Trimethylamine, (CH3)3N, Is A Weak Base (Kb = 6.4 × 10–5) That ...

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.

Chemistry Acids and Bases pH

1 Answer

Stefan V. May 27, 2016

#"pH" = 4.40#

Explanation:

Your starting point here will be to write the balanced chemical equation that describes the ionization of the trimethylammonium cation, #("CH"_3)_3"NH"^(+)#, the conjugate acid of trimethylamine, #("CH"_3)_3"N"#.

Next, use an ICE table to determine the equilibrium concentration of the hydronium cations, #"H"_3"O"^(+)#, that result from the ionization of the conjugate acid.

The trimethylammonium cation will react with water to reform some of the weak base and produce hydronium cations, both in a #1:1# mole ratio.

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))^(+)#

#color(purple)("I")color(white)(aaaaacolor(black)(0.1)aaaaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaacolor(black)(0)# #color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaaaaaaacolor(black)((+x))aaaaacolor(black)((+x))# #color(purple)("E")color(white)(aaacolor(black)(0.1-x)aaaaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaacolor(black)(x)#

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#, for the trimethylammonium cation

#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 #K_a# has such a small value when compared with the initial concentration of the conjugate acid, you can use the approximation

#0.1 - x ~~ 0.1#

This will get you

#1.56 * 10^(-10) = x^2/0.1#

Solve for #x# to find

#x = sqrt((1.56 * 10^(-10))/0.1) = 3.95 * 10^(-5)#

Since #x# represents the equilibrium concentration of the hydronium cations, you will have

#["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)|)))#

Answer link

Related questions

• How do you determine pH from pKa?
• How do pH values of acids and bases differ?
• How do you calculate something on a pH scale?
• How do you calculate pH diprotic acid?
• How do you calculate pH from acid dissociation constant?
• How do you calculate pH of acid and base solution?
• Why is the pH scale 0-14?
• Is pH a measure of the hydrogen ion concentration?
• How does solubility affect pH?
• How does pH relate to pKa in a titration?

See all questions in pH

Impact of this question

91196 views around the world You can reuse this answer Creative Commons License