Question #5e0ee

This one is pretty straightforward - you need to use an ICE table and the known value of the base dissociation constant, #K_b#, to find the equilibrium concentration of hydroxide ions in solution.

So, ethylamine, #"CH"_3"CH"_2"NH"_2#, will accept a proton from water to form the ethylammonium ion, #"CH"_3"CH"_2"NH"_3^(+)#, and hydroxide ions, #"OH"^(-)#.

Use an ICE table to help you with the calculations

#"C"_2"H"_5"NH"_text(2(aq]) + "H"_2"O"_text((l]) -> "C"_2"H"_5"NH"_text(3(aq])^(+) " "+" " "OH"_text((aq])^(-)#

#color(purple)("I")" " " " " 0.075" " " " " " " " " " " " " " " " " 0" " " " " " " " " " " "0#
#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " " "(+x)#
#color(purple)("E")" "(0.075-x)" " " " " " " " " " " " " "x" " " " " " " " " " " "x#

By definition, the base dissociation constant, #K_b#, will be equal to

#K_b = (["C"_2"H"_5"NH"_3^(+)] * ["OH"^(-)])/(["C"_2"H"_5"NH"_2])#

#K_b = (x * x)/(0.075 - x) = 6.4 * 10^(-4)#

Because #K_b# is so small, you can say that

#0.075 - x ~~ 0.075#

The equation becomes

#x^2/0.075 = 6.4 * 10^(-4)#

#x = sqrt(0.075 * 6.4 * 10^(-4)) = 0.006928#

The concentration of hydroxide ions will thus be

#["OH"^(-)] = x = "0.006928 M"#

To get the pH of the solution, calculate the #"pOH"# first

#"pOH" = -log(["OH"^(-)])#

#"pOH" = -log(0.006928) = 2.16#

The pH of the solution will thus be

#"pH" = 14 - "pOH"#

#"pH" = 14 - 2.16 = color(green)(11.84)#