Question #acfee

[Part1] Determine the mass of diethylamine.
The reaction between diethylamine and hydrochloric acid is
#(C_2H_5)_2NH + HCl -> (C_2H_5)_2NH_2^+Cl^-#.

Let #m# (g) the mass of diethylamine.
Acording to the reaction formula, the amount of substance is the same for the acid and the base.
#m/73.14 * 1= 0.075 * 15.90/1000 * 1#.
#m=8.72xx10^-2# g.

[Part2] Then, what is the pH at the equivalent point?
The amount of substance of diethylamine chloride(#(C_2H_5)_2NH_2^+Cl^-#) after the titration is
#0.075*15.90/1000 = 1,193xx10^-3# moles, and the molarity is
#(1.193xx10^-3)/((100.00+15.90)/1000)=1.029xx10^-2# M.

The equilibrium state of diethylamine is
#(C_2H_5)_2NH + H_2O ⇄ (C_2H_5)_2NH_2^+ + OH^-#. [1]
The hydrosis of #(C_2H_5)_2NH_2^+# is
#(C_2H_5)_2NH_2^+-##⇄ (C_2H_5)_2NH + H^+# [2]

And the equilibrium constant is
#K_b=([(C_2H_5)_2NH_2^+][OH^-])/([(C_2H_5)_2NH])=3.09xx10^-4#. [3]

Then, let's solve [3].

[a] #[(C_2H_5)_2NH_2^+]=1.029xx10^-2# M(since #(C_2H_5)_2NH_2^+Cl^-# is a salt)
[b] Let #[H^+]= x# M. Then, the equation [2] tells us that
#[(C_2H_5)_2NH]=x#.
[c] #[OH^-]=K_w/([H^+]) = (1.00xx10^-14)/x# M.

Substitute [a],[b] and [c] to [3].
#(1.029xx10^-2*1.00xx10^-14)/x^2=3.09xx10^-4#
#x^2=(1.029xx10^-2*1.00xx10^-14)/(3.09xx10^-4)=3.330xx10^-13#
#x= 5.770 xx10^-7# M #=[H^+]#.

Therefore, the pH is #-log[H^+]=-log(5.770xx10^-7)~=6.24#.

We've effectively "neutralized" a weak base. Unlike traditional neutralization reactions, the pH at the equivalence point isn't equal to #7#. An #"ICE"# table is useful, here,

Note, I've assumed volumes are additive.

#(C_2H_5)_2NH(aq) + H^(+)(aq) rightleftharpoons (C_2H_5)_2NH_2^(+)(aq)#

are the concentrations at the equivalence point. Now, let's examine another equilibrium that is occurring due to the conjugate acid dissociating:

#(C_2H_5)_2NH_2^(+)(aq) rightleftharpoons H^(+)(aq) + (C_2H_5)_2NH(aq)#

We know the concentration of the conjugate acid, and using the equilibrium expression,

#K_a = ([H^+][(C_2H_5)_2NH])/([(C_2H_5)_2NH_2^(+)]) = (1.0*10^-14)/K_b approx 3.24*10^-11#

Hence,

#3.24*10^-11 = x^2/(0.0120M)#
#x = [H^+] approx 6.23*10^-7M#
#therefore pH = -log[H^+] approx 6.21#

This is reasonable, because the conjugate acid will make the solution slightly acidic at the equivalence point, the weak base having been consumed.