Grade 11

Advanced

9 min

Lesson 11 of 12

Buffer Systems and Capacity

Not Started

Exploring how certain solutions resist changes in pH when acids or bases are added.

Your blood is a chemical superhero: even if you drink a liter of acidic soda, your internal pH barely budges. How does this 'chemical shock absorber' work to keep you alive?

Learning Objectives

1.Explain the common ion effect and how it suppresses the ionization of weak acids or bases.
2.Calculate the pH of a buffer solution using the Henderson-Hasselbalch equation.
3.Analyze the factors, such as concentration and component ratios, that determine a buffer's capacity.

The Common Ion Effect

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. This magic happens because of the common ion effect. Imagine a solution of a weak acid, acetic acid (

CH_3COOH

). It partially dissociates:

CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)

If we add a salt like sodium acetate (

NaCH_3COO

), we flood the solution with the common ion,

CH_3COO^-

. According to Le Chatelier's Principle, this excess product shifts the equilibrium to the left. This suppresses the ionization of the acid, keeping the concentration of

H^+

low and stable. A buffer must contain both a weak conjugate acid-base pair to neutralize both incoming

H^+

and

OH^-

ions.

Quick Check

If you add sodium fluoride (NaF) to a solution of hydrofluoric acid (HF), what happens to the concentration of H+ ions?

Think about Le Chatelier's Principle and the direction of the shift.

Answer

The concentration of H+ ions decreases because the equilibrium shifts to the left to consume the added fluoride ions.

The Henderson-Hasselbalch Equation

To calculate the pH of these systems, we use the Henderson-Hasselbalch equation. It is derived directly from the acid dissociation constant (

K_a

) expression. By taking the negative log of both sides of the

K_a

formula, we get:

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

Where

[A^-]

is the concentration of the conjugate base and

[HA]

is the concentration of the weak acid. This equation shows that the pH is determined by the

pK_a

of the acid and the ratio of the two components. If the concentrations of the acid and base are equal, the

\log(1)

term becomes zero, and

pH = pK_a

Calculate the pH of a buffer solution that is $0.50\,M$ in acetic acid ( $CH_3COOH$ ) and $0.50\,M$ in sodium acetate ( $NaCH_3COO$ ). The $K_a$ for acetic acid is $1.8 \times 10^{-5}$ .

1. Find the $pK_a$ : $pK_a = -\log(1.8 \times 10^{-5}) = 4.74$ . 2. Identify concentrations: $[A^-] = 0.50\,M$ and $[HA] = 0.50\,M$ . 3. Apply the equation: $pH = 4.74 + \log\left(\frac{0.50}{0.50}\right)$ . 4. Solve: $pH = 4.74 + \log(1) = 4.74 + 0 = 4.74$ .

Quick Check

What happens to the pH of a buffer if you double the concentration of both the weak acid and the conjugate base?

Look at the log term in the Henderson-Hasselbalch equation.

Answer

The pH remains the same because the ratio [A-]/[HA] does not change.

Buffer Capacity and Range

Not all buffers are created equal. Buffer capacity refers to the amount of acid or base a buffer can neutralize before the pH begins to change significantly. Two factors determine this: Concentration and the Ratio. A buffer with $1.0\,M$ components has a much higher capacity than one with $0.1\,M$ components, even if their pH is the same. Furthermore, a buffer is most effective when the ratio of $[A^-]/[HA]$ is close to $1$ . Ideally, the pH of the buffer should be within $\pm 1$ unit of the $pK_a$ of the weak acid used.

Buffer A: $0.10\,M\,HF$ and $0.10\,M\,NaF$ . Buffer B: $0.50\,M\,HF$ and $0.50\,M\,NaF$ .

1. Both buffers have the same pH because the ratio is $1:1$ ( $pH = pK_a$ ). 2. If $0.05\,mol$ of $HCl$ is added to $1\,L$ of each: 3. Buffer A will see a massive pH drop because the $H^+$ added is half the concentration of the base available. 4. Buffer B will see a very small pH change because the $H^+$ added is only $1/10$ th of the base available. Conclusion: Buffer B has a higher capacity.

Calculate the pH change when $0.01\,mol$ of $NaOH$ is added to $1.0\,L$ of a buffer containing $0.10\,M\,CH_3COOH$ and $0.10\,M\,CH_3COO^-$ ( $pK_a = 4.74$ ).

1. Stoichiometry: The $OH^-$ reacts with the acid: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$ . 2. New $[HA] = 0.10 - 0.01 = 0.09\,M$ . 3. New $[A^-] = 0.10 + 0.01 = 0.11\,M$ . 4. Henderson-Hasselbalch: $pH = 4.74 + \log\left(\frac{0.11}{0.09}\right)$ . 5. $pH = 4.74 + 0.087 = 4.827$ . 6. The pH only changed by $0.087$ units!

Key Takeaways

1Buffers consist of a weak acid and its conjugate base (or vice versa) to neutralize added H+ or OH-.
2The common ion effect shifts equilibrium to prevent large changes in H+ concentration.
3The Henderson-Hasselbalch equation is $pH = pK_a + \log\left(\frac{[base]}{[acid]}\right)$ .
4Higher concentrations of buffer components lead to a higher buffer capacity.

Quick Check

Multiple Choice

Which of the following pairs would make the best buffer solution?

Multiple Choice

If a buffer has a $pK_a$ of 3.75 and the ratio of $[base]/[acid]$ is 10:1, what is the pH?

True/False

A buffer's capacity is exhausted when the amount of added strong acid exceeds the amount of conjugate base present in the buffer.

What's Next?

Review Tomorrow

In 24 hours, try to write down the Henderson-Hasselbalch equation from memory and explain why a strong acid/strong base pair cannot form a buffer.

Practice Activity

Find the pKa of blood's bicarbonate buffer system ( $H_2CO_3/HCO_3^-$ ) and calculate the ratio needed to maintain a pH of 7.4.

Browse

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Buffer Systems and Capacity

Learning Objectives

The Common Ion Effect

The Henderson-Hasselbalch Equation

Buffer Capacity and Range

Key Takeaways

Quick Check

What's Next?

Buffer Systems and Capacity

Learning Objectives

The Common Ion Effect

The Henderson-Hasselbalch Equation

Buffer Capacity and Range

Key Takeaways

Quick Check

What's Next?