Aniline, (C6H5NH2, Kb = 4.3 × 10-10 at 25°C) is an industrially important amine used in the making of dyes. Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution to achieve a concentration of 0.42 M.Group of answer choices9.749.1310.744.87

To solve this problem, we need to use the formula for the base ionization constant (Kb) and the definition of pH. Here are the steps:

1. Calculate the concentration of aniline ([C6H5NH2]) in the solution. The problem states that the solution is 0.42 M, so [C6H5NH2] = 0.42 M.

2. Write the ionization reaction for aniline: C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

3. Write the expression for Kb: Kb = [C6H5NH3+][OH-] / [C6H5NH2]

4. Since we're looking for the pH of the solution, we're interested in the concentration of OH- ions. Let's assume that x M of C6H5NH2 ionizes, so we have: [C6H5NH3+] = x [OH-] = x [C6H5NH2] = 0.42 - x

5. Substitute these values into the Kb expression: 4.3 × 10^-10 = x^2 / (0.42 - x)

6. Since Kb is very small, we can assume that x is much smaller than 0.42, so the equation simplifies to: 4.3 × 10^-10 = x^2 / 0.42

7. Solve for x (which is [OH-]): x = sqrt(4.3 × 10^-10 * 0.42) = 1.35 × 10^-5 M

8. Calculate the pOH of the solution using the formula pOH = -log[OH-]: pOH = -log(1.35 × 10^-5) = 4.87

9. Finally, calculate the pH of the solution using the formula pH = 14 - pOH: pH = 14 - 4.87 = 9.13

So, the pH of the aniline solution is approximately 9.13.