Question

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with \(0.200 \mathrm{M} \mathrm{HBr}\) : (a) sodium hydroxide \((\mathrm{NaOH})\), (b) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right)\), (c) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\right)\).

Short answer

The pH at the equivalence point for titrating 0.200 M solutions of the following bases with 0.200 M HBr are: (a) Sodium hydroxide (NaOH) - pH = 7.00, (b) Hydroxylamine (NH2OH) - pH = 7.96, and (c) Aniline (C6H5NH2) - pH = 4.70.

Step-by-step solution

Write the balanced equation for the reaction with HBr

(a) Sodium hydroxide (NaOH) is a strong base. The reaction with HBr is as follows: \( NaOH + HBr \rightarrow NaBr + H_{2}O \) (b) Hydroxylamine (NH2OH) is a weak base. The reaction with HBr is as follows: \( NH_{2}OH + HBr \rightarrow NH_{3}OH^{+} + Br^{-} \) (c) Aniline (C6H5NH2) is a weak base. The reaction with HBr is as follows: \( C_{6}H_{5}NH_{2} + HBr \rightarrow C_{6}H_{5}NH_{3}^{+} + Br^{-} \)

Calculate the moles of base and HBr at the equivalence point

Let's assume there's 1 L of the base solution. In each case, the concentration of the base solution is 0.2 M and the same for the HBr solution. Since at the equivalence point, the number of moles of base equals the number of moles of HBr: Moles of base, NaOH: \( 0.2 \mathrm{M} × 1 \mathrm{L} = 0.2 \mathrm{mol} \) Moles of HBr needed: \( 0.2 \mathrm{mol} \) The same applies to both NH2OH and C6H5NH2.

Calculate the pH for each base at the equivalence point

(a) As NaOH is a strong base, the reaction goes to completion. Therefore, after the reaction, we have 0.2 moles of water, whose pH is 7. (b) For NH2OH, it forms NH3OH+ after the reaction with HBr. Since NH2OH is a weak base, we need to use the Ka value for NH3OH+ to calculate the [H+] concentration. The Ka value for NH3OH+ is 1.1x10^-8. Let x be the [H+] concentration: \(K_{a} = \frac{[H^{+}][NH_{2}OH]}{[NH_{3}OH^{+}]} \) \(1.1 \times 10^{-8} = \frac{x \times (0.2 - x)}{x} \) Since x is very small, we can approximate: \(1.1 \times 10^{-8} = \frac{0.2x}{x} \) \(x = 1.1 \times 10^{-8}\) Since [H+]=x, and pH =-log[H+] pH = -log(1.1 x 10^{-8}) = 7.96 (c) For C6H5NH2, similarly, after the reaction, it forms C6H5NH3+. We need to use the Ka value for C6H5NH3+ to calculate the [H+] concentration. The Ka value for C6H5NH3+ is 2.0x10^-5. Let x be the [H+] concentration: \(K_{a} = \frac{[H^{+}][C_{6}H_{5}NH_{2}]}{[C_{6}H_{5}NH_{3}^{+}]} \) \(2.0 \times 10^{-5} = \frac{x \times (0.2 - x)}{x} \) Since x is very small, we can approximate: \(2.0 \times 10^{-5} = \frac{0.2x}{x} \) \(x = 2.0 \times 10^{-5}\) Since [H+]=x, and pH = -log[H+] pH = -log(2.0 x 10^{-5}) = 4.70

Final Answers

(a) Sodium hydroxide (NaOH) - pH at equivalence point = 7.00 (b) Hydroxylamine (NH2OH) - pH at equivalence point = 7.96 (c) Aniline (C6H5NH2) - pH at equivalence point = 4.70

Key concepts

Equivalence Point

The equivalence point in a titration is a critical concept in chemistry. It refers to the stage in the titration process where the number of moles of the titrant equals the number of moles of the substance being titrated. At this point, the solution is neither acidic nor basic. This balancing act achieves a neutralization reaction. For example, when titrating a weak base like hydroxylamine (forms NH3OH\(^{+}\)) with a strong acid like HBr, the equivalence point indicates complete neutralization, resulting in the formation of a weak conjugate acid.

Titration

Titration is a quantitative chemical analysis method used to determine the concentration of a given reactant. It involves the gradual addition of a titrant to a solution known as the analyte until the reaction reaches the equivalence point. The titration between a strong acid and a weak base results in three scenarios, depending on the strength of the reactants:

• For strong acids like HBr and strong bases like NaOH, the result is a neutral solution with a pH of 7 at the equivalence point.
• With weak bases, the pH at the equivalence point is determined by the conjugate acid formed after neutralization.
• The calculation of pH requires a consideration of the equilibrium constants (Ka) for the resulting conjugate acids.

The choice of indicator is crucial in titration experiments as it must change color close to the equivalence point.

Acid-Base Reaction

An acid-base reaction is a chemical reaction that occurs between an acid and a base. In an aqueous solution, it typically results in the production of water and a salt. The overall process involves the transfer of protons (H\(^+\)) from the acid to the base. The type of base and acid involved heavily influences the product:

• In the reaction between NaOH and HBr, both strong, the complete dissociation forms a neutral solution.
• For weak bases like aniline and hydoxylamine reacting with a strong acid, the products include the weak conjugate acid rather than water, often resulting in a solution with a pH different from 7.

Weak Base

A weak base is a chemical base that does not fully ionize in an aqueous solution. Unlike strong bases, which dissociate completely, weak bases like hydroxylamine (NH2OH) and aniline (C6H5NH2) only partially dissociate. When they react with a strong acid like HBr, the neutralization forms a conjugate acid which can partially ionize:

• In the case of NH2OH, the reaction with HBr forms NH3OH\(^{+}\).
• For aniline, the result is the formation of C6H5NH3\(^{+}\).

The pH at the equivalence point for titrations involving weak bases is determined by the ionization of these conjugate acids, with calculations involving their respective equilibrium constants (Ka).

Strong Base

Strong bases are substances that completely dissociate into its ions in an aqueous solution. Sodium hydroxide (NaOH) is a classic example of a strong base. When NaOH is titrated against a strong acid like HBr, the outcome is a complete neutralization reaction, producing water and salts like NaBr. This leads to a neutral pH, which at the equivalence point is precisely 7. This complete dissociation results in a simpler calculation for the pH at the equivalence point, requiring no consideration of a Ka, as no weak conjugate acids are formed.