What is the #K_a# for a #"1.0-mol L"^(-1)# #"NH"_4^(+)# solution with a #"pH"# of #4.64# ?

1 Answer
Feb 15, 2018

Answer:

#K_a = 5.3 * 10^(-10)#

Explanation:

The ammonium cation will act as a weak acid in aqueous solution, as you can undoubtedly tell from the fact that the #"pH"# of the solution is #<7#.

The balanced chemical equation that describes the partial ionization of the ammonium cation looks like this

#"NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (3(aq)) + "H"_ 3"O"_ ((aq))^(+)#

By definition, the acid dissociation constant, #K_a#, will take the form

#K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])#

Now, the #"pH"# of the solution will give you the equilibrium concentration of the hydronium cations, since

#["H"_3"O"^(+)] = 10^(-"pH") quad "M"#

Notice that every mole of ammonium cations that ionizes produces #1# mole of ammonia and #1# mole of hydronium cations. This tells you that, at equilibrium, the concentration of ammonia will be equal to the concentration of hydronium cations.

#["NH"_3] = ["H"_3"O"^(+)] = 10^(-"pH") quad "M"#

The equilibrium concentration of the ammonium cations will be equal to

#["NH"_ 4^(+)] = ["NH"_ 4^(+)]_ 0 - 10^(-"pH")#

This is the case because in order for the reaction to produce #["H"_3"O"^(+)] =10^(-"pH") quad "M"#, the initial concentration of the ammoniumc cations, #["NH"_4^(+)]_0#, must decrease by #10^(-"pH") quad "M"#

The acid dissociation constant will thus be equal to--I'll do the calculations without added units!

#K_ a = (10^(-"pH") * 10^(-"pH"))/(["NH"_ 4^(+)]_ 0 - 10^(-"pH"))#

#K_a = 10^(-2"pH")/(["NH"_ 4^(+)]_ 0 - 10^(-"pH")#

In your case, you have

#K_a = 10^(- 2 * 4.64)/(1.0 - 10^(-4.64)) = color(darkgreen)(ul(color(black)(5.3 * 10^(-10))))#

The answer is rounded to two sig figs.