We’re being asked to calculate the pH of a 0.24 M aniline solution given the K_{b} of 3.8x10^{-10}

In this case, we have to follow the steps:

**Step 1. **Setup the equilibrium and ICE chart

**Step 2.** Calculate x or OH^{-}

**Step 3. **Calculate pH

**Step 1. **The formula for aniline is C_{6}H_{5}NH_{2} and its reaction with water is:

** C _{6}H_{5}NH_{2} + H_{2}O ⇌ C_{6}H_{5}NH_{3}^{+} + OH**

We need to write an ICE chart to determine the concentration of OH^{-} which we will use to get the pOH, then the pH.

We know that the equilibrium constant expression, K_{b} is:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

What is the pH of a 0.24 M solution of aniline (C_{6}H_{5}NH_{2}, K_{b} = 3.8 × 10^{–10}) at 25°C?

A) 2.00

B) 5.02

C) 10.04

D) 13.38

E) 8.98