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Strike441 [17]
3 years ago
13

PLEASE HELP ASAP I NEED IT

Chemistry
1 answer:
Lapatulllka [165]3 years ago
3 0

the answer is true the first one

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Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×10−11. It is used to contr
VLD [36.1K]

Answer:

1.27 x 10⁻⁴M

Explanation:

For a 2 to 1 ionization ratio, solubility in pure water can be calculated using the formula S = ∛(Ksp/27) = ∛(5.61 x 10⁻¹¹/27 = 1.27 x 10⁻⁴M.

Mg(OH)₂ ⇄ Mg⁺² + 2OH⁻ => Ksp = [Mg⁺²][OH⁻]² = (x)(2x)² = 4x³

Solve for 'x' => x = Solubility = ∛Ksp/4

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4 years ago
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