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:
Note that solids are ignored and not included in equilibrium expressions.
How would you calculate the solubility of Mn(OH) 2 in grams per liter when buffered at ph = 9.5?
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