

Titration is a technique in chemistry used to determine the concentration of an unknown substance by adding a solution of known concentration (titrant) until the reaction between them is just complete. The point at which this occurs is known as the equivalence point.
02

Understanding Reactions of Strong and Weak Acids with Bases

Strong acids fully ionize in water, producing H+ ions. Weak acids only partially ionize, producing fewer H+ ions. When a strong base like NaOH is added, it removes H+ ions from the solution by turning them into water (H2O). The base will do this whether the acid is strong or weak, thus the volume of NaOH needed to react with all the H+ ions (i.e., reach the equivalence point) is the same for a given volume and concentration of any acid.
03

Differentiating the pH at the Equivalence Point

While the volume of NaOH to reach the equivalence point is the same for strong and weak acids, the pH at the equivalence point is different. This is due to the reaction of the strong base with the dissociated (for strong acid) or residual, non-dissociated (for weak acid) acid particles. For strong acid, all particles are H+ ions and all are neutralized by the base, leaving only water at the equivalence point, which has a pH of 7. For weak acid, besides H+ ions, there are also residual non-dissociated acid molecules that react with water to establish an equilibrium yielding H+ ions and making the solution acidic and hence the pH of the solution at the equivalence point is less than 7.

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Most popular questions from this chapter

To convert \(\mathrm{NH}_{4}^{+}(\text {aq })\) to \(\mathrm{NH}_{3}(\mathrm{aq}),\) (a) add \(\mathrm{H}_{3} \mathrm{O}^{+}\) (b) raise the \(\mathrm{pH} ;\) (c) add \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (d) add \(\mathrm{NaCl}\).

For the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(0.100 \mathrm{M} \mathrm{HCl},\) calculate the \(\mathrm{pOH}\) at a few representative points in the titration, sketch the titration curve of pOH versus volume of titrant, and show that it has exactly the same form as Figure \(17-9 .\) Then, using this curve and the simplest method possible, sketch the titration curve of pH versus volume of titrant.

Is a solution that is \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}(\mathrm{aq})\) likely to be acidic, basic, or pH neutral? Explain.

Calculate the change in pH that results from adding (a) \(0.100 \mathrm{mol} \mathrm{NaNO}_{2}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{2}(\mathrm{aq})\) (b) \(0.100 \mathrm{mol} \mathrm{NaNO}_{3}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}(\mathrm{aq})\) Why are the changes not the same? Explain.

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.132 \mathrm{M} \mathrm{HNO}_{2}\) when (a) \(10.00 \mathrm{mL}\) and (b) 20.00 mL of 0.116 M \(\mathrm{HNO}_{2}\) when \((\mathrm{a}) 10.00 \mathrm{mL}\) and (b) \(20.00 \mathrm{mL}\) of \(0.116 \mathrm{M}\) NaOH have been added. For \(\mathrm{HNO}_{2}, K_{\mathrm{a}}=7.2 \times 10^{-4}\) \(\mathrm{HNO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{-}\)
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