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Calculate the \(\mathrm{pH}\) at the equivalence point for titrating 0.200 \(\mathrm{M}\) solutions of each of the following bases with 0.200 \(M \mathrm{HBr} :(\mathbf{a})\) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\)

Short Answer

Expert verified
The pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr is as follows: (a) For sodium hydroxide (NaOH), pH = 15.74 (b) For hydroxylamine (NH₂OH), pH = 5.91 (c) For aniline (C₆H₅NH₂), pH = 4.60

Step by step solution

01

Identify the conjugate acid of each base

For each base, we need to identify its conjugate acid, which is formed by the addition of a proton (H⁺) to the base. We will do this for all three bases: (a) NaOH → Conjugate acid: H2O (b) NH2OH → Conjugate acid: NH₃OH⁺ (c) C6H5NH2 → Conjugate acid: C6H5NH₃⁺
02

Determine the pKₐ values for each conjugate acid

Next, we need to determine the pKₐ values for the conjugate acids. These values can be found in tables or other resources, such as a textbook or the internet. For our examples: (a) H2O → pKₐ = 15.74 (b) NH₃OH⁺ → pKₐ = 5.91 (c) C6H5NH₃⁺ → pKₐ = 4.60
03

Calculate the pH at the equivalence point

At the equivalence point, the number of moles of acid equals the number of moles of base in the solution. Since the concentrations of the acid and base solutions are equal (0.200 M), the pH is the same as the pKₐ of the conjugate acid. Therefore, we can directly use the pKₐ values to determine the pH at the equivalence point for each base: (a) For NaOH, pH = pKₐ = 15.74 (b) For NH₂OH, pH = pKₐ = 5.91 (c) For C₆H₅NH₂, pH = pKₐ = 4.60

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalence Point
Understanding the equivalence point is essential for mastering acid-base titrations. It refers to the moment in a titration when the amount of added titrant (typically an acid or a base) is exactly enough to neutralize the analyte (the substance being titrated).

At the equivalence point, the moles of acid are equal to the moles of base in the solution. This is a key concept because it marks the point where the number of hydrogen ions (H⁺) equals the number of hydroxide ions (OH⁻) for acid-base reactions. The pH value at this point can give us valuable information about the nature of the solution, which might be neutral, acidic, or basic depending on the acid and base involved in the titration.
Conjugate Acids and Bases
The concept of conjugate acids and bases is a central theme in the study of acid-base chemistry. A conjugate acid is what you get when a base gains a hydrogen ion. Conversely, a conjugate base is what remains after an acid donates a hydrogen ion. These conjugate pairs are always found together in solution and are directly related to each other through this loss or gain of a hydrogen ion.

In titrations, when a base is neutralized by an acid, it converts into its conjugate acid. For example, when sodium hydroxide (NaOH) reacts with hydrogen bromide (HBr), water (H₂O) forms, which is the conjugate acid of the base NaOH. Understanding this conversion helps in predicting the composition of the solution at the equivalence point and subsequently allows for accurate calculations of the pH.
pKa Values
pKa values are integral to understanding the strengths of acids and their conjugate bases. The pKa is the negative base-10 logarithm of the acid dissociation constant (Ka) of a solution, which gives an indication of the acid's strength. The lower the pKa value, the stronger the acid and consequently, the weaker its conjugate base.

When it comes to pH calculations at the equivalence point of a titration, the pKa of the conjugate acid formed after neutralization is particularly important. In the provided exercise, knowledge of the pKa value allows us to calculate the pH directly at the equivalence point because, at this specific point, the pH of the solution is equal to the pKa of the conjugate acid, assuming a monoprotic acid and a 1:1 stoichiometry between acid and base.
Titration
Titration is a laboratory technique used to determine the concentration of an unknown substance in a solution. It involves the gradual addition of a known reagent, called the titrant, into the solution containing the analyte. During this process, measurements are taken until the reaction reaches the equivalence point, where stoichiometrically equivalent quantities of reactants have reacted.

To measure the progress of the titration and pinpoint the equivalence point, indicators or pH meters are often used. For strong acid-strong base titrations, the equivalence point typically falls around a pH of 7, however, in titrations involving weak acids or weak bases, the equivalence point pH can be less than 7 (for a weak base and strong acid titration) or greater than 7 (for a weak acid and strong base titration), as demonstrated in the examples of the exercise.

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Most popular questions from this chapter

Which of the following solutions is a buffer? (a) A solution made by mixing 100 \(\mathrm{mL}\) of 0.100 \(\mathrm{MCH}_{3} \mathrm{COOH}\) and 50 mL of \(0.100 M \mathrm{NaOH},(\mathbf{b})\) a solution made by mixing 100 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and 500 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{NaOH}\) , (c) A solution made by mixing 100 \(\mathrm{mL}\) of 0.100 \(\mathrm{MCH}_{3} \mathrm{COOH}\) and 50 \(\mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HCl},\) \((\mathbf{d} ) \) A solution made by mixing of 0.100 \(\mathrm{MCH}_{3} \mathrm{COOK}\) and 50 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{KCl}\) .

A 20.0 -mL sample of 0.150 \(\mathrm{M} \mathrm{KOH}\) is titrated with 0.125 \(\mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(20.0 \mathrm{mL},\) (b) 23.0 \(\mathrm{mL}\) \((\mathbf{c}) 24.0 \mathrm{mL},(\mathbf{d}) 25.0 \mathrm{mL},(\mathbf{e}) 30.0 \mathrm{mL}\)

Use values of \(K_{s p}\) for AgI and \(K_{f}\) for \(A g(C N)_{2}^{-}\) to (a) calculate the molar solubility of Agl in pure water, (b) calculate the equilibrium constant for the reaction \(\operatorname{AgI}(s)+2 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{I}^{-}(a q), \quad(\mathbf{c})\) determine the molar solubility of AgI in a 0.100 \(\mathrm{MNaCN}\) solution.

(a) True or false: “solubility” and “solubility-product constant” are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: MnCO \(_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2} .\)

A sample of 7.5 \(\mathrm{L}\) of \(\mathrm{NH}_{3}\) gas at \(22^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50 -L solution of 0.40 \(\mathrm{M}\) HCl. Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

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