Question

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

Answer 1

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)`.

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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.