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)
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)
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Correct Answer: 3
Approach Solution - 1
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).
Approach Solution -2
- 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.
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