Problem 45 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The pH at the equivalence point for titrating the given bases with HBr are: for NaOH, pH = 7; for NH2OH, pH ≈ 1.33; and for C6H5NH2, pH ≈ 2.51.

Step 1: 1. Write the chemical equation for the reaction with HBr

NaOH (aq) + HBr (aq) -> NaBr (aq) + H2O (l) Since NaOH is a strong base, the reaction will go to completion. At the equivalence point, all the NaOH will have reacted with HBr.

Step 2: 2. Determine the resulting species at the equivalence point

At the equivalence point, the solution will consist of NaBr and H2O. NaBr is a neutral salt formed by the complete reaction of a strong acid (HBr) and a strong base (NaOH). Therefore, the hydrolysis of the salt is not significant.

Step 3: 3. Calculate the pH at the equivalence point

The presence of NaBr doesn't affect the pH of the solution, and the only other species is water. So, the pH at the equivalence point will be the pH of the pure water, which is 7. \(b) Hydroxylamine (NH2OH)\)

Step 4: 1. Write the chemical equation for the reaction with HBr

NH2OH (aq) + HBr (aq) -> NH3OH+ (aq) + Br- (aq) In this case, hydroxylamine is a weak base. At the equivalence point, all the weak base will have reacted with HBr.

Step 5: 2. Determine the resulting species at the equivalence point

At the equivalence point, the solution will consist of NH3OH+ (a weak acid) and Br- (a spectator ion). The contribution of NH3OH+ to the pH should be considered.

Step 6: 3. Calculate the pH at the equivalence point

NH3OH+ is a weak acid, and its dissociation can be represented as one of the following equilibrium reactions: NH3OH+ (aq) <-> NH2OH (aq) + H+ (aq) We can use Ka, the acid dissociation constant, to calculate the concentration of H+ in the solution at the equivalence point. Ka for NH3OH+ is approximately 1.1 x 10^-2. \[ Ka = \frac{[NH2OH][H^+]}{[NH3OH^+]} \] At the equivalence point, we can assume the concentration of NH2OH and H+ to be equal, and the initial concentration of NH3OH+ to be 0.200 M, the same as the starting concentration of the titrating base. Let x be the concentration of H+ and NH2OH at equilibrium. \[ 1.1 \times 10^{-2} = \frac{x^2}{0.200-x} \] The concentration of H+ is very small compared to the initial concentration of NH3OH+, so we can assume 0.200 - x ≈ 0.200: \[ 1.1 \times 10^{-2} ≈ \frac{x^2}{0.200} \] Solve for x: \[ x = \sqrt{1.1 \times 10^{-2} \times 0.200} ≈ 4.69 \times 10^{-2} \] Now, we can plug in the concentration of H+ to find the pH: \[ pH = -\log_{10}(4.69 \times 10^{-2}) \approx 1.33 \] \(c) Aniline (C6H5NH2)\)

Step 7: 1. Write the chemical equation for the reaction with HBr

C6H5NH2 (aq) + HBr (aq) -> C6H5NH3+ (aq) + Br- (aq) Aniline is also a weak base that reacts with the strong acid HBr.

Step 8: 2. Determine the resulting species at the equivalence point

At the equivalence point, the solution will consist of C6H5NH3+ (a weak acid) and Br- (a spectator ion). The contribution of C6H5NH3+ to the pH should be considered.

Step 9: 3. Calculate the pH at the equivalence point

C6H5NH3+ is a weak acid that will dissociate according to the following equilibrium reaction: C6H5NH3+ (aq) <-> C6H5NH2 (aq) + H+ (aq) We'll use Ka, the acid dissociation constant, to find the concentration of H+ ions in the solution at the equivalence point. The Ka for C6H5NH3+ is approximately 2.4 x 10^-5. \[ Ka = \frac{[C6H5NH2][H^+]}{[C6H5NH3^+]} \] At the equivalence point, the concentration of C6H5NH2 and H+ will be equal. Let x be the concentration of H+ and C6H5NH2 at equilibrium. The initial concentration of C6H5NH3+ is 0.200 M, the same as the starting concentration of the titrating base. \[ 2.4 \times 10^{-5} = \frac{x^2}{0.200-x} \] The concentration of H+ is very small compared to the initial concentration of C6H5NH3+, so we can assume 0.200 - x ≈ 0.200: \[ 2.4 \times 10^{-5} ≈ \frac{x^2}{0.200} \] Solve for x: \[ x = \sqrt{2.4 \times 10^{-5} \times 0.200} ≈ 3.07 \times 10^{-3} \] Now, we can plug in the concentration of H+ to find the pH: \[ pH = -\log_{10}(3.07 \times 10^{-3}) \approx 2.51 \] In conclusion, the pH at the equivalence point for titrating the given bases with HBr are: for NaOH, pH = 7; for NH2OH, pH ≈ 1.33; and for C6H5NH2, pH ≈ 2.51.

Key Concepts

Titration of Bases

The process of titration involves adding a known concentration of acid to a solution containing a base until the reaction reaches the equivalence point, where the number of moles of acid equals the number of moles of base.

For a strong base like sodium hydroxide (NaOH), the reaction with a strong acid such as hydrobromic acid (HBr) will go to completion, leaving us with a neutral solution at the equivalence point with a pH of 7. However, in the case of weak bases like hydroxylamine (NH₂OH) or aniline (C₆H₅NH₂), the situation is different. After reacting with HBr, the resulting solution contains the conjugate acid of the weak base, which affects the pH of the solution.

Strong and Weak Electrolytes

Electrolytes are substances that dissolve in water to form ions, which makes the solution capable of conducting electricity. Strong electrolytes dissociate completely into ions in solution. Sodium hydroxide, for instance, dissociates completely to form Na⁺ and OH^- ions in water, making it a strong base and strong electrolyte.

In contrast, weak electrolytes dissociate only partially, maintaining an equilibrium between the undissociated molecule and the ions. Weak bases like aniline do not completely ionize in solution, and that's why they require careful consideration during a titration to determine their effect on the pH.

Acid-Base Equilibria

Understanding acid-base equilibria is essential in titrations. It is the balance between the forward and reverse reactions of acids and bases in water. For strong acids and bases, this equilibrium lies far to the right (products favored), indicating complete ionization.

For weak acids and bases, however, the equilibrium is more central, and we need to carefully consider the dynamic between the weak acid (conjugate acid of a weak base) and its ionization in water. The extent to which these weak acids influence the pH depends on the equilibrium constant, known as the acid dissociation constant, or Ka.

Acid Dissociation Constant (Ka)

The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution. It is defined as the equilibrium constant for the dissociation of the acid into its ions. For weak acids, which are the conjugate acids of weak bases following a titration, the Ka value is critical in calculating the pH.

The smaller the Ka value, the weaker the acid and the less it dissociates in solution. In the given exercise, the Ka values of the weak acids formed at the equivalence points for hydroxylamine and aniline are provided, allowing us to calculate the concentration of hydrogen ions and subsequently the pH of the solution.

pH Calculation

To calculate the pH of a solution, one needs to know the concentration of hydrogen ions (H⁺). The pH is defined as the negative logarithm of the hydrogen ion concentration: pH = -log₁₀[H⁺].

An assumption often made during these calculations is that the concentration of H⁺ is much less than the initial concentration of the acid, simplifying the arithmetic. The calculations become more intricate with weak acids, where the equilibrium concentration of H⁺ is found by solving the equilibrium expression involving Ka. Once these concentrations are determined, finding the pH is straightforward using logarithmic calculations, as demonstrated with hydroxylamine and aniline in the exercise.