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Benzoic acid (C6H5COOH) and aniline (C6H5NH2) are both derivatives of benzene. Benzoic acid is an acid with Ka = 6.3 × 10–5 and aniline is a base with Ka = 4.3 × 10–10.

(a) What are the conjugate base of benzoic acid and the conjugate acid of aniline? (b) Anilinium chloride (C6H5NH3Cl) is a strong electrolyte that dissociates into anilinium ions (C6H5NH3+) and chloride ions. Which will be more acidic, a 0.10 M solution of benzoic acid or a 0.10 M solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium?

C6H5COOH(aq) + C6H5NH2(aq) C6H5COO-(aq) + C6H5NH3+(aq)

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Nestor Rutherford
Nestor RutherfordLv2
25 May 2020

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Related questions

Aniline hydrochloride, [C6H5NH3]Cl, is a weak acid. Its conjugate base is the weak base aniline, C6H5NH2. Ka= 2.40 X 10-5.

C6H5NH3+(aq) + OH-(aq) → C6H5NH2(aq) +H2O(l)

Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.200 M NaOH. What is the pH at the equivalence point?

Aniline hydrochloride, (C6H5NH3)Cl, is a weak acid. (Its conjugate base is the weak base aniline, C6H5NH2.) The acid can be titrated with a strong base such as NaOH.

C6H5NH3+(aq) + OH-(aq) → C6H5NH2(aq) + H2O(ℓ)

Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (Ka for aniline hydrochloride is 2.4 × 10-5.)

(a) What is the pH of the (C6H5NH3)Cl solution before the titration begins?

(b) What is the pH at the equivalence point?

(c) What is the pH at the halfway point of the titration?

(d) Which indicator in Figure 17.11 could be used to detect the equivalence point?

(e) Calculate the pH of the solution after adding 10.0, 20.0, and 30.0 mL of base.

(f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

Aniline hydrochloride, (C6H5NH3)Cl, is a weak acid. (Its conjugate base is the weak base aniline, C6H5NH2.) The acid can be titrated with a strong base such as NaOH.

C6H5NH3+(aq) + OH-(aq) → C6H5NH2(aq) + H2O(ℓ)

Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (Ka for aniline hydrochloride is 2.4 × 10-5.)

(a) What is the pH of the (C6H5NH3)Cl solution before the titration begins?

(b) What is the pH at the equivalence point?

(c) What is the pH at the halfway point of the titration?

(d) Which indicator in Figure 17.11 could be used to detect the equivalence point?

(e) Calculate the pH of the solution after adding 10.0, 20.0, and 30.0 mL of base.

(f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

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