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 \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\).

Short answer

The pH at the equivalence point for each titration is: (a) Sodium hydroxide (\(\mathrm{NaOH}\)): pH = 7 (b) Hydroxylamine (\(\mathrm{NH_2OH}\)): pH = 3.94 (c) Aniline (\(\mathrm{C_6H_5NH_2}\)): pH = 9.30

Step-by-step solution

Determine the volume of HBr required for equivalence point

For all three cases, the concentrations of the base and \(\mathrm{HBr}\) are the same (\(0.200 \mathrm{M}\)). Since they have a 1:1 stoichiometric ratio, the volume of \(\mathrm{HBr}\) required to reach the equivalence point will be equal to the volume of the base in each case. Let's assume we are titrating 50 mL of each base.

Find salt formed for each case and the equation for the acidic or basic species release

(a) Sodium hydroxide (\((NaOH)\)) will react with \((HBr)\) to form \(NaBr\) and water. \(NaBr\) is a neutral salt, so the pH at the equivalence point will be 7. (b) Hydroxylamine (\((NH_2OH)\)) will react with \((HBr)\) to form \(NH_3OBr\) and water. The salt formed, \(NH_3OBr\), can release hydroxylamine ions as acidic species according to the equation: \[ NH_3O^+ + H_2O \rightleftharpoons NH_2OH + H_3O^+ \] (c) Aniline (\((C_6H_5NH_2)\)) will react with \((HBr)\) to form \(C_6H_5NH_3Br\) and water. The salt formed, \(C_6H_5NH_3Br\), can release aniline ions as basic species according to the equation: \[ C_6H_5NH_3^+ + H_2O \rightleftharpoons C_6H_5NH_2 + H_3O^+ \]

Calculate the concentration of the acidic or basic species for each case

(a) Since there are no acidic or basic species present for this case, our work is done and the pH is simply 7. (b) At the equivalence point, the concentration of \(NH_3O^+\) will be the same as the initial concentration of the base, since the volume doubles at the equivalence point: \[ [\mathrm{NH}_3 \mathrm{O}^+] = \frac{0.200 M \times 50 mL}{100 mL} = 0.100 M \] (c) Similarly, at the equivalence point, the concentration of \(C_6H_5NH_3^+\) will be: \[ [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^+] = \frac{0.200 M \times 50 mL}{100 mL} = 0.100 M \]

Use calculations to determine pH

(a) The pH is 7. (b) For hydroxylamine, we use the given dissociation equation and the concentration of the acidic species to write an expression for the concentration of \(H_3O^+\): \[ K_a = \frac{[\mathrm{NH_2OH}][\mathrm{H_3O^+}]}{[\mathrm{NH_3O^+}]} \] We can look up the \(K_a\) value for hydroxylamine, which is approximately \(1.3 \times 10^{-8}\). Since the concentration of \(H_3O^+\) is very small compared to the concentration of the acidic species, we can write: \[ \frac{x^2}{0.100\ \mathrm{M}} = 1.3 \times 10^{-8} \rightarrow x = [\mathrm{H_3O^+}] \approx 1.14 \times 10^{-4}\ \mathrm{M} \] Finally, we can calculate the pH: \[ \mathrm{pH} = -\log([\mathrm{H_3O^+}]) \approx 3.94 \] (c) For aniline, we use the given dissociation equation and the concentration of the basic species to write an expression for the concentration of \(OH^-\): \[ K_b = \frac{[\mathrm{C_6H_5NH_2}][\mathrm{OH^-}]}{[\mathrm{C_6H_5NH_3^+}]} \] We can convert the \(pK_a\) value of aniline, which is approximately 4.63, to a \(K_b\) value using the formula: \[ K_b = \frac{K_w}{K_a} \] where \(K_w\) is the ion product of water, equal to \(1 \times 10^{-14}\). Therefore, \(K_b\) is approximately \(3.98 \times 10^{-10}\). Similar to the previous case, we can write: \[ \frac{x^2}{0.100\ \mathrm{M}} = 3.98 \times 10^{-10} \rightarrow x = [\mathrm{OH^-}] \approx 2.00 \times 10^{-5}\ \mathrm{M} \] We can now calculate the \(pOH\) and finally the pH: \[ \mathrm{pOH} = -\log([\mathrm{OH^-}]) \approx 4.70 \] \[ \mathrm{pH} = 14 - \mathrm{pOH} \approx 9.30 \] The pH at the equivalence point for each titration is: (a) Sodium hydroxide: pH = 7 (b) Hydroxylamine: pH = 3.94 (c) Aniline: pH = 9.30

Key concepts

Equivalence Point in Acid-Base Titration

The equivalence point in an acid-base titration is a critical concept to grasp. This is the point at which the amount of acid equals the amount of base during the titration process. Imagine it as the precise moment when the acid's protons have completely reacted with the base's hydroxides. Often, this results in a neutralization reaction:

• For strong acid and strong base titrations, the pH will be around 7.
• For weak acid paired with strong base titrations, the pH at the equivalence point is greater than 7.
• Conversely, for weak base and strong acid titrations, the pH is less than 7.

Identifying the equivalence point ensures that the stoichiometric quantities necessary for complete reaction have been achieved. During titrations, the pH is measured to detect this point, often accompanied by an indicator that changes color near the equivalence point.

Acid-Base Titration

Acid-base titration involves a controlled addition of a solution of known concentration, called the titrant, to a solution of analyte until the chemical reaction between the two is complete. It's a practical method used in many labs to determine the concentration of an unknown solution. Here's how it works in detail:

• The titrant is added gradually to the analyte until the equivalence point is reached.
• An indicator or a pH meter is used to determine when the equivalence point occurs.
• The analysis relies on the stoichiometry between the acid and base and the volume of titrant used.

Each titration curve will have its unique shape, reflecting the pH changes as the titration progresses. The sharp rise or drop in pH at the equivalence point is essential for precise determination of the analyte concentration.

Understanding Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They enable us to convey what happens during a reaction in a concise manner. For titration problems, they help illustrate the reaction between the titrant and the analyte. Here are a few basics:

• Reactants are shown on the left, products on the right.
• Balanced equations ensure that the same number of atoms of each element appears on both sides.
• Coefficients are used to balance the number of each atom involved in the reaction.

For instance, in the titration of NaOH with HBr, the reaction is simply: \[ \text{NaOH} + \text{HBr} \rightarrow \text{NaBr} + \text{H}_2\text{O} \]. Understanding and balancing chemical equations is vital for predicting the outcomes of reactions and understanding the stoichiometric relations used in calculations, like during titrations.