How would you calculate the pH of the #"0.39 M"# #"NH"_3# #"/"# #"0.73 M"# #"NH"_4"Cl"# buffer system?
1 Answer
STARTING FROM ACIDIC SIDE OF THE EQUILIBRIUM
Because it's a buffer system, which in this case consists of a weak base (
#\mathbf("pH" = "pKa" + log\frac(["B"])(["BH"^(+)]))# ,
wherein ammonium (
#\mathbf("NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "OH"^(-)(aq))# .
The
#color(blue)("pH") = 9.26 + log(("0.39 M")/("0.73 M"))#
#~~ color(blue)(8.99)#
STARTING FROM BASIC SIDE OF THE EQUILIBRIUM
Another way to do it is to use the
#-log(K_b) = "pKb"#
#= -log(1.8xx10^(-5))#
#= 4.74#
Now, the equation becomes
#\mathbf("pOH" = "pKb" + log\frac(["BH"^(+)])(["B"]))# ,
wherein ammonia (
#\mathbf("NH"_3(aq) + "H"_2"O"(l) rightleftharpoons "NH"_4^(+)(aq) + "OH"^(-)(aq))# .
We get:
#color(green)("pOH") = 4.74 + log(("0.73 M")/("0.39 M"))#
#~~ color(green)(5.01)#
Then, the
#color(blue)("pH") = 14 - color(green)("pOH")#
#= 14 - 5.01 ~~ color(blue)(8.99)# .
From this, we can conclude that:
- A base and its conjugate acid, or an acid and its conjugate base participate in an equilibrium consistent with each other.
- Using the correct equilibrium constants (from sources of similar, comparable accuracies) for the correct equilibrium reaction, whether you assume a forward reaction using one substance or its conjugate, allows you to achieve the correct answer.
So either way you should get the same answer, so long as you use
Use whichever way, but because of the above reason regarding source consistency, don't be surprised if one way somehow gives a slightly different answer.