What is a Buffer Solution?

Hi, and welcome to this video on buffers.

Buffered solutions resist changes in pH when acids and bases are added, so before we jump into their chemistry, let’s take a quick look at acids, bases, and pH.

Remember that when an acid, HA, is added to water, some of it dissociates into the conjugate base, A, and a proton, H+ (which is picked up by a water molecule to form the hydronium ion, H3O+).

HA + H2O ⇌ H3O+ + A

The effect of adding acid to a solution is measured by the pH, which is the negative log of the concentration of H3O+

pH = -log[H3O+]

The larger the concentration of H3O+, the lower the pH.

Similarly, when a base, B, is added to water, some of it picks up a hydrogen from water to form the conjugate acid, BH+, and a hydroxide ion, OH.

B + H2O ⇌ BH+ + OH

The addition of OH- to the solution reduces the concentration of H3O+ and raises the pH of the solution.

So we know that when we add an acid or base to an aqueous solution, the concentration of H3O+ changes, which changes the pH. But why do we care? Well, it turns out that the pH greatly affects the chemistry that’s possible in a solution. The pH impacts the solubility of some compounds, the electrical conductivity of the solution, and the structure and activity of biological macromolecules. For example, the enzyme that digests proteins in your intestine works twice as fast at a pH of 8 versus a pH of 7 or 9! Thus, there are instances, especially in biological systems, where it is critical to maintain a specific pH range. In such cases, buffered solutions are used to resist changes in pH so small additions of acids and bases don’t wreak havoc on the system.

Buffers are prepared by making a solution of a weak conjugate acid/base pair- so either a weak acid, HA, and its conjugate base, A, or a weak base, B, and its conjugate acid, BH+.

Remember, weak acids and bases do not significantly dissociate in water.

Their equilibrium lies to the left. This is quantified by the acid dissociation constant, Ka, and base dissociation constant, Kb: Ka is equal to the concentration of ‘H3O plus’, times the concentration of the conjugate base, ‘A minus’, divided by the concentration of the acid, ‘HA’ . Kb is equal to the concentration of the conjugate acid, ‘BH plus’, times the concentration of the hydroxide ion, ‘OH minus’, divided by the concentration of the base, B.

These constants are simplified by taking their negative log, which provides a number typically between 2-14 to quantify the strength of weak acids and bases- the pKa and pKb.

pKa = -log Ka

pKb = -log Kb

So when we prepare a buffer and begin by adding a weak acid or base, we know that not much dissociates or reacts. Furthermore, when the conjugate pair is added, the equilibrium shifts back to the reactants per Le Chatelier’s Principle. Thus, the concentration of HA/A- in the buffer remains relatively constant until a new acid or base is added. Now let’s consider that scenario.

Imagine we’ve made a buffer with a weak acid HA and its conjugate base A. When a strong acid, HX, is added to our buffer, it will completely dissociate to form X and H3O+. However, instead of sticking around to decrease the pH, the H3O+ will react with A to make HA and water. So in total, HX will be consumed by A, resulting in a decrease of A and an increase of HA:

Similarly, when a strong base, B, is added to our buffer, it will completely dissociate to form BH+ and OH. But rather than sticking around to increase the pH, OH will react with HA to form A and water. In this case, B has been consumed by HA, decreasing HA and increasing A:

In summary, buffers work by having a reserve of weak acid and base to neutralize small amounts of added acids and bases. Now that we have the basics down, let’s explore an example:

Suppose we make a buffer by combining 22.88 mL pure acetic acid (a weak acid with pKa of 4.76), 32.81 grams sodium acetate (the salt of its conjugate base), and water to make 1 L of solution. This gives us a 0.4 M concentration of both acetic acid and the acetate ion.

Let’s start by calculating the expected pH of this solution. To do this, we use the Henderson-Hasselbalch equation, the primary equation for solving buffer problems, which is easily derived from the equation for Ka that we reviewed earlier.

We start with our definition of Ka

\(K_a = \frac{[H^+][A^-]}{[HA]}\)

then take the log of both sides: The log of ‘Ka’ is equal to the log of ‘H3O plus’ times the concentration of the ‘A minus’ divided by the concentration of HA.

\(log K_a = log(\frac{[H^+][A^-]}{[HA]})\)

With some rearrangement and use of logarithm rules, this becomes: The negative log of ‘H3O plus’ is equal to the negative log of ‘Ka’ plus the log of the concentration of ‘A minus’ divided by the concentration of the ‘HA’.

\(-log[H_30^+] = -logK_a+ log(\frac{[A^-]}{[HA]})\)

Notice that some of these terms are familiar to us! The negative log of the concentration of H+ is the pH and the negative log of Ka, is the pKa. This gives us the Henderson-Hasselbalch equation: The pH of a solution is equal to the ‘pKa’ of the acid plus the log of the concentration of ‘A minus’ divided by the concentration of the HA.

\(pH = pK_a + log(\frac{[A-]}{[HA]})\)

So, to calculate the pH of the buffer we made, we simply need to know the pKa of acetic acid and the concentrations of HA and A-, which we do! The pH of our buffer is equal to 4.76 (the ‘pKa’ of acetic acid) plus the log of 0.4 divided by 0.4, which equals 4.76 plus the log of 1. The log of 1 is 0, so our pH is 4.76.

\(pH = 4.76 + log(\frac{[0.4 M]}{[0.4 M]}) = 4.76 + log(1) = 4.76 + 0 = 4.76 \)

Notice, that when the concentrations of the weak acid and conjugate base are equal, the pH of the solution is simply equal to the pKa of the acid.

Now that our buffer is made and we have an estimate for the pH, let’s see what happens when we add a small amount of HCl, let’s say, 0.1 moles.

When HCl is added to the buffer, because it is a strong acid with a pKa of -6.3, it completely dissociates into 0.1 moles of H+ and 0.1 moles of chloride. The chloride doesn’t do much in solution, it’s happy on its own, but the H+ reacts completely with the acetate ions to form acetic acid. Let’s see how this impacts the concentrations of HA and A.

Acetic Acid (HA)Acetate ion (A)
Initial moles1L X 0.4 M = 0.4 moles1L x 0.4 M = 0.4 moles
Change in moles+0.1 moles-0.1 moles
Final moles0.5 moles0.3 moles

We started with 0.4 moles of acetic acid, but then gained 0.1 moles when HCl was added, giving us 0.5 moles total. We also started with 0.4 moles of the acetate ion, but then lost 0.1 moles, giving us 0.3 moles total. Now, let’s see how much this impacted the pH of our solution.

\(pH=4.76+log(\frac{[0. 3 M]}{[0.5 M]})=4.76-0.22=4.54\)

Because our solution was buffered, the addition of 0.1 moles of HCl only decreased our solution by 0.22 pH units.

If we had simply added 0.1 moles of HCl to 1 liter of water,

\(pH=-log[0.1 M]=1\)

the pH would have gone from 7 to 1, a decrease of 6 pH units!

While our buffer worked quite well when a small amount of acid had been added, note that if we had added 0.5 moles of HCl, all of the acetate ion would have been consumed and the additional moles of H+ would have drastically decreased the pH of the solution.

In other words, buffers have a capacity and can only consume so much additional acid or base before the pH will begin to change drastically. We can show this graphically by plotting the moles of acid or base added to the buffer against the pH.

Before the buffer is consumed, the change in pH is small, but once outside that region, the pH changes significantly. We can increase buffer capacity by starting with higher concentrations of the conjugate acid/base pair. It’s also helpful to have similar starting concentrations of the acid and base, which, per the Henderson-Hasselbalch equation, means that buffers work best at a pH similar to their pKa.

Finally, let’s wrap up with a review of what we’ve learned today. We began with a review of acids, bases, and pH, and then considered why pH is an important metric of a chemical system. There are many instances, especially in biochemistry, when a small pH window is necessary. Thus, chemists (and your body!) use buffered solutions to reduce the impact of acids and bases on pH. We wrapped up by working through an example of a buffer, using the Henderson-Hasselbalch equation to calculate the pH change when a strong acid was added. And lastly we considered buffer capacity.

Thanks for watching, and happy studying!

 

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by Mometrix Test Preparation | This Page Last Updated: February 21, 2022