Question

Question #5e0ee

Answer 1

`"pH" = 11.84`

Explanation:

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)`