Answer:

So for M2X3 we have, s =[ ksp/108 ]^1/5

It gives us S which can be used for the 2nd rxn,

i.e.for M2X , we have ,

S= [ksp/4]^1/3

OR

ksp=4S^3

It will give u Ksp=3×10^-12

Hope this will help you!

Explanation:

Explanation:

M2X3(s) ===> 2M(aq) + 3X(aq)

Ksp = [M]^2[X]^3 = 2.2 x 10^-10

Let [M] = x. Then [X] = (3/2)x

(x)^2(3x/2)^3 = (27/8)x^3 = 2.2 x 10^-10

x^3 = 6.5 x 10^-11, so x = 4.0 x 10^-4 = [M]

According to the balanced equation, [M] in solution = (1/2)[M2X3] that dissolves molar solubility). So molar solubility of M2X3 = (1/2)(4.0x10^-4) = 2.0 x 10^-4

M2X(s) ===> 2M(aq) + X(aq)

Ksp = [M]^2[X]

This time, let [X] = y. Then [M] = 2y. According to the balanced equatiion, [X] in solution = [M2X] that dissolved (molar solubility). According to the problem, y = 2x, so y = 4.0 x 10^-4

Ksp = (2y)^2(y) = 4y^3 = (4)(4.0x10^-4)^3 = 2.6 x 10^-10 to two significant figures..

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