Buffers

A buffer is a solution that contains substantial amounts of a compound in both its protonated and deprotanted forms.  As such, it is "resistant" to pH change upon the addition of strong acid or strong base.  This is because the protonated form can neutralize any strong base and the deprotonated form can neutralize any strong acid.  When this happens the ratio of protonated to deprotonated changes, but if their initial amounts are significant, then these changes are small.  The total amount of acid or base that can be added to the buffer before neutralizing all of one of the forms of the compound is its buffer capacity.  The pH of a buffer is generally in a range around the pKa of the acidic form of the compound (or you can think of this as the pOH is around the pKb of the basic form).  So if you want to maintain the pH in a specific region, you should use a buffer where the conjugate pair has a pKa near that value.

For example, let's imagine we wanted to make a buffer with a pH around 3.2, you would then like a compound with a pKa near this value.  Benzoic acid has a pKa of 3.18 so this would be a good choice.  Now you need to make a solution that contains both benzoic acid (the protonated form) and the ion benzoate (the deprotonated form).

There are three ways that we can accomplish this.  The first is to mix a solution of benzoic acid with a solution of a salt that contains the benzoate ions (such as sodium benzoate).  If we put in comparable numbers of moles of benzoic acid and the benzoate ion we'll end up with a buffer solution.  For the pH to exactly match the pKa the concentrations of these two need to be equal.  However, the buffer does not need to have exactly equal concentrations of these two.  We can make a slightly more acidic buffer with more acid (benzoic acid). We can make a slight more basic buffer with more base (benzoate).  Alternatively we can make the buffer by starting with a solution that is benzoic acid and partial neutralizing it to generate the benzoate.  So if I have a 1 mole of benzoic acid in solution and I add 0.5 moles of OH^-, the resulting solution will have 0.5 moles of benzoic acid and 0.5 moles of the benzoate ion.   The third way is the opposite of this, I can start with a solution that essentially has all of the compound in the deprtonated from of the benzoate ion and I can partially neutralize this with a strong acid.

Note: In neutralizing the solutions to form the buffer, I cannot neutralize all of the acid or all of the base or I won't make a buffer.

Finally, I can make buffers of different capacity by using different amounts of the compounds.  Let's take the example above of a solution with 1 mole of benzoic acid that is neutralized with 0.5 moles of OH^-.  The resulting buffer contains 0.5 moles of both the protonated and deprotonated species.  It has a buffer capacity of 0.5 moles for both acid and base.  Conversely, if I started with a solution with 0.1 moles of benzoic acid and neutralized it with 0.05 moles of OH^-, the resulting buffer would have 0.05 moles of each benzoic acid and the benzoate ion.  This buffer would had the same pH as the previous example since there are equal concentrations of protonated (acid) and deprotonated (base) forms.  However, the capacity of this buffer is 10 times less, since there are 10 times fewer moles of each.  The addition of 0.05 moles of strong acid to the first buffer would only make a small change as it would neutralize only ten percent of the base.  The addition of 0.05 moles to the second buffers would neutralize all of the base.

• 100 mL of 1M HFand 100 mL of 2M NaOH
• 100 mL of 1M HCland 50 mL of 1M NaOH
• 100 mL of 1M NH3and 50 mL of 1M NaOH
• 100 mL of 1M NH3and 50 mL of 2M HCl
• 100 mL of 1M NH3and 50 mL of 1M HCl

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the concentrations of a protonated form of a weak acid (base) to the deprotonated form.  It can be used to easily calculate (predict) the pH of a buffer where we have substantial concentrations of both the protonated and deprotonated forms of the compound.  For simplicity, we'll write it here for the acid, but it could just as easily be written for the base.

By taking the log of the equilibrium expression we find

\[\rm{pH = pK_a + log({[A^-] \over [HA]})}\]

where the pKa = -log(K_a).  Thus we can see that in a situation in which we have equal amounts of the protonated and deprotonated forms of a conjugate pair, the pH = pKa.  Alternatively, you could say that at a pH that is equal to the pKa, these concentrations are equal.   This means that for any compound, you can determine if the majority will be in the protonated or deprotonated form at a given pH by comparing the pH of the solution to the pKa.  If the pH is less than the pKa, the solution is "too acidic" and the majority of the compound will be in the protonated (acidic) form.  If the pH is greater than the pKa, the majority will be in the deprotonated (basic) form.  The exact ratio can be obtained from the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation can also be approximated to determine the pH of a buffer from the initial concentrations that I use to make the solution.  If one makes a solution with both the protonated and deprotonated forms of a compound, the differences in concentration at equilibrium compared to the initial concentrations will be very small.  Thus one can approximate the concentrations at equilibrium as the initial concentrations, in this case you can predict the pH is

\[\rm{pH \approx pK_a + log({[A^-]_0 \over [HA]_0})}\]

where the concentrations of [HA]₀+ and [A^-]₀ are the initial concentrations which we have placed into the equilibrium expression (assuming the change is small).

The same idea can be written out for K_b and OH^-

\[\rm{pOH = pK_b + log({[BH^+] \over [B]})}\]

Approximate pH formulas

You will begin to notice that we keep solving the same equilibrium problem over and over again with the same approximations.

It is easy at some point to remember these simple results (along with the approximations we made to get here).

We general make two approximations when working acid/base problems.  First, we assume the amount of H₃O⁺ and OH^- in pure water are significantly small that we can ignore them.  This works well as long as we don't have extremely dilute solutions.  Second, we assume that the changes in concentrations for weak acids and bases are small compared to the concentrations.  This is the step where we say at equilibrium we have some concentration minus a change that is approximately equal to the initial concentration.   This works best when the change is small.  That is the equilibrium constant is small.  And when the concentration is high.  That is the solution is not dilute.

Using these approximations, we arrive at the following.

For a strong acid,

\[\rm{[H_3O^+] \approx C_a}\]

That is the H₃O⁺ concentration is simply the concentration of the initial acid, C_a.  To see how this might break down think about the pH of a super dilute solution (10^-35 M HCl is not pH = 35).

Likewise, for a strong base, we get

\[\rm{[OH^-] \approx C_b}\]

That is the OH^- concentration is just the concentration of the base.

For a weak acid, if we work through the equilibrium with approximations, we get

\[\rm{[H_3O^+] \approx \sqrt{K_a C_a}}\]

For a weak base, we get

\[\rm{[OH^-] \approx \sqrt{K_b C_b}}\]

Finally, for a buffer, we get

\[\rm{[H_3O^+] \approx K_a \times {C_a \over C_b}}\]

or

\[\rm{[OH^-] \approx K_b \times {C_b \over C_a}}\]

© 2013 mccord/vandenbout/labrake

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