Answer:
[tex]\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}[/tex]
Explanation:
a) Balanced equation
The balanced chemical equation for the titration is
[tex]\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq)}[/tex]
b) pH at start
For simplicity, let's use B as the symbol for NH₃.
The equation for the equilibrium is
[tex]\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}[/tex]
(i) Calculate [OH]⁻
We can use an ICE table to do the calculation.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.150 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.150 - x x x
[tex]K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}[/tex]
Check for negligibility:
[tex]\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150[/tex]
(ii) Solve for x
[tex]\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}[/tex]
(iii) Calculate the pH
[tex]\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}[/tex]
(c) pH at equivalence point
(i) Calculate the moles of each species
[tex]\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}[/tex]
B + HI ⇌ BH⁺ + I⁻
I/mol: 3.00 3.00 0
C/mol: -3.00 -3.00 +3.00
E/mol/: 0 0 3.00
(ii) Calculate the concentration of BH⁺
At the equivalence point we have a solution containing 3.00 mmol of NH₄I
Volume = 20.00 mL + 20.00 mL = 40.00 mL
[tex]\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}[/tex]
(iii) Calculate the concentration of hydronium ion
We can use an ICE table to organize the calculations.
BH⁺+ H₂O ⇌ H₃O⁺ + B
I/mol·L⁻¹: 0.0750 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.0750 - x x x
[tex]K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ $\ll$ 0.0750}[/tex]
[tex]\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}[/tex]
(iv) Calculate the pH
[tex]\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}[/tex]
The titration curve below shows the pH at the beginning and at the equivalence point of the titration.
