The cell $Ag ∣ AgX (s) ∣ KX (0.1 M) ∣∣ AgNO X_{3} (0.1 M) ∣ Ag$ , where AgX is a sparingly soluble silver salt, has EMF $0.480 V$ at 298 K with the left electrode as anode. The value of $p K_{s p} = - lo g K_{s p}$ of AgX is (one decimal; take $2.303 RT / F = 0.0591 V$ ).

Numerical answer type, tolerance 0.1.

Answer: 10.1

On the left,

[Ag X^{+}] = K_{s p} / [X X^{-}] = K_{s p} /0.1

. The cell is a silver concentration cell:

E = 0.0591 lo g \frac{0.1}{K _{s p} /0.1} = 0.0591 lo g \frac{0.01}{K _{s p}}

lo g \frac{0.01}{K _{s p}} = 0.480/0.0591 = 8.12

, so

lo g K_{s p} = - 2 - 8.12 = - 10.12

and

p K_{s p} = 10.1

(

K_{s p} \approx 7.6 \times 1 0^{- 11}