Trimethylamine, (CH3)3N, is a weak base (Kb = 6.4 × 10–5) that hydrolyzes by the following equilibrium: (CH3)3N + H2O → (CH3)3NH+ + OH– What is the pH of a 0.1 M solution of (CH3)3NH+? (Enter pH to 2 decimal places; hundredth's.
1 Answer
Explanation:
Your starting point here will be to write the balanced chemical equation that describes the ionization of the trimethylammonium cation,
Next, use an ICE table to determine the equilibrium concentration of the hydronium cations,
The trimethylammonium cation will react with water to reform some of the weak base and produce hydronium cations, both in a
This means that for every mole of conjugate acid that ionizes, you get one mole of weak base and one mole of hydronium cations.
The ICE table will thus look like this
#("CH"_ 3)_ 3"NH"_ ((aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons ("CH"_ 3)_ 3"N"_ ((aq)) + "H"_ 3"O"_ ((aq))^(+)#
Now, you know that an aqueous solution at room temperature has
#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_a xx K_b = K_W)color(white)(a/a)|)))#
where
#K_w = 10^(-14) -># the ionization constant of water
Use this equation to calculate the acid dissociation constant,
#K_a = K_W/K_b#
#K_a = 10^(-14)/(6.4 * 10^(-5)) = 1.56 * 10^(-10)#
By definition, the acid dissociation constant will be equal to
#K_a = ([("CH"_3)_3"N"] * ["H"_3"O"^(+)])/([("CH"_3)_3"NH"^(+)])#
In your case, you will have
#K_b = (x * x)/(0.1 - x) = 6.4 * 10^(-5)#
Since
#0.1 - x ~~ 0.1#
This will get you
#1.56 * 10^(-10) = x^2/0.1#
Solve for
#x = sqrt((1.56 * 10^(-10))/0.1) = 3.95 * 10^(-5)#
Since
#["H"_3"O"^(+)] = 3.95 * 10^(-5)"M"#
As you know, the pH of the solution is defined as
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
Plug in your value to find
#"pH" = - log(3.95 * 10^(-5)) = color(green)(|bar(ul(color(white)(a/a)4.40color(white)(a/a)|)))#