Question

what is the ph of a 0.270 m solution of aniline (c6h5nh2, kb = 4.3 × 10^-10)?

Explanation:

Step1: Set up the equilibrium expression

Aniline (\(C_6H_5NH_2\)) reacts with water: \(C_6H_5NH_2 + H_2O
ightleftharpoons C_6H_5NH_3^+ + OH^-\). Let \(x\) be the concentration of \(OH^-\) and \(C_6H_5NH_3^+\) at equilibrium. The concentration of \(C_6H_5NH_2\) at equilibrium is \(0.270 - x\). Since \(K_b\) is very small (\(4.3\times 10^{-10}\)), \(0.270 - x\approx0.270\). The \(K_b\) expression is \(K_b=\frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]}\), so \(4.3\times 10^{-10}=\frac{x\cdot x}{0.270}\).

Step2: Solve for \(x\) (concentration of \(OH^-\))

From \(4.3\times 10^{-10}=\frac{x^2}{0.270}\), we get \(x^2 = 4.3\times 10^{-10}\times0.270\). Then \(x=\sqrt{4.3\times 10^{-10}\times0.270}\).

$$x=\sqrt{1.161\times 10^{-10}} = 1.077\times 10^{-5}\,M$$

Step3: Calculate \(pOH\)

Using the formula \(pOH=-\log[OH^-]\), substitute \(x = [OH^-]=1.077\times 10^{-5}\).

$$pOH=-\log(1.077\times 10^{-5})= 4.97$$

Step4: Calculate \(pH\)

Since \(pH + pOH=14\), then \(pH = 14 - pOH\).

$$pH=14 - 4.97=9.03$$

Answer:

\(9.03\)