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

1 Answer
Dec 22, 2015

Answer:

#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.