Question: if 10.0 mg of Mg(OH)2 is placed in 100 ml of pure water, will all of the sample dissolve before equilibrium can become established, or will some sample remain undissolved?

Question

Question: if 10.0 mg of Mg(OH)2 is placed in 100 ml of pure water, will all of the sample dissolve before equilibrium can become established, or will some sample remain undissolved?

i got the Molarity of Mg(OH)2 to be .00172 M. Where does that exactly go into the ice table. the equation is Mg(OH)2 -> Mg + 2OH. How do i set up the ice table. Will the sample remain undisolved?

Answer 1

M = moles/L

So if the M = 0.00172, then the moles in a liter will be 0.00172. Since the problem tells you it is in 100 mL, then 1/10 that will be the moles that will dissolve in 0.100 L. To convert that to grams, moles = g/molar mass, then convert grams to mg.

Answer 2

Don't you want to set up an ice table and find X so you can determine if the sample dissolves or if the sample remaines undissolved?

Answer 3

I don't think an ICE table is needed. You have the solubility at 0.00172 M so 1.72 x 10^-3 M is the # moles in 1 L, then 1.72 x 10^-4 is the # moles in 0.100 L, and grams = moles x molar mass.

Answer 4

To determine whether the sample will remain undissolved or if all of it will dissolve, you can use the solubility product constant (Ksp) of magnesium hydroxide (Mg(OH)2) and compare it to the ion product (IP).

First, let's set up the balanced chemical equation:
Mg(OH)2 → Mg2+ + 2OH-

Now, let's consider the initial conditions. We have 10.0 mg of Mg(OH)2, which is equal to 0.010 g. To convert this to moles, we need to divide by the molar mass of Mg(OH)2, which is approximately 58.33 g/mol.

0.010 g / 58.33 g/mol = 0.000172 mol

Next, let's calculate the concentration (Molarity) of Mg(OH)2. We know that we have 100 mL of pure water, which is equivalent to 0.100 L. Using the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

Molarity = 0.000172 mol / 0.100 L = 0.00172 M

To set up the ICE table, we know that the initial concentration of Mg2+ and OH- is zero since Mg(OH)2 dissociates completely.

┌───────────────────┬───────────────┬────────┬───────────────┐
│ Species │ Initial (M) │ Change │ Equilibrium (M) │
├───────────────────┼───────────────┼────────┼───────────────┤
│ Mg(OH)2 │ 0.00172 │ - │ - │
│ Mg2+ │ 0 │ + │ ? │
│ OH- │ 0 │ + │ ? │
└───────────────────┴───────────────┴────────┴───────────────┘

The change row for Mg2+ and OH- would be +x because they are both products with a stoichiometric coefficient of 1.

Now, we need to consider the solubility product constant (Ksp) for Mg(OH)2, which is approximately 1.8 × 10^-11 at 25°C. The Ksp expression for Mg(OH)2 is:

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

Let's substitute the equilibrium concentrations into the Ksp expression:

1.8 × 10^-11 = (x)(2x)^2
1.8 × 10^-11 = 4x^3

Now, we can solve for x (which represents the concentration of Mg2+ and OH-) by taking the cube root of both sides:

x = (1.8 × 10^-11 / 4)^(1/3)

Calculating this expression will give you the concentration of Mg2+ and OH- at equilibrium, which will allow you to determine if all of the sample will dissolve before equilibrium or if some will remain undissolved.