# Question #1935e

Jan 12, 2015

The balanced equation is:

${H}_{2} S {O}_{4 \left(a q\right)} + 2 N {H}_{4} O {H}_{\left(a q\right)} \rightarrow {\left(N {H}_{4}\right)}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} {O}_{\left(l\right)}$

Sulfuric acid is diprotic so is able to give up 2 H+ ions. $N {H}_{4} O H$ is not a compound that can actually exist as a discrete substance. It is better described as ammonia solution which is formed from dissolving ammonia gas in water:

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

Jan 12, 2015

You are dealing with a reaction between a strong acid and a weak base, which results in the formation of salt and water. The reaction's general equation is

${H}_{2} S {O}_{4 \left(a q\right)} + 2 N {H}_{4} O {H}_{\left(a q\right)} \to {\left(N {H}_{4}\right)}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} {O}_{\left(l\right)}$

The complete ionic equation looks like this

$2 {H}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -} + 2 N {H}_{4 \left(a q\right)}^{+} + 2 O {H}_{\left(a q\right)}^{-} \to 2 N {H}_{4 \left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -} + 2 {H}_{2} {O}_{\left(l\right)}$

The net ionic equation you'll often see is

${H}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} \to {H}_{2} {O}_{\left(l\right)}$

However, sulfuric acid, which is a strong acid, reacts with ammonium hydroxide, which is a weak base, to produce salt, ammonium sulfate, and water.

Since it is a weak base, ammonium hydroxide will dissociate only slightly in water, so another version of the net ionic equation could be:

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

However, this does not change the balanced general reaction, which still stands as it was written in the beginning of the post.

Jan 12, 2015

${H}_{2} S {O}_{4} + 2 N {H}_{4} O H \to {\left(N {H}_{4}\right)}_{2} S {O}_{4} + 2 {H}_{2} O$

You see that at the right of the equation there are 2 $N {H}_{4}$-groups, while at the left there is only one.

To even this out you may only double the whole $N {H}_{4} O H$
which works out right, because then the 2 $H$'s from the ${H}_{2} S {O}_{4}$ can combine with the 2 $O H$'s from the $N {H}_{4} O H$
to form 2 ${H}_{2} O$'s

In reality it all happens a bit differently, with ions and stuff like that, but for balancing the equation this is enough.