We’re being asked to determine the solubility of Mn(OH)2 in a solution buffered at pH = 9.6.
Since the Mn(OH)2 is an ionic compound, it forms ions when dissociating in water.
The dissociation of Mn(OH)2 in water is as follows:
The hydroxide ion, OH–, has a charge of –1. Mn then has a charge of +2:
Mn(OH)2(s) ⇌ Mn2+(aq) + 2 OH–(aq)
Since the solution is buffered, we can calculate an initial OH– concentration from the given pH.
Compute [OH-] from the given pH:
pOH = 4.4
[OH-] = 3.98 x 10-5
Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH= 9.6.
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