What is the molar solubility of MgF₂ in a 0.36 M Mg(NO₃)₂ solution? For MgF₂,K_sp = 8.4 × 10^-8.

Question

Quizplus · Accepted Answer

First, write the balanced equation for the dissolution of MgFS1U1B12S1U1B0:

MgFS1U1B12S1U1B0(s) ⇌ MgS1U1B12S1U1B0+(aq) + FS1U1B1-(aq)

The equilibrium expression for this reaction can be written as:

KS1U1B1spS1U1B0 = [MgS1U1B12S1U1B0+][FS1U1B1-]

We are given that KS1U1B1spS1U1B0 = 8.4 × 10S1U1P1-8S1S1P0. Let x be the molar solubility of MgFS1U1B12S1U1B0 in the solution. Then, at equilibrium:

[MgS1U1B12S1U1B0+] = x
[FS1U1B1-] = x

We can then substitute these values into the equilibrium expression:

KS1U1B1spS1U1B0 = x^2

Solving for x, we get:

x = √(KS1U1B1spS1U1B0) = √(8.4 × 10S1U1P1-8S1S1P0) = 2.4 × 10S1U1P1-4S1S1P0 M

Therefore, the answer is B, 2.4 × 10S1U1P1-4S1S1P0 M.

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