Question

x mg of Mg(OH)$_2$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ____ mg. (Nearest integer) (Given: Mg(OH)$_2$ is assumed to dissociate completely in H$_2$O)

Hint:

When calculating the mass of a compound from its concentration, remember that dissociation of salts like Mg(OH)$_2$ can provide multiple moles of ions for each mole of the compound. The number of moles of the compound is related to the ion concentration through the dissociation stoichiometry.

Answer 1

Correct Answer: 3

Solution:

To find the amount of Mg(OH)₂ needed to produce a pH of 10.0, we first determine the corresponding hydroxide ion concentration [OH⁻].

1. Calculate pOH from the given pH:

pOH = 14.0 - 10.0 = 4.0

2. Calculate [OH⁻]:

[OH⁻] = 10^−pOH = 10^−4.0 = 1.0 × 10⁻⁴ M

3. Because Mg(OH)₂ dissociates completely in water as follows:

Mg(OH)₂(s) → Mg²⁺(aq) + 2OH⁻(aq)

The concentration of OH⁻ is twice the molarity of Mg(OH)₂ in solution:

Let the concentration of Mg(OH)₂ = [Mg(OH)₂].

Thus, 2[Mg(OH)₂] = [OH⁻]

[Mg(OH)₂] = [OH⁻] / 2 = 0.5 × 10⁻⁴ M

4. Calculate the amount (in moles) of Mg(OH)₂ needed for 1.0 L solution:

Moles of Mg(OH)₂ = [Mg(OH)₂] × Volume (L) = 0.5 × 10⁻⁴ mol/L × 1.0 L = 0.5 × 10⁻⁴ mol

5. Convert moles to grams:

Molar mass of Mg(OH)₂ = 58 g/mol

Mass = moles × molar mass = 0.5 × 10⁻⁴ mol × 58 g/mol = 2.90 × 10⁻³ g

6. Convert grams to milligrams:

Mass = 2.90 × 10⁻³ g × 1000 mg/g = 2.90 mg

Rounded to the nearest integer, x = 3 mg.

Thus, the value of x is 3 mg, which fits within the provided range (3,3).

Answer 2

Given:
- pH = 10.0
- pOH = 4.0
- [OH$^-$] = $10^{-4}$ M
The concentration of OH$^-$ is $10^{-4}$ M, and since Mg(OH)$_2$ dissociates completely in water, the number of moles of OH$^-$ will be equal to the number of moles of Mg(OH)$_2$. Step-by-step solution:
1. Number of moles of OH$^-$ = $10^{-4}$ moles (from the concentration of OH$^-$).
2. Number of moles of Mg(OH)$_2$ = $\frac{10^{-4}}{2}$ = $5 \times 10^{-5}$ moles, because one mole of Mg(OH)$_2$ gives two moles of OH$^-$.
3. The mass of Mg(OH)$_2$ is then calculated as: \[ \text{mass of Mg(OH)$_2$} = 5 \times 10^{-5} \times 58 \times 10^3 \, \text{mg} \] \[ = 2.9 \, \text{mg} \]
Thus, the value of x is 2.9 mg, which is approximately 3 mg.