Question

Question #01170

Answer 1

The `"pH"` is about `5.1`

Explanation:

Because ammonium `("NH"_4^+)` is a weak acid, you need to determine the amount of `"H"^+` in the solution by setting an equilibrium problem.

Step 1: find `K_"a"` of ammonium from the `K_"b"` of ammonia (acid / base conjugates).

`K_"a" * K_"b" = 10^-14`

`K_"a"` of `("NH"_4)^(+)=10^-14/(K_text(b))=5.6 xx 10^-10`

Step 2: Set up the dissociation of ammonium equilibrium, and find the amount of `text(H)^+` in solution at equilibrium.

#{: (," ",bb("NH"_4^+),bb->,bb"NH"_3,+,bb("H"^+)), ("initial:",,"0.100 M",,"0 M",,"0 M"), ("change:",,-x,,+x,,+x), ("equilibrium:",,0.1-x,,x,,x) :}#

You can approximate that

`0.1 - x ~~ 0.1`

b/c `K_"a"` is very small relative to `"0.1 M"`.

`K_"a" = (["NH"_3] * ["H"^+])/(["NH"_4^+]) = (x * x) /0.1 = 5.6 xx 10^-10`

`x^2 = 0.1 * 5.6 xx 10^-10`

`x = sqrt(5.6 xx 10^-11) =7.5 xx 10^-6`

`["H"^+] = x = 7.5 xx 10^-6`

Step 3: calculate `"pH"` from the concentration of `"H"^+` found in step 2.

`"pH" = -log["H"^+] = -log(7.5xx10^-6) = 5.1`