Question

Question #a8da2

Answer 1

The pH of the solution is `11.22`.

Explanation:

To solve this problem, you need the value of the base dissociation constant of ammonia, `K_b`, which is listed as being equal to `1.8 * 10^(-5)` `->` see here.

Since ammonia is a weak base, it will increase the concentration of `"OH"^(-)` ions in the solution, so you would expect the solution to have a pH greater than `7`.

Use the ICE table (more here) to solve for the concentration of `"OH"^(-)` ions, which are labeled here as `x`

`" " " ""NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH" _ (4(aq))^(+) + "OH" _((aq))^(-)`

`"I"color(white)(aaaaacolor(black)(0.15)aaaaaaaaaaaaaaaa color(black)(0) aaaaaaa color(black)(0)`
`"C"color(white)(aaaaacolor(black)(-x)aaaaaaaaaaaaaaa color(black)(+x) aaaaa color(black)(+x)`
`"E"color(white)(aaacolor(black)(0.15-x)aaaaaaaaaaaaaa color(black)(x) aaaaaaa color(black)(x)`

Use the definition of the base dissociation constant

`K_b = (["OH"^(-)] * ["NH"_4^(+)])/(["NH"_3])`

`1.8 * 10^(-5) = ( x * x)/(0.15 - x) = x^2/(0.15-x)`

Because `K_b` is so small compared with the initial concentration of ammonia, you can use the approximation

`0.15 - x ~~ 0.15`

You will have

`1.8 * 10^(-5) = x^2/0.15`

`x = sqrt(0.15 * 1.8 * 10^(-5))`

`x = 0.001643`

This means that you have

`["OH"^(-)] = "0.001643 M"`

The solution's `"pOH"` will be

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

`"pOH" = -log(0.001643) = 2.78`

Therefore, the `"pH"` of the solution will be

`"pH" = 14 - "pOH"`

`"pH" = 14 - 2.78 = color(green)(11.22)`