# Question #5e0ee

##### 1 Answer

#### Answer:

#### Explanation:

This one is pretty straightforward - you need to use an **ICE table** and the known value of the *base dissociation constant*,

So, *ethylamine*, *ethylammonium ion*, *hydroxide ions*,

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])^(-)#

By definition, the *base dissociation constant*,

#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

#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" = -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)#