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 H₂ O ]

Solution

Chemistry ma’am · 4 steps

Go from pH to [OH⁻], then halve it, because each Mg(OH)₂ releases two hydroxide ions.

1. pH to pOH. At 298 K, pH + pOH = 14. So pOH = 14 - 10 = 4, and [OH⁻] = 10⁻⁴ mol L⁻¹.

   Mg(OH)₂ → Mg²⁺ + 2OH⁻

2. Moles of base. One mole of Mg(OH)₂ gives two moles of OH⁻. So the base needed is 10⁻⁴/2 = 5 × 10⁻⁵ mol in 1.0 L.
3. Convert to mass. Mass = 5 × 10⁻⁵ × 58 = 2.9 × 10⁻³ g = 2.9 mg.
4. So. x = 2.9 mg, which rounds to 3.

The common slip: 6. Forgetting that each Mg(OH)₂ gives two OH⁻ ions uses 10⁻⁴ mol of base and gives 5.8 mg, which rounds to 6, not 3.

The idea to keep. For a base like M(OH)ₙ, the base concentration is [OH⁻] divided by n. Always check how many hydroxide ions one formula unit releases.

Answer from the NTA final answer key. Solution by PrepWiser.