The solubility product of Mg(OH)2 is 1.2 x 10^-11. What minimum OH- concentration must be attained (for example, by adding NaOH) to decrease the Mg concentration in a solution of Mg(NO3)2 to less than 1.0 x 10^-10 M (molar)?

Mg(OH)2 ==> Mg^+2 + 2OH^-

Ksp = (Mg^+2)(OH^-)^2 = 1.2 x 10^-11

You know (Mg^+) and Ksp, solve for (OH^-).
It's that simple.

0.109

Well, in order to decrease the Mg concentration in a solution of Mg(NO3)2, we need to increase the OH- concentration. It's like calling in the cavalry to save the day, but in this case, we're calling in our trusty friend NaOH!

Now, let's do some math, but don't worry, I won't bore you with all the gory details. We can use the solubility product (Ksp) expression to get an idea of what's going on here.

The balanced equation for the dissolution of Mg(OH)2 is:

Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)

According to the solubility product expression, Ksp = [Mg2+][OH-]^2.

We want to decrease the Mg concentration to less than 1.0 x 10^-10 M, so let's call that [Mg2+] ≤ 1.0 x 10^-10 M.

Now, rearranging the Ksp expression, we can solve for [OH-]:

[OH-]^2 = Ksp / [Mg2+]

[OH-]^2 = (1.2 x 10^-11) / (1.0 x 10^-10)

Taking the square root of both sides:

[OH-] = √((1.2 x 10^-11) / (1.0 x 10^-10))

So, the minimum OH- concentration needed would be approximately equal to the square root of 0.12, divided by 1. Let me use my handy-dandy calculator here...

*Takes out a calculator and starts pressing buttons*

Alright, after some intense calculations, the minimum OH- concentration needed would be...drumroll please...approximately 0.1095 M (molar)!

So, we would need to add enough NaOH to achieve an OH- concentration of at least 0.1095 M to decrease the Mg concentration in our solution of Mg(NO3)2 to less than 1.0 x 10^-10 M.

Remember, though, this answer is no joke, unlike my comedy.

To solve this problem, we need to use the concept of the solubility product (Ksp) and the equation for the dissolution of Mg(OH)2 in water:

Mg(OH)2 ⇌ Mg²⁺ + 2OH⁻

Given:
Solubility product (Ksp) of Mg(OH)2 = 1.2 x 10^-11
Desired Mg²⁺ concentration = 1.0 x 10^-10 M

The Ksp expression for the dissolution of Mg(OH)2 is given by:

Ksp = [Mg²⁺][OH⁻]²

Using the stoichiometry of the balanced equation, we know that [Mg²⁺] = [OH⁻], so we can substitute [OH⁻] for [Mg²⁺] in the Ksp expression:

Ksp = (OH⁻)(OH⁻)²
1.2 x 10^-11 = OH⁻³

Rearranging the equation to solve for [OH⁻], we have:

OH⁻ = (1.2 x 10^-11)^(1/3)
OH⁻ ≈ 1.78 x 10^-4 M

Therefore, the minimum OH⁻ concentration (or the minimum concentration of NaOH) needed to decrease the Mg concentration in a solution of Mg(NO3)2 to less than 1.0 x 10^-10 M is approximately 1.78 x 10^-4 M.

To answer this question, we need to use the concept of solubility product (Ksp) and the balanced chemical equation for the dissolution of Mg(OH)2.

The balanced chemical equation for the dissolution of Mg(OH)2 is:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

The solubility product equation is:
Ksp = [Mg2+][OH-]^2

Given the solubility product (Ksp) for Mg(OH)2 as 1.2 x 10^-11, we can substitute the given values into the equation:

1.2 x 10^-11 = [Mg2+][OH-]^2

Since we want to decrease the Mg concentration in a solution of Mg(NO3)2 to less than 1.0 x 10^-10 M, the Mg2+ concentration must be less than 1.0 x 10^-10 M.

Let's assume the OH- concentration as x M. Substituting these values into the solubility product equation, we get:

1.2 x 10^-11 = (1.0 x 10^-10)(x)^2

Now, we can solve for x:

x^2 = (1.2 x 10^-11) / (1.0 x 10^-10)
x^2 = 0.12
x ≈ √0.12
x ≈ 0.346

Therefore, the minimum OH- concentration that must be attained is approximately 0.346 M.