Question

the ph of an aqueous solution of 0.413 m ethylamine (a weak base with the formula c₂h₅nh₂) is

Explanation:

Step1: Write the base - dissociation reaction and \(K_b\) expression

Ethylamine (\(C_2H_5NH_2\)) dissociates in water as \(C_2H_5NH_2 + H_2O
ightleftharpoons C_2H_5NH_3^++OH^-\). The \(K_b\) for ethylamine is \(5.6\times 10^{-4}\). Let \(x\) be the concentration of \(OH^-\) and \(C_2H_5NH_3^+\) at equilibrium, and \((0.413 - x)\) be the concentration of \(C_2H_5NH_2\) at equilibrium. Then \(K_b=\frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]}\), so \(5.6\times 10^{-4}=\frac{x\cdot x}{0.413 - x}\). Since \(K_b\) is small, \(0.413 - x\approx0.413\).

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

\(5.6\times 10^{-4}=\frac{x^{2}}{0.413}\), then \(x^{2}=5.6\times 10^{-4}\times0.413\), \(x^{2}=2.3128\times 10^{-4}\), \(x = [OH^-]=\sqrt{2.3128\times 10^{-4}}\approx0.0152\ M\).

Step3: Calculate \(pOH\)

\(pOH=-\log[OH^-]=-\log(0.0152)\approx1.82\).

Step4: Calculate \(pH\)

Since \(pH + pOH=14\), then \(pH = 14 - pOH\), \(pH=14 - 1.82 = 12.18\).

Answer:

\(12.18\)