# A 0.10 M NH3 solution has a degree of dissociation 1.3% at temperature T. Calculate Kb for NH3 at this temperature?

Dec 22, 2015

${K}_{b} = 1.7 \cdot {10}^{- 5}$

#### Explanation:

In order to solve this problem, you need to know two things

• the balanced chemical equation for the dissociation of ammonia in aqueous solution
• the equilibrium concentrations of the three chemical species that are of interest for this reaction

As you know, ammonia, ${\text{NH}}_{3}$, is a weak base, which means that it does not dissociate completely in aqueous solution to form ammonium ions, ${\text{NH}}_{4}^{+}$, its conjugate acid, and hydroxide anions, ${\text{OH}}^{-}$.

Instead, only a fraction of the number of molecules of ammonia will actually dissociate. That fraction depends on ammonia's base dissociation constant , ${K}_{b}$.

You are told that your ammonia solution has a degree of dissociation of 1.3% at a given temperature $T$. What that means is that for every $100$ molecules of ammonia present in solution, only $1.3$ will dissociate.

Alternatively, you can think about it like this - for every $1000$ molecules of ammonia, only $13$ will ionize to form the aforementioned ions. The rest will remain as molecules of ammonia.

The balanced chemical equation for this reaction looks like this

${\text{NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

BY definition, the base dissociation constant will be equal to

${K}_{b} = \left(\left[{\text{NH"_4^(+)] * ["OH"^(-)])/(["NH}}_{3}\right]\right)$

You know that

["NH"_3] = "0.10 M"

Now, use the degree of dissociation to find the concentrations of the two ions. You have $1 : 1$ mole ratios between all chemical species, so you can say that

$\left[{\text{NH"_4^(+)] = ["OH"^(-)] = 1.3/100 * ["NH}}_{3}\right]$

["NH"_4^(+)] = ["OH"^(-)] = 1.3 * 10^(-2) * "0.10 M" = 1.3 * 10^(-3)"M"

This means that you get

${K}_{b} = \frac{1.3 \cdot {10}^{- 3} \cdot 1.3 \cdot {10}^{- 3}}{0.10} = \textcolor{g r e e n}{1.7 \cdot {10}^{- 5}}$

The answer is rounded to two sig figs, the number of sig figs you have for the initial concentration of ammonia.