# Question #f11d0

Aug 6, 2015

$M n {\left(O H\right)}_{2 \left(a q\right)} + 2 {H}_{3} {O}_{\left(a q\right)}^{+} \to 4 {H}_{2} {O}_{\left(l\right)} + M {n}_{\left(a q\right)}^{2 +}$

#### Explanation:

You're actually dealing with a neutralization reaction that takes place between the dissolved hydroxide ions and the added hydronium ions.

Manganese (II) hydroxide is actually insoluble in aqueous solution, meaning that very, very little amounts actually dissolve to give

$M n {\left(O H\right)}_{2 \left(s\right)} r i g h t \le f t h a r p \infty n s M {n}_{\textrm{\left(a q\right]}}^{2 +} + 2 O {H}_{\left(a q\right)}^{-}$

However, you can significantly increase the solubility of manganese (II) hydroxide by decreasing the solution's pH, i.e. by adding acid.

In essence, this is where the hydronium ions come from. Here's how this reaction takes place.

A solution of manganese (II) hydroxide will have a pH of about 10. When you increase the concentration of the hydronium ions by adding an acid to the solution, the manganese (II) hydroxide will actually dissolve more.

That happens because the hydronium ions react with the excess hydroxide ions that are coming from the solid.

This will force the equilibrium to shift to the right in order to compensate for this reduction in the concentration of hydroxide ions - think Le Chatelier's Principle.

As a result, more of the solid will dissolve.

So, the reaction will look like this

$M n {\left(O H\right)}_{2 \left(a q\right)} + 2 {H}_{3} {O}_{\left(a q\right)}^{+} \to 4 {H}_{2} {O}_{\left(l\right)} + M {n}_{\left(a q\right)}^{2 +}$