Question

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08160 M NaI with 0.05190 M AgNO3. Calculate pAg after the following volumes of AgNO3 are added: (a) at 35.10 mL (b) at Ve (volume at equilibrium) (c) at 47.10 mL

Answer 1

Answer:

(a) 13.64; (b) 8.04; (c) 2.25

Explanation:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

$K_{\text{sp}} = {\text{[Ag ^{+} ][I ^{-} ]} = 8.3\times 10^{-17}$

(a) pAg at 35.10 mL

$\text{Moles of I ^{-} } = \text{0.02500 L} \times \dfrac{\text{0.08160 mol}}{\text{1 L}} = 2.040 \times 10^{-3}\text{ mol/L }\\\text{Moles of Ag ^{+} } = \text{0.03510 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 1.822 \times 10^{-3}\text{ mol/L}$

             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

$\text{[I ^{-} ]} = \dfrac{0.218 \times 10^{-3}\text{ mol}}{\text{0.0610 L}} = 3.57 \times 10^{-3}\text{ mol/L}$

                     AgI(s) ⇌ Ag⁺(aq)    +    I⁻(aq)

E/mol·L⁻¹:                         s       3.57 × 10⁻³ + s

$K_{\text{sp}} = s(3.57 \times 10^{-3} + s) = 8.3\times 10^{-17}$

Check for negligibility:

$\dfrac{3.57 \times 10^{-3}}{8.3\times 10^{-17}} = 4.3 \times 10^{13} > 400\\\\\therefore s \ll 3.63 \times 10^{-3}\\K_{\text{sp}} = s\times 3.63 \times 10^{-3}= 8.3\times 10^{-17}\\\\s = \text{[Ag ^{+} ]} = \dfrac{8.3\times 10^{-17}}{3.63 \times 10^{-3}} =2.29 \times 10^{-14}\\\\\text{pAg} = -\log \left (2.29\times 10^{-14} \right) = \mathbf{13.64}$

(b) At equilibrium

                AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹:                    s             s

$K_{\text{sp}} = s\times s = s^{2} = 8.3\times 10^{-17}\\s = \sqrt{8.3\times 10^{-17}} = 9.11 \times 10^{-9}\\\text{pAg} = -\log \left (9.11 \times 10^{-9} \right) = \mathbf{8.04}$

(c) At 47.10 mL

$\text{Moles of Ag ^{+} } = \text{0.04710 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 2.444 \times 10^{-3}\text{ mol}$

              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

$\text{[Ag ^{+} ]} = \dfrac{0.404 \times 10^{-3}\text{ mol}}{\text{0.0721 L}} = 5.61 \times 10^{-3}\text{ mol/L}\\\text{pAg} = -\log(5.61 \times 10^{-3}) = \mathbf{2.25}$