Question

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

Answer 1

`K_b = 1.7 * 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, `"NH"_3`, is a weak base, which means that it does not dissociate completely in aqueous solution to form ammonium ions, `"NH"_4^(+)`, its conjugate acid, and hydroxide anions, `"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

`"NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH"_text((aq])^(-)`

BY definition, the base dissociation constant will be equal to

`K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])`

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

`["NH"_4^(+)] = ["OH"^(-)] = 1.3/100 * ["NH"_3]`

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

This means that you get

`K_b = (1.3 * 10^(-3) * 1.3 * 10^(-3))/0.10 = color(green)(1.7 * 10^(-5))`

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