Mass of Sr(OH)₂ should be added to a 400 ml volumetric flask is 24.34 gm.
Let the mass of Sr(OH)₂ = P g
Molar mass of Sr(OH)₂ = 121.63 g/mol
Mole of Sr(OH)₂ = P gm/ 121.63 g mol⁻¹ = 0.00822 × P mol
[Sr(OH)₂] = (0.00822 × P mol) / 0.400 L = (0.2055 × P) M
[OH⁻] = 2 × [Sr(OH)₂] = 0.0411 P M
pH = 14
pOH = 14 - 14 = 0
[OH⁻] = 10⁰ = 1
∴ 0.0411 P M = 1
⇒ P = 24.34 gm
Hence, mass of Sr(OH)₂ should be added to a 400 ml volumetric flask is 24.34 gm.
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