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

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.