Answer:
(a) 13.64; (b) 8.04; (c) 2.25
Explanation:
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
(a) pAg at 35.10 mL
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
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
E/mol·L⁻¹: s 3.57 × 10⁻³ + s
Check for negligibility:
(b) At equilibrium
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
E/mol·L⁻¹: s s
(c) At 47.10 mL
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