JEE MAIN - Chemistry (2025 - 4th April Evening Shift - No. 25)
$x \mathrm{mg}$ of $\mathrm{Mg}(\mathrm{OH})_2($ 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 : $\mathrm{Mg}(\mathrm{OH})_2$ is assumed to dissociate completely in $\mathrm{H}_2 \mathrm{O}$ ]
Explanation
To determine the mass of Mg(OH)₂ that needs to be dissolved to achieve a pH of 10.0, follow these steps:
Given:
pH = 10
Therefore, pOH = 14 - pH = 4
[OH⁻] = 10⁻⁴ M
First, calculate the number of moles of OH⁻ ions:
Number of moles of OH⁻ = 10⁻⁴
Since Mg(OH)₂ dissociates completely in water, from one mole of Mg(OH)₂, you get two moles of OH⁻:
Number of moles of Mg(OH)₂ = (10⁻⁴) / 2 = 5 × 10⁻⁵ moles
Next, calculate the mass of Mg(OH)₂:
Molar mass of Mg(OH)₂ = 58 g/mol
Convert the number of moles to mass:
Mass of Mg(OH)₂ = (5 × 10⁻⁵ moles) × (58 g/mol) = 2.9 × 10⁻³ g
Convert to milligrams: 2.9 × 10⁻³ g = 2.9 mg
Therefore, the calculated mass of Mg(OH)₂ required is approximately 2.9 mg.
Comments (0)
