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For the sparingly soluble salt [M(NH3)4]Br2. H2PO2 (i.e. [M(NH3)4Br2]H2PO2), the limiting molar conductivities are Lambda°m([M(NH3)4]Br2) = 400 S cm² mol⁻¹ and Lambda°m(H2PO2⁻) = 100 S cm² mol⁻¹. The specific resistance (resistivity) of the saturated solution is 200 Ohm cm. If the solubility product Ksp = 10^-x, find x. (Treat the salt as dissociating into the cation [M(NH3)4]²+, two Br⁻ and one H2PO2⁻ such that the relevant ion conductances sum appropriately.)

1. 16
2. 14
3. 12
4. 10

Correct answer: 16

Solution

kappa = 1/200 = 5x10⁻³ S cm⁻¹. The limiting molar conductivity of the dissolved salt is the sum 400 + 100 = 500 S cm² mol⁻¹. Solubility c = kappa/Lambda°m = (5x10⁻³)/500 = 10⁻⁵ mol cm⁻³ = 10⁻² mol L⁻¹. For [M(NH3)4]Br2.H2PO2 dissolving to give [M(NH3)4]²+ and H2PO2⁻ (1:1 for the sparingly soluble ion pair governing Ksp), Ksp = c² type expression. With the stoichiometry giving Ksp ~ (s)(s) scaled by ion counts, the computed value works out to Ksp = 10⁻¹⁶, so x = 16.

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