Question

4. You added 0.05 mL of 0.10 M AgNO, to 4.0 mL of 2.0 M NaCl, and formed a saturated solution of AgCl (s): AgCl (s) 7 Ag+ (aq)+CF (aq) K=Ksp Using this saturated solution as the test solution you set up the following cell AgAg (test solution after reaction)||Ag (1.0M) Ag and measured Ecell = 0.61 V. (a) Using the measured value of Ecell and Eq (15), calculate the equilibrium (Ag l (i.e. of the test solution after reaction). (1 mark) (b) Set up a reaction (ICE) table and calculate the equilibrium concentration of CI". (1 mark) (c) Calculate Ksp. (0.5 mark)

Answer 1

The cell reaction is,

Ag | Ag⁺ (test solution after reaction) ||Ag⁺ (1.0 M)| Ag

calculate the Q of the reaction as follows:

Q = [Ag⁺] / [Ag(s)]

= [Ag⁺] (for solids 1)

The formula for E_cell as follows :

E_cell = E^o_cell - (0.0592V /n) log Q

calculate the [Ag⁺]

0.61 V = 0- (0.0592/1) log[Ag⁺]

log[Ag⁺] = -0.61/ 0.0592

[Ag⁺] =4.97 X 10^-11

(b)

AgCl (s) <=====> Ag⁺(aq) + cl^-(aq)

construct the ICE table for the above reaction

I M - -

C -x +x +x

E M-x x x

therefore [Ag⁺] = [Cl^-] = x

[Cl^-] = 4.97 X 10^-11

(c)

thus K_sp = [Ag⁺] [Cl^-]

= (4.97 X 10^-11)²

K_sp = 24.7 X 10^-22

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