Question #6dd6f

May 1, 2015

A reversible reaction is said to contain two acids and two bases, that's why the term conjugate pairs was coined.

That means that the forward reaction will have an acid and a base, the products being the conjugate base of the acid and the conjugate acid of the base, which in turn will act as base and acid for the reverse reaction.

$N {H}_{4}^{+} + O {H}^{-} r i g h t \le f t h a r p \infty n s N {H}_{3} + {H}_{2} O$

The forward reaction is

${\underbrace{N {H}_{4}^{+}}}_{\textcolor{b l u e}{\text{acid")) + overbrace(OH^(-))^(color(green)("base")) -> underbrace(NH_3)_(color(blue)("conjugate base")) + overbrace(H_2O)^(color(green)("conjugate acid}}}$

The ammonium ion, $N {H}_{4}^{+}$, will donate a proton, hence act as an acid, to the hydroxide ion, $O {H}^{-}$, which will be the base.

Ammonia will thus be the conjugate base of ammonium, and water will be the conjugate acid of hydroxide.

The reverse reaction is

${\underbrace{N {H}_{3}}}_{\textcolor{b l u e}{\text{base")) + overbrace(H_2O)^(color(green)("acid")) -> underbrace(NH_4^(+))_(color(blue)("conjugate acid")) + overbrace(OH^(-))^(color(green)("conjugate base}}}$

Now it's the other way around. Water will donate a proton, hence act as an acid, to ammonia, which will be the base.

As a result, the ammonium ion will be the conjugate acid of ammonia, and the hydroxide ion will be the conjugate base of water.

So, if you take into account the way the equilibrium reaction was given to you, you'll get

$N {H}_{4}^{+} - N {H}_{3}$ $\to$ acid - conjugate base

$O {H}^{-} - {H}_{2} O$ $\to$ base - conjugate acid