Question

Consider a hypothetical salt of low solubility, "MX", where M⁺ is a metal cation and X^- is a non-metal anion.

A voltaic cell was set up with a cathode containing metal "M" immersed in a solution containing M⁺ at a concentration of 1.00 mol L^-1. The anode consisted of metal "M" immersed in a saturated solution of "MX".

The measured cell voltage was 0.186 V.

What is the K_sp of "MX" at 25°C ?

Answer 1

You have constructed a concentration cell, with one compartment containing a 1.0 M solution of M⁺ and the other containing a dilute solution of M⁺ in 1.0 mol L^-1 of MX. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation.The first step is to relate the concentration of M⁺ in the dilute solution to K_sp:

[M⁺][X^-] = K_sp

[M⁺] = K_sp / [X⁻] = K_sp / 1.0 M = K_sp

[M⁺] = K_sp

The reduction of M⁺ to M is a one-electron process and proceeds according to the following reaction:

M⁺(aq, concentrated) → M⁺(aq, dilute)

so

E_cell = E^∘_cell − (0.0591 / n) log ([M⁺]_dilute / [M⁺]_concentrated)

0.186 V = 0 V − (0.0591 / 1 ) log (K_sp / 1.0 M)

0.186 V = −0.0591 log(K_sp / 1.0)

-3.15 = log K_sp

K_sp = 7.1 x 10^-4