Question #21e81

Jul 5, 2016

Ammonia gas in water.

Explanation:

Ammonia liquor is actually another name used for aqueous ammonium hydroxide, $\text{NH"_4"OH}$, i.e. aqueous ammonia, which can be made by dissolving ammonia gas, ${\text{NH}}_{3}$, in water.

When dissolved in water, ammonia reacts with water to form ammonium cations, ${\text{NH}}_{4}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$. You can thus say that you have

${\text{NH"_ (3(g)) rightleftharpoons "NH}}_{3 \left(a q\right)}$

and

${\text{NH"_ (3(aq)) + "H"_ 2 "O"_ ((l)) rightleftharpoons "NH"_ (4(aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

This equilibrium will lie mostly to the left, as shown by the base dissociation constant of ammonia, ${K}_{b} = 1.8 \cdot {10}^{- 5}$.

As you can see, ammonium hydroxide doesn't exist in aqueous solution, which is why it's referred to here as ammonia liquid.

It's worth mentioning that adding some sodium hydroxide, $\text{NaOH}$, to an aqueous solution of ammonia will actually decrease the solubility of the gas.

Since sodium hydroxide is a soluble ionic compound, dissolving the salt in solution will increase the concentration of hydroxide anions. This will cause the equilibrium to shift even further to the left.

As a result, the concentration of aqueous ammonia will increase, which will in turn affect the solubility of the gas, i.e. some of the excess aqueous ammonia will go back to the gaseous state as shown by the first equilibrium.