Calculate the pH of a solution that is 0.124 M C2H5NH2 and 0.124 M C2H5NH3Cl. (Assume that the solution is at 25°C.)
To calculate the pH of a solution, we need to determine the concentration of hydrogen ions (H+) in the solution. In this case, we have a mixture of two substances: C2H5NH2 (ethylamine) and C2H5NH3Cl (ethylammonium chloride).
First, let's identify the species that can contribute to the pH of the solution. Ethylamine (C2H5NH2) is a weak base that can accept a proton to form ethylammonium ion (C2H5NH3+), while ethylammonium chloride (C2H5NH3Cl) is its conjugate acid. The equilibrium equation is:
C2H5NH2 + H2O ⇌ C2H5NH3+ + OH-
Since we have both ethylamine and its conjugate acid in the solution, they will undergo the following reactions:
C2H5NH2 ⇌ C2H5NH3+ + OH- (Reaction 1)
C2H5NH3Cl ⇌ C2H5NH3+ + Cl- (Reaction 2)
Since Reaction 1 involves the production of hydroxide ions (OH-), it will affect the pH of the solution. Therefore, we need to determine the concentration of hydroxide ions (OH-) to calculate the pH.
For Reaction 1, we need to consider the base dissociation constant (Kb) of ethylamine, which can be given as:
Kb = [C2H5NH3+][OH-] / [C2H5NH2]
In this case, [C2H5NH3+] = [OH-] because the reaction is 1:1. Therefore:
Kb = [OH-]^2 / [C2H5NH2]
We know that the Kb value for ethylamine is 6.4 x 10^-4. We can rearrange the equation to solve for [OH-]:
[OH-]^2 = Kb * [C2H5NH2]
[OH-]^2 = (6.4 x 10^-4) * (0.124 M)
Now, we can solve for [OH-]:
[OH-]^2 = 8.96 x 10^-5
[OH-] = √(8.96 x 10^-5)
[OH-] ≈ 9.47 x 10^-3 M
To calculate the pOH (negative logarithm of [OH-]), we can use the formula:
pOH = -log[OH-]
pOH = -log(9.47 x 10^-3)
pOH ≈ 2.02
Finally, to find the pH of the solution, we can use the formula:
pH = 14 - pOH
pH = 14 - 2.02
pH ≈ 11.98
Therefore, the pH of the solution containing 0.124 M C2H5NH2 and 0.124 M C2H5NH3Cl at 25°C is approximately 11.98.