How would you calculate the pH of the #"0.39 M"# #"NH"_3# #"/"# #"0.73 M"# #"NH"_4"Cl"# buffer system?

1 Answer
Jul 13, 2016

STARTING FROM ACIDIC SIDE OF THE EQUILIBRIUM

Because it's a buffer system, which in this case consists of a weak base (#"NH"_3#) and its conjugate weak acid (#"NH"_4"Cl"#), you could use the Henderson-Hasselbalch equation

#\mathbf("pH" = "pKa" + log\frac(["B"])(["BH"^(+)]))#,

wherein ammonium (#"BH"^(+)#) acts as a Bronsted acid (proton donor) for the equilibrium process described by the reaction written as

#\mathbf("NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "OH"^(-)(aq))#.

The #"pKa"# of #"NH"_4^(+)#, the ion in #"NH"_4"Cl"#, is about #9.26#, so we can proceed to calculate the #"pH"#:

#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 #K_b# of the weak-base ammonia (#1.8xx10^(-5)#) to get the #"pKb"# of ammonia and use the opposite version of the Henderson-Hasselbalch equation.

#-log(K_b) = "pKb"#

#= -log(1.8xx10^(-5))#

#= 4.74#

Now, the equation becomes

#\mathbf("pOH" = "pKb" + log\frac(["BH"^(+)])(["B"]))#,

wherein ammonia (#"B"#) acts as a Bronsted base (proton acceptor) for the equilibrium process described by the reaction written as

#\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")# becomes:

#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 #"pKa"#'s or #"pKb"#'s from consistent sources in terms of their accuracy.

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.