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Sagot :
SOLUTION:
Step 1: Summarize the concentrations using the ICE table.
[tex]\begin{array}{lccccccc} & \text{CH}_3\text{NH}_2(aq) & + & \text{H}_2\text{O}(l) & \rightleftharpoons & \text{CH}_3\text{NH}_3^+(aq) & + & \text{OH}^-(aq) \\ \text{Initial} \: (M): & 0.072 & & & & 0 & & 0 \\ \text{Change} \: (M): & -x & & & & +x & & +x \\ \hline \text{Equilibrium} \: (M): & 0.072-x & & & & x & & x \\ \end{array}[/tex]
Step 2: Write the Kb expression.
[tex]K_{\text{b}} = \dfrac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]}[/tex]
Step 3: Substitute the given values to the Kb expression.
[tex]\begin{aligned} 4.4 \times 10^{-4} & = \frac{(x)(x)}{0.10 - x} \\ 4.4 \times 10^{-4} & = \frac{x^2}{0.072 - x} \end{aligned}[/tex]
Step 4: Apply approximation method to solve for the hydroxide ion concentration at equilibrium.
Note: 0.072 - x ≈ 0.072
[tex]\begin{aligned} 4.4 \times 10^{-4} & = \frac{x^2}{0.072} \\ x^2 & = (4.4 \times 10^{-4})(0.072) \\ x^2 & = 3.168 \times 10^{-5} \\ x & = \sqrt{3.168 \times 10^{-5}} \\ x & = 5.6285 \times 10^{-3} \: M\end{aligned}[/tex]
Testing the approximation
[tex]\begin{aligned} \text{value} & = \frac{[\text{OH}^-]}{[\text{CH}_3\text{NH}_2]_{\text{initial}}} \times 100\% \\ & = \frac{5.6285 \times 10^{-3} \: M}{0.072 \: M} \times 100\% \\ & = 7.8\% \end{aligned}[/tex]
Since the value is greater than 5%, the approximation is invalid. We must use quadratic formula. Therefore,
[tex]\begin{aligned} 4.4 \times 10^{-4} & = \frac{x^2}{0.072 - x} \\ x^2 & = (4.4 \times 10^{-4})(0.072 - x) \\ x^2 & = 3.168 \times 10^{-5} - 4.4 \times 10^{-4}x \\ x^2 + 4.4 \times 10^{-4}x - 3.168 \times 10^{-5} & = 0 \end{aligned}[/tex]
[tex]\begin{aligned} a = 1 \quad b = 4.4 \times 10^{-4} \quad c = -3.168 \times 10^{-5} \end{aligned}[/tex]
[tex]\begin{aligned} x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ x & = \frac{-4.4 \times 10^{-4} \pm \sqrt{(4.4 \times 10^{-4})^2 - 4(1)(-3.168 \times 10^{-5})}}{2(1)} \\ x & = 5.4 \times 10^{-3} \: M \: (accepted) \\ x & = -5.85 \times 10^{-3} \: M \: (rejected) \end{aligned}[/tex]
It follows that the value of OH⁻ concentration is
[tex][\text{OH}^-] = x = 5.4 \times 10^{-3} \: M[/tex]
Step 5: Calculate the pOH and pH of the solution.
• For pOH
[tex]\begin{aligned} \text{pOH} & = -\text{log} \: [\text{OH}^-] \\ & = -\text{log} \: (5.4 \times 10^{-3}) \\ & = \boxed{2.27} \end{aligned}[/tex]
• For pH
[tex]\begin{aligned} \text{pH} & = 14 - \text{pOH} \\ & = 14 - 2.27 \\ & = \boxed{11.73} \end{aligned}[/tex]
Hence, the pOH and pH of a 0.072-M methylamine solution are 2.27 and 11.73, respectively.
[tex]\\[/tex]
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