Question

use the references to access important values if needed for this question. the ph of an aqueous solution of 0.413 m phenol (a weak acid), (kₐ (c₆h₅oh) = 1.00 × 10⁻¹⁰) is

Explanation:

Step1: Write the dissociation equation and \(K_a\) expression

Phenol (\(C_6H_5OH\)) dissociates as \(C_6H_5OH
ightleftharpoons C_6H_5O^-+H^+\). The \(K_a\) expression is \(K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}\). Let \(x = [H^+]=[C_6H_5O^-]\), and \([C_6H_5OH]=0.413 - x\). Since \(K_a = 1.00\times10^{-10}\) is very small, \(0.413 - x\approx0.413\).

Step2: Substitute into \(K_a\) formula

Substitute into \(K_a=\frac{x\times x}{0.413}\), so \(1.00\times 10^{-10}=\frac{x^{2}}{0.413}\).

Step3: Solve for \(x\)

Cross - multiply: \(x^{2}=1.00\times 10^{-10}\times0.413\), \(x^{2}=4.13\times 10^{-11}\), \(x=\sqrt{4.13\times 10^{-11}}\). Using a calculator, \(x = [H^+]=6.42\times 10^{-6}\space M\).

Step4: Calculate pH

Use the formula \(pH=-\log[H^+]\). So \(pH =-\log(6.42\times 10^{-6})\). Using the logarithm property \(\log(ab)=\log a+\log b\), \(pH=-( \log6.42+\log(10^{-6}))\). Since \(\log(10^{-6})=- 6\) and \(\log6.42\approx0.807\), \(pH=-(0.807 - 6)=5.193\approx5.19\).

Answer:

\(5.19\)