Question

use the references to access important values if needed for this question.
calculate the hydronium ion concentration of an aqueous solution of 0.413 m triethylamine (a weak base with the formula (c₂h₅)₃n).
k_b for triethylamine is 5.20 × 10⁻⁴.
h₃o⁺ = m
submit answer retry entire group 1 more group attempts remaining

Explanation:

Step1: Write the base ionization equation and \(K_b\) expression

Triethylamine \((C_2H_5)_3N\) ionizes in water as \((C_2H_5)_3N + H_2O
ightleftharpoons (C_2H_5)_3NH^+ + OH^-\).
The \(K_b\) expression is \(K_b=\frac{[(C_2H_5)_3NH^+][OH^-]}{[(C_2H_5)_3N]}\). Let \(x = [OH^-]=[(C_2H_5)_3NH^+]\), and \([(C_2H_5)_3N]=0.413 - x\). Since \(K_b = 5.20\times10^{-4}\) is small, \(0.413 - x\approx0.413\).
So \(K_b=\frac{x\cdot x}{0.413}\), i.e., \(5.20\times 10^{-4}=\frac{x^{2}}{0.413}\).

Step2: Solve for \(x = [OH^-]\)

From \(5.20\times 10^{-4}=\frac{x^{2}}{0.413}\), we can solve for \(x\).

$$x^{2}=5.20\times 10^{-4}\times0.413$$

$$x^{2}=2.1476\times 10^{-4}$$

$$x=\sqrt{2.1476\times 10^{-4}} = 0.01466$$

So \([OH^-]=0.01466\space M\)

Step3: Use \(K_w=[H_3O^+][OH^-]\) to find \([H_3O^+]\)

Since \(K_w = 1.0\times10^{-14}\) at \(25^{\circ}C\), and \(K_w=[H_3O^+][OH^-]\)

$$ [H_3O^+]=\frac{K_w}{[OH^-]}$$

Substitute \(K_w = 1.0\times 10^{-14}\) and \([OH^-]=0.01466\space M\)

$$ [H_3O^+]=\frac{1.0\times 10^{-14}}{0.01466}$$

$$ [H_3O^+]=6.82\times 10^{-13}\space M$$

Answer:

\(6.82\times 10^{-13}\)