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

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 (${\text{NH}}_{3}$) and its conjugate weak acid ($\text{NH"_4"Cl}$), you could use the Henderson-Hasselbalch equation

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

wherein ammonium (${\text{BH}}^{+}$) acts as a Bronsted acid (proton donor) for the equilibrium process described by the reaction written as

$\setminus m a t h b f \left({\text{NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "OH}}^{-} \left(a q\right)\right)$.

The $\text{pKa}$ of ${\text{NH}}_{4}^{+}$, the ion in $\text{NH"_4"Cl}$, is about $9.26$, so we can proceed to calculate the $\text{pH}$:

color(blue)("pH") = 9.26 + log(("0.39 M")/("0.73 M"))

$\approx \textcolor{b l u e}{8.99}$

STARTING FROM BASIC SIDE OF THE EQUILIBRIUM

Another way to do it is to use the ${K}_{b}$ of the weak-base ammonia ($1.8 \times {10}^{- 5}$) to get the $\text{pKb}$ of ammonia and use the opposite version of the Henderson-Hasselbalch equation.

$- \log \left({K}_{b}\right) = \text{pKb}$

$= - \log \left(1.8 \times {10}^{- 5}\right)$

$= 4.74$

Now, the equation becomes

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

wherein ammonia ($\text{B}$) acts as a Bronsted base (proton acceptor) for the equilibrium process described by the reaction written as

$\setminus m a t h b f \left({\text{NH"_3(aq) + "H"_2"O"(l) rightleftharpoons "NH"_4^(+)(aq) + "OH}}^{-} \left(a q\right)\right)$.

We get:

color(green)("pOH") = 4.74 + log(("0.73 M")/("0.39 M"))

$\approx \textcolor{g r e e n}{5.01}$

Then, the $\textcolor{b l u e}{\text{pH}}$ becomes:

$\textcolor{b l u e}{\text{pH") = 14 - color(green)("pOH}}$

$= 14 - 5.01 \approx \textcolor{b l u e}{8.99}$.

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 $\text{pKa}$'s or $\text{pKb}$'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.