Problem: How would you calculate the solubility of Mn(OH) 2 in grams per liter when buffered at ph = 9.5?

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We have to calculate the solubility of Mn(OH)2 in a solution that is buffered at a pH of 9.5.

Mn(OH)2 is a sparingly soluble substance and dissociates reversibly in its aqueous solution.

Mn(OH)2(s)  Mn2+(aq) + 2 OH-(aq)

The Ksp for this equilibrium can be written as:

Ksp = productsreactants = Mn2+OH-2

Note that solids are ignored and not included in equilibrium expressions.

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How would you calculate the solubility of Mn(OH) 2 in grams per liter when buffered at ph = 9.5?

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