ABC - Buffers (Lesson)

Buffers

In previous lessons, the solution properties of either weak acids or weak bases in water were explored. While the pH of these solutions was calculated, the pH changes upon addition of a strong acid (H₃O⁺) or strong base (OH^-) were not investigated. As can be imagined, the pH change upon the addition of even a slight amount of strong base to a weak acid solution will result in a significant lowering of the pH. In this lesson the properties of buffer solutions will be explored, including how they mitigate these drastic changes in pH.

Buffers

By definition, buffers are solutions that contain both a weak acid and its conjugate base OR a weak base and its conjugate acid simultaneously. It has already been demonstrated how weak acids and bases produce a small amount of their conjugates after reaching equilibrium; however, the concentration of these conjugates is very small, and these solutions would not qualify as buffer solutions. A buffer solution must contain significant amounts of both substances in order to truly behave as a buffer solution.

One important example of a buffer solution is one that every person relies on to maintain the pH of blood. As CO₂ is produced through cellular respiration, it is converted into carbonic acid (H₂CO₃) and moved to the bloodstream.

CO₂ (g) + H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ H₂CO₃ (aq)

Because carbonic acid is a Bronsted-Lowry acid, an increase in its concentration in the blood would result in a decrease in the pH. This carbonic acid rapidly dissociates into hydrogen carbonate (HCO3-) and hydronium as shown below. Because carbonic acid is continually being produced and continually dissociating into hydrogen carbonate, there is a consistent amount of this weak acid/conjugate base pair present which results in a buffer solution that prevents drastic changes in the pH of the bloodstream.

H₂CO₃ (aq) + H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ HCO₃^- (aq) + H₃O⁺ (aq)

Image is showing a red blood cell carrying oxygen.

Buffers derive their name from the fact that they are able to resist, or buffer, against changes in pH upon the addition of either a strong acid or strong base. Take for example an acetic acid solution that also contains a significant amount of sodium acetate. Because acetate is the conjugate base of the weak acetic acid, this solution would qualify as a buffer solution. The equilibrium describing this solution is shown below:

CH₃COOH (aq) +H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ CH₃COO^- (aq) + H₃O⁺ (l)

Upon the addition of a strong acid, there is the presence of acetate (weak base) to react with this acid and prevent changes to the pH of the overall solution. Likewise, the addition of strong base is met with the presence of the acetic acid and is also neutralized. This is also illustrated in the diagram below.

Calculating the pH of Buffer Solutions

The pH of buffer solutions can be calculated in much the same way that solutions of weak acids or weak bases are calculated. The only difference in these situations is that the initial concentration of the conjugate acid or base is not zero. To illustrate this, assume an HF/F^- buffer solution where the concentrations of HF and F^- are 0.25 M and 0.40 M respectively. Also note that the K_a value for hydrofluoric acid is 6.6 x 10^-4.

As in all other cases the first step is to write the chemical equilibrium and equilibrium expression:

HF (aq) + H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ F^- (aq) + H₃O⁺ (aq)

$LaTeX: K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack F^-\right\rbrack}{\lbrack HF]}$ $K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack F^-\right\rbrack}{\lbrack HF]}$

Now the values provided can be added to the ICE table. Notice that the only real difference between buffer calculations and calculations involving weak acids only is the fact that the [CH₃COO^-] is not equal to zero initially.

Table
	HF (aq)	H₂O (l)	F^- (aq)	H₃O⁺ (aq)
Initial	0.25 M	--	0.40	0
Change	- x	--	+ x	+ x
Equilibrium	0.25 - x	--	0.40 + x	x

Substitute the values at equilibrium into the equilibrium expression:

$LaTeX: K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack F^-\right\rbrack}{\lbrack HF]}=6.6\times10^{-4}$ $K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack F^-\right\rbrack}{\lbrack HF]}=6.6\times10^{-4}$

$LaTeX: K_a=\frac{\left(x\right)\left(0.40+x\right)}{0.25-x}=6.6\times10^{-4}$ $K_a=\frac{\left(x\right)\left(0.40+x\right)}{0.25-x}=6.6\times10^{-4}$

At this point, the x in the denominator would be ignored to avoid the necessity of using the quadratic formula to solve the equation. (Doing so in this case would not alleviate this problem, as there would still be an x in the numerator that would lead to a quadratic equation). The justification for ignoring the x in the denominator was that it is so small compared to the number that it is subtracted from that it is negligible and can therefore be ignored. The same holds true for the x in the numerator as well. This value of x can also be ignored to simplify the problem and avoid the need for the quadratic formula:

$LaTeX: K_a=\frac{\left(x\right)\left(0.40\right)}{0.25}=6.6\times10^{-4}$ $K_a=\frac{\left(x\right)\left(0.40\right)}{0.25}=6.6\times10^{-4}$

$LaTeX: 0.40x=1.7\times10^{-4}$ $0.40x=1.7\times10^{-4}$

$LaTeX: x=4.3\times10^{-4}$ $x=4.3\times10^{-4}$

As with weak acids, the x value in this case also refers to the [H₃O⁺] at equilibrium. Using this value and substituting it into the pH equation allows pH to be calculated:

pH = -log [H₃O⁺]

pH = -log (4.3 x 10^-4)

pH = 3.37

The Henderson-Hasselbalch Equation

The pH calculations for buffers can be performed using the same exact techniques that have been utilized for weak acid or base equilibria; however, a shortcut of sorts exists in the case of buffers. Derived from the equations related to weak acid equilibria and pH, the Henderson-Hasselbalch Equation allows for the calculation of the pH of buffer solutions without the need for ICE tables. There are two forms of this equation as shown below. The first form is for acids and is related to Ka:

$LaTeX: pH=pK_a+\log\frac{\left\lbrack A^-\right\rbrack}{\left\lbrack HA\right\rbrack}$ $pH=pK_a+\log\frac{\left\lbrack A^-\right\rbrack}{\left\lbrack HA\right\rbrack}$

$LaTeX: pK_a=-\log K_a$ $pK_a=-\log K_a$

In this equation [A^-] refers to the concentration of the conjugate base present in solution and [HA] corresponds to the concentration of the weak acid.

In addition to this form, there is an alternative form in terms of pOH. Depending on what values are provided or what is being solved for, it could be more helpful.

$LaTeX: pOH=pK_b+\log\frac{\left\lbrack BH^+\right\rbrack}{\left\lbrack B\right\rbrack};where$ $pOH=pK_b+\log\frac{\left\lbrack BH^+\right\rbrack}{\left\lbrack B\right\rbrack};where$

$LaTeX: pK_b=-\log K_b$ $pK_b=-\log K_b$

Calculating the pH of a Buffer Using the Henderson-Hasselbalch Equation

Using the Henderson-Hasselbalch equation provides a much more simple and straightforward way to calculate the pH of a buffer solution. Utilizing one simple formula instead of relying on ICE tables to organize information greatly shortens the calculation process.

For example, assume a sodium lactate/lactic acid buffer system where the concentrations of lactate and lactic acid are 0.15 M and 0.11 M respectively. The K_a for lactic acid is 1.4 x 10^-4. The first step in this process would be to convert K_a to pK_a in preparation for using the Henderson Hasselbalch equation:

$LaTeX: pK_a=-\log K_a$ $pK_a=-\log K_a$

$LaTeX: pK_a=-\log\left(1.4\times10^{-4}\right)=3.85$ $pK_a=-\log\left(1.4\times10^{-4}\right)=3.85$

Now all values can be substituted in:

$LaTeX: pH=pK_a+\log\frac{\left\lbrack A^-\right\rbrack}{\left\lbrack HA\right\rbrack}$ $pH=pK_a+\log\frac{\left\lbrack A^-\right\rbrack}{\left\lbrack HA\right\rbrack}$

$LaTeX: pH=3.85+\log\frac{(0.15)}{(0.11)}$ $pH=3.85+\log\frac{(0.15)}{(0.11)}$

LaTeX: pH=3.98 $pH=3.98$

You Try It!

In the self-assessment activity below, calculate the pH of the buffer system. Click on the plus sign to check your answer!

Buffer Capacity

A buffer's ability to guard against changes in pH has been explained above; however, not all buffer solutions have the same ability to protect against pH changes. Buffers have varying degrees of effectiveness, referred to as buffer capacity, based on the concentrations of the substances used. For example, suppose two different HF/F^- buffer solutions are made such that one solution has an HF concentration of 1.0 M and F^- M whereas the second buffer only has HF and F^- concentrations of 0.1 M respectively. Because the concentrations of the two components in the first solution are ten times higher than the concentrations in the second solution, it will have ten times the buffer capacity. Quite simply, because the concentrations of the components are higher, there is a greater amount to react with an addition of strong base or strong acid and would therefore guard against drastic pH changes to a greater extent.

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