Answer:
[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;
pH = 11.4; pOH = 2.6
Explanation:
The chemical equation is
[tex]\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \,\text{OH}$^{-}$[/tex]
For simplicity, let's re-write this as
[tex]\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$[/tex]
1. Calculate [OH]⁻
(a) Set up an ICE table.
B + H₂O ⇌ BH⁺ + OH⁻
0.310 0 0
-x +x +x
0.310-x x x
[tex]K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}[/tex]
Check for negligibility:
[tex]\dfrac{0.310}{1.8 \times 10^{-5}} = 17 000 > 400\\\\x \ll 0.310[/tex]
(b) Solve for [OH⁻]
[tex]\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}[/tex]
2. Calculate the pOH
[tex]\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}[/tex]
3. Calculate the pH
[tex]\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.6 = \mathbf{11.4}[/tex]
4 Calculate [H₃O⁺]
[tex]\text{H$_{3}$O$^{+}$} = 10^{-\text{pH}} = 10^{-11.4} = \mathbf{4.2 \times 10^{-12}} \textbf{ mol/L}[/tex]
