Problem: Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH= 9.6.

FREE Expert Solution

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. 

Recall that:


pH + pOH = 14


Compute [OH-] from the given pH:


pH + pOH = 14pOH = 14 - 9.6

 pOH = 4.4


pOH = -log [OH-]pOH log= -log [OH-] log [OH-] = 10-pOH [OH-] = 10-4.4 

  [OH-] = 3.98 x 10-5

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Problem Details

Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH= 9.6.

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