# How do you use the Henderson-Hasselbalch to calculate the pH of a buffer solution that is .50 M in NH_3 and . 20 M in NH_4Cl?

## For ammonia, $p {K}_{b} = 4.75$.

Aug 4, 2016

That buffer solution has a pH of $9.65$

#### Explanation:

Before I introduce Henderson-Hasselbalch's equation, we should identify the acid and base. Ammonia $\left(N {H}_{3}\right)$ is always a base and the ammonium ion $\left(N {H}_{4}^{+}\right)$ is the conjugate acid of ammonia. A conjugate acid has one more proton $\left({H}^{+}\right)$than the base you started with.

Now, we can use this equation:

As you can see, we are given a pKb instead of a pKa. But, no worries we can use the following equation that relates both constants to each other:

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$pKa + pKb = 14

We can solve for the pKa by subtracting the given pKb from 14:

$14 - 4.75 = 9.25$

Next, we can obtain the [base] and [acid] from the question.

[$N {H}_{3}$]= .50 M [$N {H}_{4}$] =.20 M

We're not really concerned with the chloride anion that attached to the ammonium ion because it's a spectator ion and it has no effect on the buffer system.

Now, we have all of the information to determine the pH. Let's plug our values into the equation:

$p H = 9.25 + \log \left(\frac{0.50}{0.20}\right)$

$p H = 9.65$