Final answer:
To calculate the pH of the solution resulting from the addition of 25.0 mL of 0.20 M HCl to 50.0 mL of 0.10 M aniline, we use the equilibrium constant Kb to find the concentration of hydroxide ions and then convert it to pH. The pH is approximately 23.72.
Step-by-step explanation:
To calculate the pH of the solution resulting from the addition of 25.0 mL of 0.20 M HCl to 50.0 mL of 0.10 M aniline (C6H5NH2), we can use the concept of acid-base reactions and the equilibrium constant, Kb. The reaction between HCl and aniline can be represented as:
C6H5NH2 + HCl → C6H5NH3+ + Cl-
Since aniline is a weak base and HCl is a strong acid, the reaction will proceed to the right. We can calculate the concentrations of aniline and its conjugate acid at equilibrium, and then use them to find the pH of the solution.
First, we need to find the concentrations of aniline and its conjugate acid. Since we are provided with the volumes of the solutions, we can use the formula:
molarity = moles/volume
Concentration of aniline (C6H5NH2) = (0.10 M) * (50.0 mL / 1000 mL/L) = 0.0050 M
Concentration of anilinium ion (C6H5NH3+) = (0.10 M) * (25.0 mL / 1000 mL/L) = 0.0025 M
Now, we can use the equilibrium constant, Kb, to find the concentration of hydroxide ions (OH-) and then convert it to pH.
Kb = [C6H5NH2][OH-] / [C6H5NH3+]
We can rearrange this equation to solve for [OH-]:
[OH-] = (Kb * [C6H5NH3+]) / [C6H5NH2]
Substituting the given values:
[OH-] = (3.8 x 10-10 * 0.0025 M) / 0.0050 M = 1.9 x 10-10 M
Finally, we can calculate the pH using the equation:
pH = 14 - pOH = 14 - (-log10[OH-]) = 14 - (-log10(1.9 x 10-10)) = 14 + 10 - log10(1.9)
pH = 24 + log10(1/1.9) = 24 + log10(1.9-1) = 24 + (-0.28) = 23.72
Therefore, the pH of the solution resulting from the addition of 25.0 mL of 0.20 M HCl to 50.0 mL of 0.10 M aniline is approximately 23.72.