$\require{\mhchem}$

Buffer systems are important part of of organic and inorganic chemistry. In short, a buffer is a solution that is resistant to pH changes, such as the addition of an acid or a base.

Let's go over an example problem you might encounter on a homework assignment or exam.

What is the pH of an 80 mL .30M NH₃/0.36M NH₄⁺ buffer system after 20.0 mL of 0.050M NaOH has been added? The Kₐ value for $$\ce{NH4+}$$ is $$5.8\text{x}10^{-10}$$

Lets break this problem down. We have 80 mL of a solution made up of $$\ce{NH3}$$ (ammonia) and $$\ce{NH4+}$$ (ammonium).

We know the concentrations of each of these species, denoted by brackets.

$$[\ce{NH3}] = \frac{\text{0.30 moles } \ce{NH3}}{\text{1 Liter}}$$ $$[\ce{NH4+}] = \frac{\text{0.36 moles } \ce{NH4+}}{\text{1 Liter}}$$

Then we add 20.0 mL of $$\ce{NaOH}$$, which has the following concentration.

$[\ce{Na+}] = \frac{\text{0.050 moles } \ce{Na+}}{\text{1 Liter}}$

To proceed with this problem, we need to figure out what will happen to the $$\ce{NaOH}$$ (sodium hydroxide) when it is added to the solution.

Sodium hydroxide is a strong base, and will completely disassociate into the Sodium and Hydroxide ions like so.

$\ce{NaOH ->[H₂O] Na+ + OH- }$

Theoretically $$\ce{Na+}$$ would react with $$\ce{H2O}$$ to form more $$\ce{NaOH}$$, but this will not happen, because $$\ce{NaOH}$$ is a strong base and will not be formed. For this reason, we can ignore it in our further calculations. This leaves us only the $$\ce{OH-}$$ to worry about.

Since $$\ce{OH-}$$ is a base, it will react with the acid in our buffer solution. Buffer solutions are always made up of conjugate acid/base pairs. In this case, $$\ce{NH4+}$$ is the acid, because it is the conjugate acid of $$\ce{NH3}$$.

$\ce{NH4+(aq) + OH- (aq) <-> NH3(aq) + H2O(l)}$

In this reaction, hydroxide acts as a base and accepts the proton from ammonium.

To further solve this problem, we are going to need to use a RICE table.

$$\begin{array}{c|c@{}c@{}c@{}c@{}c} \hline R & \ce{NH4+(aq) &$${}+{}$$& OH- (aq) & <-> & NH3(aq) &$${}+{}$$& H2O(l)} \\ \hline I & ?.?? && ?.?? && ?.?? && \text{---} \\ C & ?.?? && ?.?? && ?.?? && \text{---} \\ E & ?.?? && ?.?? && ?.?? && \text{---} \\ \hline \end{array}$$

Let's start by figuring out $$I$$, the initial amount of moles for each species. Since water is in such high abundance, we can leave it out of the equation.

$\text{moles } \ce{NH4+} = \frac{0.36 \text{ moles } \ce{NH4+}}{1 \text{ L}} * 0.0800 \text{ L} = 0.0288 \text{ moles}$

$\text{moles } \ce{OH-} = \frac{0.050 \text{ moles } \ce{OH-}}{1 \text{ L}} * 0.0200 \text{ L} = 0.0010 \text{ moles}$

$\text{moles } \ce{NH3} = \frac{0.30 \text{ moles } \ce{NH3}}{1 \text{ L}} * 0.0800 \text{ L} = 0.0240 \text{ moles}$

Here is our updated rice table with the initial amounts of each species.

$$\begin{array}{c|c@{}c@{}c@{}c@{}c} \hline R & \ce{NH4+(aq) &$${}+{}$$& OH- (aq) & <-> & NH3(aq) &$${}+{}$$& H2O(l)} \\ \hline I & 0.0288 && 0.0010 && 0.0240 && \text{---} \\ C & ?.?? && ?.?? && ?.?? && \text{---} \\ E & ?.?? && ?.?? && ?.?? && \text{---} \\ \hline \end{array}$$

We can see that all species in this reaction are one to one ratios. This means that $$\ce{OH-}$$ is the limiting reactant, as it has the least amount of moles present.

Additionally, this means the initial number of moles of $$\ce{OH-}$$ is the maximum number of moles that can be reacted of either of the other chemical species.

Now we can fill in $$C$$ and $$E$$, or the change in moles during the reaction and the final amount at equilibrium.

$$\begin{array}{c|c@{}c@{}c@{}c@{}c} \hline R & \ce{NH4+(aq) &$${}+{}$$& OH- (aq) & <-> & NH3(aq) &$${}+{}$$& H2O(l)} \\ \hline I & 0.0288 && 0.0010 && 0.0240 && \text{---} \\ C & -0.0010 && -0.0010 && -0.0010 && \text{---} \\ E & 0.0278 && 0 && 0.0250 && \text{---} \\ \hline \end{array}$$

We can see that all of the hydroxide was depleted in this reaction. Now that we have the number of moles of each species at equilibrium, we can preform the following calculations to find the concentration at equilibrium.

$$[\ce{NH3}]_{eq} = \frac{\text{0.0250 moles } \ce{NH3}}{\text{0.1000 Liter}}$$ = 0.250 M

$$[\ce{NH4+}]_{eq} = \frac{\text{0.0278 moles } \ce{NH4+}}{\text{0.1000 Liter}}$$ = 0.278 M

We use the volume of .1 L because we had an initial volume of 80 mL and then we added an additional 20 mL of $$\ce{NaOH}$$.

We can use these equilibrium molarities as inputs to the Henderson–Hasselbalch equation, which takes the following form:

$pH = pK_a + \text{log}\left( \frac{[A^-]}{[HA]} \right)$

In this example, $$\ce{HA}$$ and $$\ce{A-}$$ represent the acid and it's conjugate base respectively.

Let's start substituting in values. We know that the Kₐ of $$\ce{NH4+}$$ is $$5.8\text{x}10^{-10}$$, and from that we can derive the pKₐ value, which is equal to $$-\text{log}\left( K_a \right)$$.

We also have the equilibrium concentrations for the conjugate acid/base pair. After filling in our values, the HH equation should look like:

$pH = 9.24 + \text{log}\left( \frac{0.250}{0.278} \right)$

Plug this into your calculator, and you should get a resulting pH of $$9.19$$.

# Conclusion⌗

Please let me know if you have any questions or corrections for me! Writing these exercises helps me practice for my exams 😄