# How can I calculate Ka of NH_4^+?

Dec 6, 2014

$N {H}_{4}^{+}$ is a weak acid, which would give us an idea if not about the actual value, then at least about its dissociation constant's dimension. Knowing that the ammonium ion ($N {H}_{4}^{+}$) is a weak acid, we should expect ${K}_{a}$ to be less than $1$.

That is, a very small quantity of products ( ammonia and hydronium cation ) will be formed, the reaction favoring the reactans to a large degree.

$N {H}_{4}^{+} \left(a q\right) + {H}_{2} O \left(l\right) \iff N {H}_{3} \left(a q\right) + {H}_{3}^{+} O \left(a q\right)$

$\left[{K}_{a}\right]$ is defined as

${K}_{a} = \frac{\left[N {H}_{3}\right] \cdot \left[{H}_{3}^{+} O\right]}{\left[N {H}_{4}^{+}\right]}$, and can be seen as a measure of an acid's strenght in solution. The best way to go about actually solving a problem is by using the ICE method (I've linked a wikipedia article on this).

$N {H}_{4}^{+}$ has an actual dissociation constant (I've provided an acid strenght table) of ${K}_{a} = 5.8 \cdot {10}^{- 10}$.