# C2h5nh2

Ethylamine, C2H5NH2, has a Kb value of 6.4 x 10-4. What concentration of C2H5NH2  is required to produce an ethylamine solution with a pH of 11.875.

Please explain step by step. :(

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The basicity equation is: C2H5NH2 + H2O <-----> C2H5NH3+ + OH-

Let, x be the degree of dissociation of the amine, and c its molar concentration at the beginning, then concentration terms of all the species in solution will be:

C2H5NH2 + H2O <-----> C2H5NH3+ + OH-
c(1-x)                                    cx                   cx

Kb = cx . cx/c(1-x) = cx˄2/(1-x) = cx˄2 {as x is too small compared to 1, so (1-x) = 1}

Or, x = √(Kb/c)

[OH-] = cx = c. √(Kb/c) = √(Kb.c)

Here, pH = 11.875, pOH = 14 – 11.875 = 2.125; -log[OH-] = 2.125, or, [OH-] = 7.5×10˄-3

By condition, 7.5×10˄-3 = √(6.4 x 10˄-4.c)

Or, c = (7.5×10˄-3)˄2/(6.4 x 10˄-4) = 8.79×10˄-2 = 0.0879

Hence the concentration of C2H5NH2  required to produce an ethylamine solution with a pH of 11.875 is 0.0879 M.

Sources: