Question

calculate the percent ionization of a 0.536 m solution of phenol (a weak acid). (assume that ( k _ { a } ( c _ { 6 } h _ { 5 } o h ) = 1.0 \times 10 ^ { - 10 } ))
% ionization =
%

Explanation:

Step1: Set up the ionization equation and equilibrium expression

Phenol (\(C_6H_5OH\)) ionizes as \(C_6H_5OH
ightleftharpoons C_6H_5O^-+H^+\). Let \(x\) be the concentration of \(H^+\) (and \(C_6H_5O^-\)) at equilibrium. The initial concentration of \(C_6H_5OH\) is \(c = 0.536M\). The acid - dissociation constant \(K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}\), and \(K_a = 1.0\times10^{-10}\). At equilibrium, \([C_6H_5OH]=c - x\approx c\) (since \(K_a\) is very small, \(x\ll c\)). So \(K_a=\frac{x\cdot x}{c}\).

Step2: Solve for \(x\)

Substitute the values into the \(K_a\) formula: \(1.0\times 10^{-10}=\frac{x^{2}}{0.536}\). Then \(x^{2}=1.0\times 10^{-10}\times0.536\), and \(x=\sqrt{1.0\times 10^{-10}\times0.536}\approx7.32\times 10^{-6}M\).

Step3: Calculate the percent ionization

The formula for percent ionization is \(\text{Percent Ionization}=\frac{[H^+]_{equilibrium}}{[HA]_{initial}}\times100\%\). Substitute \(x = [H^+]_{equilibrium}\) and \(c=[HA]_{initial}\): \(\text{Percent Ionization}=\frac{7.32\times 10^{-6}}{0.536}\times 100\%\).

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Answer:

\(1.37\times 10^{-3}\%\)