To calculate the pH of the solution, we first need to understand that anilinium chloride is the conjugate acid of aniline. When it is dissolved in water, it donates a proton to water to form aniline and hydronium ion. The reaction can be represented as follows:

C6H5NH3+ + H2O - C6H5NH2 + H3O+

Since aniline is a weak base, we can assume that the reaction does not go to completion and that an equilibrium is established. We can represent this equilibrium using the expression for the acid dissociation constant (Ka) of anilinium ion:

Ka = [C6H5NH2][H3O+] / [C6H5NH3+]

We know the initial concentration of anilinium ion is 0.86 M and we can assume that the change in concentration is x. Therefore, at equilibrium, the concentration of anilinium ion is 0.86 - x, the concentration of aniline is x and the concentration of hydronium ion is also x. Substituting these values into the expression for Ka gives:

Ka = x^2 / (0.86 - x)

We also know that Ka and Kb are related by the equation Ka = Kw / Kb, where Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C. Substituting the given value of pKb into this equation gives:

Ka = 1.0 x 10^-14 / 10^-4.87 = 1.35 x 10^-10

Setting this equal to the expression for Ka obtained from the equilibrium concentrations and solving for x gives:

1.35 x 10^-10 = x^2 / (0.86 - x)

Solving this equation gives x = 1.05 x 10^-5 M, which is the concentration of hydronium ion in the solution.

Finally, the pH of the solution is given by the negative logarithm of the hydronium ion concentration:

pH = -log[H3O+] = -log(1.05 x 10^-5) = 4.98

Therefore, the pH of the solution is approximately 5.0 when rounded to one decimal place.

Question

To calculate the pH of the solution, we first need to understand that anilinium chloride is the conjugate acid of aniline. When it is dissolved in water, it donates a proton to water to form aniline and hydronium ion. The reaction can be represented as follows:

C6H5NH3+ + H2O -> C6H5NH2 + H3O+

Since aniline is a weak base, we can assume that the reaction does not go to completion and that an equilibrium is established. We can represent this equilibrium using the expression for the acid dissociation constant (Ka) of anilinium ion:

Ka = [C6H5NH2][H3O+] / [C6H5NH3+]

We know the initial concentration of anilinium ion is 0.86 M and we can assume that the change in concentration is x. Therefore, at equilibrium, the concentration of anilinium ion is 0.86 - x, the concentration of aniline is x and the concentration of hydronium ion is also x. Substituting these values into the expression for Ka gives:

Ka = x^2 / (0.86 - x)

We also know that Ka and Kb are related by the equation Ka = Kw / Kb, where Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C. Substituting the given value of pKb into this equation gives:

Ka = 1.0 x 10^-14 / 10^-4.87 = 1.35 x 10^-10

Setting this equal to the expression for Ka obtained from the equilibrium concentrations and solving for x gives:

1.35 x 10^-10 = x^2 / (0.86 - x)

Solving this equation gives x = 1.05 x 10^-5 M, which is the concentration of hydronium ion in the solution.

Finally, the pH of the solution is given by the negative logarithm of the hydronium ion concentration:

pH = -log[H3O+] = -log(1.05 x 10^-5) = 4.98

Therefore, the pH of the solution is approximately 5.0 when rounded to one decimal place.

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