Answer:
[tex]\large \boxed{\text{3.75}}[/tex]
Explanation:
The equation for the equilibrium is
[tex]\rm (CH_3)_{3}N + H_2O \, \rightleftharpoons \,(CH_3)_{3}NH^{+} + \text{OH}^{-}\\\text{For ease of typing, let's rewrite this equation as}\\\rm B + H_2O \, \rightleftharpoons \, BH^{+} + OH^{-}; K_{\text{b}} = 6.5 \times 10^{-5}[/tex]
Data:
Vb = 46 mL; [B] = 0.11 mol·L⁻¹
Va = 17 mL; [BH⁺] = 0.13 mol·L⁻¹
Kb = 6.5 × 10⁻⁵
Calculations:
1. Calculate the moles of B and BH⁺
(a) Moles of B
[tex]\text{moles of B} = \text{46 mL B} \times \dfrac{\text{0.11 mmoL B}}{\text{1 mL B}} = \text{5.06 mmol B}[/tex]
(b) Moles of BH⁺
[tex]\text{moles of BH}^{+} = \text{17 mL BH}^{+} \times \dfrac{\text{0.13 mmoL BH}^{+}}{\text{1 mL BH}^{+}} = \text{2.21 mmol BH}^{+}[/tex]
2. Calculate the pOH
The Henderson-Hasselbalch equation for a basic buffer is
[tex]\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}[/tex]
(a) Calculate pKb
[tex]\text{pK}_{\text{b}} = -\log(6.5 \times 10^{-5}) = 4.19[/tex]
(b) Calculate the pOH
The components are in the same solution, so the ratio of the moles is the same as the ratio of the concentrations
[tex]\text{pOH} = 4.19 + \log \dfrac{2.21}{5.06} = 4.19 + \log 0.437 = 4.19 - 0.437 = 3.75\\\\\text{The pOH of the solution is $\large \boxed{\textbf{3.75}}$}[/tex]
