Answer:
Here's what I get.
Explanation:
1. Brønsted-Lowry theory
An acid is a substance that can donate a proton to another substance.
A base is a substance that can accept a proton from another substance.
2. pH of ammonia
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]
(a) Set up an ICE table.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.335 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.335 + x x x
[tex]\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}[/tex]
Check for negligibility:
[tex]\dfrac{0.335}{1.8 \times 10^{-5}} = 28 000 > 400\\\\x \ll 0.335[/tex]
(b) Solve for [OH⁻]
[tex]\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}[/tex]
(c) Calculate the pOH
[tex]\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.46 \times 10^{-3}) = 2.61[/tex]
(d) Calculate the pH
pH = 14.00 - pOH = 14.00 - 2.61 = 11.39
