Answer:
(a) 13.64; (b) 8.04; (c) 2.25
Explanation:
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
[tex]K_{\text{sp}} = {\text{[Ag$^{+}$][I$^{-}$]} = 8.3\times 10^{-17}[/tex]
(a) pAg at 35.10 mL
[tex]\text{Moles of I$^{-}$} = \text{0.02500 L} \times \dfrac{\text{0.08160 mol}}{\text{1 L}} = 2.040 \times 10^{-3}\text{ mol/L }\\\text{Moles of Ag$^{+}$} = \text{0.03510 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 1.822 \times 10^{-3}\text{ mol/L}[/tex]
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
I/mol: 1.822 × 10⁻³ 2.040 × 10⁻³
C/mol: -1.822 × 10⁻³ -1.822 × 10⁻³
E/mol: 0 0.218 × 10⁻³
We have a saturated solution of AgI containing 0.218 × 10⁻³ mol of excess I⁻.
V = 25.00 mL + 35.10 mL = 60.10 mL
[tex]\text{[I$^{-}$]} = \dfrac{0.218 \times 10^{-3}\text{ mol}}{\text{0.0610 L}} = 3.57 \times 10^{-3}\text{ mol/L}\\[/tex]
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
E/mol·L⁻¹: s 3.57 × 10⁻³ + s
[tex]K_{\text{sp}} = s(3.57 \times 10^{-3} + s) = 8.3\times 10^{-17}\\[/tex]
Check for negligibility:
[tex]\dfrac{3.57 \times 10^{-3}}{8.3\times 10^{-17}} = 4.3 \times 10^{13} > 400\\\\\therefore s \ll 3.63 \times 10^{-3}\\K_{\text{sp}} = s\times 3.63 \times 10^{-3}= 8.3\times 10^{-17}\\\\s = \text{[Ag$^{+}$]} = \dfrac{8.3\times 10^{-17}}{3.63 \times 10^{-3}} =2.29 \times 10^{-14}\\\\\text{pAg} = -\log \left (2.29\times 10^{-14} \right) = \mathbf{13.64}[/tex]
(b) At equilibrium
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
E/mol·L⁻¹: s s
[tex]K_{\text{sp}} = s\times s = s^{2} = 8.3\times 10^{-17}\\s = \sqrt{8.3\times 10^{-17}} = 9.11 \times 10^{-9}\\\text{pAg} = -\log \left (9.11 \times 10^{-9} \right) = \mathbf{8.04}[/tex]
(c) At 47.10 mL
[tex]\text{Moles of Ag$^{+}$} = \text{0.04710 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 2.444 \times 10^{-3}\text{ mol}[/tex]
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
I/mol: 2.444 × 10⁻³ 2.040 × 10⁻³
C/mol: -2.040 × 10⁻³ -2.040 × 10⁻³
E/mol: 0.404 × 10⁻³ 0
V = 25.00 mL + 47.10 mL = 72.10 mL
[tex]\text{[Ag$^{+}$]} = \dfrac{0.404 \times 10^{-3}\text{ mol}}{\text{0.0721 L}} = 5.61 \times 10^{-3}\text{ mol/L}\\\text{pAg} = -\log(5.61 \times 10^{-3}) = \mathbf{2.25}[/tex]
