What is the pH of a 0.570 M solution of aniline? Kb = 7.4 × 10-10
To find the pH of a solution of aniline, we need to consider the basic nature of aniline and its equilibrium with water.
Aniline (C6H5NH2) is a weak base that reacts with water to form its conjugate acid, C6H5NH3+, and hydroxide ion (OH-). The equilibrium can be represented as follows:
C6H5NH2 + H2O ↔ C6H5NH3+ + OH-
To calculate the pH, we need to determine the concentration of OH- ions in the solution. This can be done by solving the equilibrium expression using the equilibrium constant (Kb) and the initial concentration of aniline.
The equilibrium constant (Kb) for aniline's reaction with water can be written as follows:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
Given that Kb = 7.4 × 10^-10 and the initial concentration of aniline is 0.570 M, we can set up the equation as follows:
7.4 × 10^-10 = [C6H5NH3+][OH-] / 0.570
Now, let's solve for [OH-]:
[C6H5NH3+][OH-] = (7.4 × 10^-10) * 0.570
[C6H5NH3+][OH-] = 4.218 × 10^-10
Since aniline is a weak base, we can assume that the concentration of [C6H5NH3+] is negligible compared to [OH-] because only a small fraction of aniline will react with water. Therefore, we can simplify the equation to:
[OH-] = 4.218 × 10^-10
To find the pOH, we take the negative logarithm of the concentration of hydroxide ions:
pOH = -log([OH-])
pOH = -log(4.218 × 10^-10)
pOH ≈ 9.375
Finally, to get the pH, we subtract the pOH from 14 (pH + pOH = 14):
pH ≈ 14 - 9.375
pH ≈ 4.625
Therefore, the pH of the 0.570 M solution of aniline is approximately 4.625.
To find the pH of a solution of aniline, we need to determine the concentration of hydroxide ions (OH-) in the solution.
Aniline is a weak base, so it reacts with water to form an anilinium ion (C6H5NH3+) and hydroxide ion (OH-).
The reaction can be represented as:
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-
The equilibrium constant for this reaction is called the base dissociation constant (Kb). In this case, Kb = 7.4 × 10-10.
To find the concentration of OH- ions, we can use the formula for Kb:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
Since we are given the concentration of aniline (0.570 M), we can assume that the concentration of the anilinium ion is negligible compared to the concentration of aniline at equilibrium.
Therefore, we can write Kb = [OH-][C6H5NH2] / [C6H5NH2].
Simplifying the equation, we get [OH-] = Kb * [C6H5NH2].
Plugging in the given values, we have [OH-] = (7.4 × 10-10) * (0.570).
Calculating this, we get [OH-] = 4.218 × 10-10 M.
To find the pH, we need to convert the concentration of hydroxide ions to concentration of hydrogen ions.
Using the equation Kw = [H+][OH-], where Kw is the ionization constant of water (1 * 10^-14 at 25 degrees Celsius), we can calculate [H+].
Kw = [H+][OH-]
1 * 10^-14 = [H+][4.218 × 10^-10]
Rearranging the equation, we have [H+] = (1 * 10^-14) / (4.218 × 10^-10).
Calculating this, we get [H+] = 2.374 × 10^-5 M.
To find the pH, we can use the formula pH = -log[H+].
Substituting the value of [H+], we get pH = -log(2.374 × 10^-5).
Calculating this, we get pH = 4.62.
Therefore, the pH of a 0.570 M solution of aniline is approximately 4.62.