A saturated solution of Mg(OH)2 is prepared having a large excess of Mg(OH)2. Sn(NO3)2 is added to the solution. Ksp = 1.9 10-11 for Mg(OH)2 and Ksp = 6.0 10-26 for Sn(OH)2.

And your question is?

To determine what happens when Sn(NO3)2 is added to the saturated solution of Mg(OH)2, we need to compare the solubility product constants (Ksp) of Mg(OH)2 and Sn(OH)2.

1. Start by writing down the balanced chemical equation representing the dissolution of Mg(OH)2 in water:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

2. According to the information given, the solution of Mg(OH)2 is saturated, which means the concentrations of Mg2+ and OH- ions are already at their maximum values.

3. Now, let's determine the equilibrium concentrations of Mg2+ and OH- ions using the Ksp expression for Mg(OH)2:
Ksp = [Mg2+][OH-]^2
1.9 * 10^-11 = [Mg2+][OH-]^2

4. Since the solution is saturated, we'll assume the concentration of Mg2+ is equal to the solubility of Mg(OH)2 (s), and the concentration of OH- is twice the solubility:
[Mg2+] = s
[OH-] = 2s

5. Substitute these values into the Ksp expression:
1.9 * 10^-11 = (s)(2s)^2
1.9 * 10^-11 = 4s^3

6. Solve the equation and find the value of s, which represents the solubility of Mg(OH)2:
s = (1.9 * 10^-11 / 4)^(1/3)
s ≈ 1.27 * 10^-4 M

Now that we know the solubility of Mg(OH)2 is approximately 1.27 * 10^-4 M, we can use this information to analyze the addition of Sn(NO3)2 to the solution.

7. When Sn(NO3)2 is added, the Sn2+ ions (from Sn(NO3)2) will react with OH- ions (from Mg(OH)2) to form a precipitate of Sn(OH)2.

The equation for the reaction is:
Sn2+(aq) + 2OH-(aq) ⇌ Sn(OH)2(s)

8. To determine whether a precipitate will form, we need to compare the value of the reaction quotient (Q) to the solubility product constant (Ksp) for Sn(OH)2.

The expression for Q is:
Q = [Sn2+][OH-]^2

9. The concentration of Sn2+ ions will depend on the amount of Sn(NO3)2 added. If we assume complete dissociation of Sn(NO3)2, then the concentration of Sn2+ ions will be equal to the initial concentration of Sn(NO3)2.

10. Substitute the values and compare Q to Ksp to determine whether a precipitate will form. If Q > Ksp, a precipitate of Sn(OH)2 will form. If Q < Ksp, no precipitate will form.

Remember, the Ksp value for Sn(OH)2 is 6.0 × 10^-26.

By following these steps, you can determine whether a precipitate of Sn(OH)2 will form when Sn(NO3)2 is added to the saturated solution of Mg(OH)2.