Answer:
[tex]\boxed{11.13}[/tex]
Explanation:
The chemical equation is
[tex]\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ +\text{OH}$^{-}$[/tex]
For simplicity, let's rewrite this as
[tex]\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$[/tex]
1. Initial concentration of NH₃
[tex]\text{[B]} = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.050 mol}}{\text{0.500 L}} = \text{0.100 mol/L}[/tex]
2. Calculate [OH]⁻
We can use an ICE table to do the calculation.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.100 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.100 - 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{0.100 }{1.8 \times 10^{-5}} = 5600 > 400\\\\ x \ll 0.100[/tex]
3. Solve for x
[tex]\dfrac{x^{2}}{0.100} = 1.8 \times 10^{-5}\\\\x^{2} = 0.100 \times 1.8 \times 10^{-5}\\\\x^{2} = 1.80 \times 10^{-6}\\\\x = \sqrt{1.80 \times 10^{-6}}\\\\x = \text{[OH]}^{-} = 1.34 \times 10^{-3} \text{ mol/L}[/tex]
4. Calculate the pH
[tex]\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.34 \times 10^{-3}) = 2.87\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.87 = \mathbf{11.13}\\\\\text{The pH of the solution at equilibrium is } \boxed{\mathbf{11.13}}[/tex]
